Physics | 13.25
Solved Examples
10−3 × 250 N× 300 JEE Main/Boards = × 23 760 22400 6×× 10 273 Example 1: An electric bulb of volume 250cm³ was sealed off during manufacture at the pressure of −3 23 10× 250 ×× 6 10 × 273 15 10-3mm of Hg at 27°C. Find the number of air molecules N = = 8.02x10 mole 760×× 22400 300 in the bulb. NR Sol: PV= RT = N T = NkT AA Example 2: One gram-mole of oxygen at 27°C and one atmospheric pressure is enclosed in a vessel. Let N be the number of air molecules in the bulb. (a) Assuming the molecules to be moving with V , find V =250cm³, P = 10-3mm of Hg, rms 1 1 the number of collisions per second which molecules T1 = 273 + 27 = 300°K make with one square meter area of the vessel wall.
As P11 V= Nk T where k is constant, then (b) The vessel is next thermally insulated and moved with a constant speed ν . It is then suddenly stopped. 10-3x 250 = N.k.300 … (i) 0 The process results in a rise of the temperature of the At N.T.P., one mole of air occupies a volume of 22.4 litre, gas by 1°C. Calculate the speed ν0 . 3 V00= 22400cm ,P= = 760mmofHg,760 mm of Hg, P 3N kT × 23 n = V = A T = 273° K and N0 = 6 10 molecules Sol: Formula based: & rms . Recall kT Mn ∴760 × 22400 = 6x1023 × k × 273 … (ii) the assumption of KTG. Kinetic energy changed to Dividing equation (i) by equation (ii), we get internal energy. 13.26 | Kinetic Theory of Gases and Thermodynamics
f (a) n = P/kT +1 2 2 where k = Boltzmann constant = Sol: (a) y = or y1= + f f 1.38 x 10-23 J/mol/K 2 nR P = 1 atmosphere = 1.01 x 105 n/m², (b) W= (T − T ) 1−γ fi T = 27°C = 300°K (a) For adiabatic process 5 1.01x10 25− 3 γ−11 γ− n = = 2.44x10 m TV= T V ; V= 5.66V and T= T / 2 1.38x10−23 x300 11 11 γ−11T γ− γ−1 ∴=TV x(5.66V) ; (5.66)= 2 The root mean square velocity Vrms is given by 2
3RT 3N kT (γ− 1)log5.66 = log2 V = = A ( R= N k) rms M A Mm m log2 0.3010 γ−1 = ; γ−1 = =0.4 × 23 log5.66 0.7528 NA= Avogadro number = 6.02 10 molecule/mole -3 ∴ γ=1.4(diatomicgas) Mm = 32gm = 32 × 10 kg.
36.02x1023 x1.38x10-23 x300 The degrees of freedom of gas molecules = 5 V = =483.4m/s. rms -3 (b) Work done during adiabatic change, W, is given by 32x10 1 W= [P V − PV] Since each molecule may be moving at a given instant, 1−γ 11 along any of the six directions (i.e. +X, +Y and +Z) only For an ideal gas equation, (1/6)th of the total molecules contained in the volume along positive and negative x, y, z directions, move PV PV11 TV1 11 P =or P1 = P ; Px1 = P = towards the wall. The number of collisions per second T T11 TV 2 5.66 11.32 with one square meter area of the vessel wall will be 1P 1125 W = x5.66V− PV =xn × V =××× 2.44 10 483.4 = 1.97 × 1027. 1− 1.4 11.32 66rms
1 2 (b) K.E. = mv PV 5.66 PV 2 0 =−=1 = 1.25 PV. 0.4 1132 0.8 Heat energy gained = C∆= T C x1 = C vv v 5 R Example 4: Two moles of helium gas with γ equal to CCR−= or C = 3 pv v γ−1 are initially at temperature 27°C and occupy a volume of 1R2 20 litres. The gas is first expanded at constant pressure ∴ γ=1.41 ; mv= C = 210vγ− until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its 2R 2x8.31x1 initial value. ν=0 = =35.6m / sec. m(γ− 1) 32x10−3 x141− 1 (a) Sketch the process on P-V diagram. (b) Find the final value of volume and pressure of the Example 3: An ideal gas having initial pressure P, gas. volume V and temperature T is allowed to expand (c) Find the work done by the gas under isobaric and adiabatically until its volume be-comes 5.66 V while its adiabatic processes and total work done. temperature fall to T/2.
(a) How many degrees of freedom do the gas molecules Sol: Relate P1 V1 and T1 with P2, V2, T2 and P3, V3, T3. have? Calculate the work alone of process AB, BC separately. (b) Obtain the work done by the gas during the (a) The sketch of P-V diagram is shown in the figure. expansion as a function of the initial pressure P and volume V. Physics | 13.27
5 PVT 2.5 x 10 111 PV22T2 Example 5: Two moles of helium gas undergo a cyclic AB process as shown in the figure. Assuming the gas to be ideal, calculate the following C quantities in this process. -2 P(Nm) PVT 333 (a) The net change in the heat energy
-3 -3 20 x 10 40 x 10 (b) The net work done -3 V(m) (c) The net change in internal energy
−33 Sol: Calculate for isobaric and isothermal process only. (b) At A, V1 = 20l = 20x10 m T1 = 27 °= C 300K ; P?1 = For an ideal gas, PV = nRT. (a) AB is isobaric process. The work done during this process from A to B: nRT 2x8.31x300 P =1 = = 2.5x1052 Nm− 1 −3 WAB= P(V 2 −= V 1 ) nR(T 2 − T 1 ) V1 20x10 52− −33 or WAB = 2x8.3x(400 −= 300) 1160 joule At B, P21= P = 2.5x10 Nm , V2 = 40x10 m Work done during isothermal process from B to C: For isobaric process, P A B TV40x10−3 2 atm 22= = = 2 T= 2T = 2x300 = 600K −3 21 TV1120x10 D C The gas now undergoes adiabatic expansion from B to C. 1 atm T γ−11 γ− TV33= TV 22 300 K 400 K
1 1/(5/3− 1) W= nRT log (V / V )= nRT log (P / P ) V T γ−1 600 3 ABCe21 Ce21 3 =2 = =2 = = 2 2 2 2.83 =2x8.3x 400x2.303log 2 V23 T 300 10 =2x8.3x 400x2.303x0.3010 = 4602.9 joule Final Volume = −−3 33 Work done during isobaric process from C to D: V32= V x2.82 = 40x10 x2.83= 113x10 m V WCD= nR(T D −= T C ) 2x8.3x(300 − 400) = −1660 joule PVγγ= PV Final pressure P3 is given by 33 22 Work done during isothermal process from D to A: γ 5/3 V −3 2 5 40x10 W= mRT log (P / P ) = nRT log 2 P32= P x = 2.5x10 x DA D e D A De V 113x10−3 3 =2x8.03x300x2.0303x0.3010= − 3452.2Joule 5 5/3 5 = 2.5x10 x(0.353)= 2.5x10 x0.176 Net workdone = WWWAB++ BC CD + W DA 52− = 4.410 N.m = 1660 + 4602.9 – 1660 – 3452.2 (c) Work done during isobaric process along AB = P = 1150.7 joule 53− (V21−= V ) 2.5x10 x20x10= 5000J (b) First law of thermodynamics gives Work done during adiabatic process along BC= ∆QUW =∆ +∆ Work done during adiabatic process along BC= nR(T23− T ) 2x831x(600− 300) As ∆=U0, in cyclic process nR(T23=− T ) 2x831x(600− 300) BC =γ− 1 = 5 γ−1 −1 5 ∴∆Q =∆ W = 1150.7 joule 3 −1 3 The heat given to the system = 1150.7 joule Total work done = 5000 + 7479 = 11479 J. (c) As the gas returns to its original state, there is no change in internal energy. 13.28 | Kinetic Theory of Gases and Thermodynamics
Example 6: An ideal gas is taken a cyclic thermodynamic (b) Is there any way of telling afterwards which sample process through four steps. The amount of heat involved of Helium went through the process ABC and which in these steps are Q12==−=− 5960 J,Q 5585J,Q 3 2980 J went through the process ADC? Write Yes or No. Q= 3645J and 4 respectively. The corresponding works (c) How much is the heat involved in each of the involved are W1 = 2200 J, W2 = − 825J, respectively. processes ABC and ADC? (a) Find the value of W 4 Sol: Work = Area under P–V curve hence, work done (b) What is the efficiency of the cycle? in ABC is more than in ADC so is the heat (Q). At C, system’s thermodynamic states are same, it can’t be ∆=U0 Sol: QTotal =WTotal as in cyclic process, determined how they are achieved. W 3 η= T 2x10 only (a) n= = 500 (Q absorb ed) 4 ∆= As the process is cyclic, U0 At A, PAA V= nRT A or TA= (P AA V / nR) Net heat absorbed by the system (5x104 )x10 ∴=TA =120.33K QQ=+++123 Q Q Q 4 500x831 = 5960 – 5585 – 2980 + 3645 = 1040 J (10x104 )x10 Similarly, TB = = 240.66K Net work performed 500x831 WWW=+++ W W (10x104 )x20 123 4 T = = 481.32K C 500x831 = 2200 – 825 – 1100 + W4 = 275 + W4 (5x104 )x20 According to the first law of thermodynamics T = = 240.66K D 500x831 Q=∆+ UW ; 1040 = 0 + 275 + W4 (b) No ∴=W4 1040 −= 275 765 joule. (c) For process ABC: Work done(W) Efficiency η= Change in internal energy ∆=U nC ∆ T Heatabsorbed(Q14+ Q ) v 275+ 765 1040 3 = = = 0.1082 ∆ = ∆ 5960+ 3645 9605 ( U)ABC nRT 2 Percentage efficiency = 10.82% 3 6 = 500x x8.3 x[481.32− 120.33] = 2.25x10 J 2 Example 7: A sample of 2 kg of monoatomic Helium 4 Work done (∆ W)ABC =10 x (10 x 10 ) (assumed ideal) is taken through the process ABC and 6 (∆ Q) = (dU) +∆ ( W) another sample of 2 kg of the same gas is taken through = 10 J ABC ABC ABC the process ADC as shown in the figure. =2.25x1066 += 10 3.25x10 6 J 42 P(10 N/m) For process ADC: B 10 C (∆ W) = 5 x 104 [20-10] = 0.5 x 106 J ADC 6 (∆ U)ADC = 2.25 x 10
6 6 ∴∆( Q)ADC = (2.25 x 10 ) + (0.5 x 10 )
6 5 D = 2.75 x 10 J A
Example 8: Pressure versus temperature graph of an 0 3 10 20 V(m ) ideal gas is as shown in figure. Density of the gas at point A is ρ0 . Density at B would be Given molecular mass of Helium = 4 (a) What is the temperature of Helium in each of the Sol: Use relation between density and temperature, states A, B, C and D? given below. Physics | 13.29
PM P the gases A and B are 4 and 32 respectively. The gases ρ= ∝ RT T A and B do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation 19/13 Pressure PV = constant, in adiabatic processes. P P (a) Find the number of gram moles of the gas B in the = 0 2 gaseous mixture. TTA 0 (b) Compute the speed of sound in the gaseous mixture P at T=300K. (c) If T is raised by 1 K from 300 K, find the percentage B 3P0 change in the speed of sound in the gaseous mixture.
th P (d) The mixture is combined adiabatically to 1/5 of 0 A its initial volume V. Find the change in its adiabatic compressibility in terms of the given quantities. T T0 2T0 Sol: γ of the mixture is known from equation of process.
Therefore CV can be known. Compare CV (mixture) n C+ n (C ) P 3P0 3P A V B VB and = = = A T 2T 4T BA0 (nAB+ n ) 33 (a) As the gaseous mixture follows the equation PV19/13 ∴ ρ= ρ = ρ 44A0 = constant, then for the mixture of the gas γ=19 / 13. C CR+ R 19 γ=PV = =1 + = Example 9: The root mean square (rms) speed of CVV C C v 13 hydrogen molecules at a certain temperature is R6 13R 300 m/s. If the temperature is doubled and hydrogen = , ∴=C C 13 V 6 gas dissociates into atomic hydrogen the rms speed V 19 will become CCR= += R PV 6 Sol: Formula based VTrms ∝ . For gas A, γ=A 5/3; For gas B, γ=B 7 / 5. 3RT 35 Vrms = (C )= R and(C )= R M VA 22VB 57 T is doubled and M is halved. Therefore, rms speed will (C )= R and(C )= R become two times or 600 m/s. PA 22PB
Let nA and nB be the number of kg moles in gas A and Example 10: The changes in temperature of an ideal gas B respectively. gas, when its volume changes from V to 2V in the −3 n= 1gm mole = 10 kg.mole process P = aV, is (Here a is a positive constant) A As the gases have fixed volume, no work is done by the Sol: Use relation between P and V. gas and vessel system. In the adiabatic process, no heat is exchanged with the surroundings, the internal energy PV∝ (P=aV) of the system will remain constant. Therefore, pressure and volume both are doubled or ∴+(nA n B )C V dT = n A (C VA ) dT + n B (C VB ) dT temperature becomes four times ( T∝ PV ) (nA+= n B )C V n A (C VA ) + n B (C V )B
JEE Advanced/Boards −−3313 3 5 (1x10+= n)BB R (1x10 )R + n R 6 26 Example 1: A gaseous mixture enclosed in a vessel of −−33 13x10+= 13nBB 9x10 + n (15) volume V consists of one gram mole of a gas A with 2n= 4x10 −3 γ=(CPV / C ) 5 / 3 and another has B with γ=7/5 at a B certain temperature T. The gram molecular weights of 13.30 | Kinetic Theory of Gases and Thermodynamics
4x10 −3 (a) For adiabatic expansion ∴=n =2x10−3 kgmole = 2gm mole B 2 γ−11 γ− γ−11γ− TV11= T 2 V 2 ; 300x(V)= T2 (2V) (b) The speed of sound in gaseous mixture is given by 1 300 T= 300x =( γ= 5 / 3) γRT 2 γ−1 2/3 ν= (2) (2) M ∴=T2 188.99K where R is gas constant and M is equivalent gram (b) Change in thermal energy is given by molecular weight of gaseous mixture. Let MA and M be the gram molecular weights of gases A and B B ∆=U nCV2 (T − T 1 ) = 2x(3R/ 2)(188.99− 300) respectively, then n M+=+ n M (n n )M 3x8.3 AA BB A B =2x x( −= 111.09) 2767.5J 2 nM+ nM (1x 4)+ (2x32) M= AA BB= or The negative sign indicates that there is a decrease in nnAB 1+ 2 the internal energy. 68 68 −3 =gm = x10 kg (c) For adiabatic process, 33 ∆+∆=W U 0 or ∆=−∆ W U 19 8.3x300x3 ∴ν = x = 400.7m / s The workdone by the gas is given by −3 13 68x10 nR(T− T ) 2x8.3x(188.99− 300) ∆=−W 21 =− γ−1 (5 / 3− 1) γRT (c) At temperature T, ν= =2764.2J M Example 3: Two moles of an ideal monoatomic gas, γRT' T'= (300 + 1)K, ν= ' initially at pressure p1 and volume V1 undergo an M adiabatic compression until its volume is V2. Then the
gas is given heat Q at constant volume V2. νν' T' 301 ' 301 ∴ = = or−= 1 −1 (a) Sketch the complete process on a P-V diagram. νν T 300 300 (b) Find the total workdone by the gas, the total change ν−ν' in its internal energy and the final temperature of the The percentage change in speed of sound = x100 ν gas. 301 =−=1 x100 0.17% [Give your answers in terms of p112 ,V ,V ,Q and R]. 300 γ−1 nR∆ T Sol: Use formula TV = constant and W = (d) Adiabatic compressibility = 1−γ 1 dv 1 (a) P-V diagram is shown in the figure where AB is β=−mt =− V dpmt γ P adiabatic compression and BC is isobaric heating
1 1 13 V 1 P ∆β=1 − =1 − = 2.487x10−3 . γγ γP(5) 3x19 RT (5) P3 C AB=Adiabatic compression P2 B Example 2: At 27°C, two moles of an ideal monoatomic BC= Isobaric heating gas occupy a volume V. The gas expands adiabatically to a volume 2V. Calculate A P1 (a) The final temperature of the gas V (b) Change in its internal erergy V2 V1 ...(l) (c) The workdone by the gas during this process (b) Let T1 and T2 be the initial temperature and the γ Sol: Formula TV = constant, ∆=U nCv ∆ T WU= −∆ temperature after adiabatic compression respectively. used. Physics | 13.31
For adiabatic compression The pressure and temperature at A, B etc., are denoted γ−11 γ− VTVT11= 22 by PAABB ,T ;P ,T etc. respectively. pV 5 T,=11 γ= for monoatomic gas Given T= 1000K,P = (2 / 3)P &P= (1 / 3)P . Calculate 1 2R 3 A BACA (a) The work done by the gas in the process AB→ 5/3 2/3 pV11 2/3 pV11 ∴=V VT ∴ T ….(i) (b) The heat lost by the gas in the process BC→ and 12R 22 2 2/3 2RV2 (c) Temperature TD given (2/3) = 0.85 and R-8.31 J/mol K. For isochoric process at temperature T3, heat supplied is Q. Sol: Use the relation for the respective processes. Such as T/P relation in adiabatic process. ∴=Q nCV dT ; Q= 2.3 / 2R(T32 − T ) (a) As for adiabatic change 5/3 γ Q QQp1V1 = =TT − ;∴TT =+= + …. (ii) PV constant 3R 32 323R 3R 2/3 2RV2 γ nRT i.e. P= constant [as PV= nRT ] The total work done by the gas is equal to the work P done in adiabatic process plus the work done in γ isochoric process when W0= T BC i.e. = constant Pγ−1 ∴WTotal =+= WW AB BC W AB nR(T− T ) 5 W =12 = where γ= Total γ−1 3 1 5/3 2/3 1− 2/5 pV pv V γ 2R 11 11 3 1 22 = −=−pV11 1 i.e. TBA= T = 1000 = 850K 2 / 3 2R 2RV 2/3 2 V 33 2 2 nR[T− T ] ∆= − if 1x8.3[1000− 850] Change in internal energy U nCV3 (T T 1 ) so W = = AB γ−1 [(5 / 3)− 1] 5/3 3QpV11 pV 11 ∴∆U = 2x R + − W= (3 / 2)x8.31x150= 1869.75J 2 3R 2/3 2R i.e. AB 2RV2 (b) For B → C, V = constant so ∆=W0 5/3 2/3 33pV 3V =+−=+Q 11 pV Q pV1 − 1 so from first law of thermodynamics 22/3 211 2V11 V2 2 ∆Q =∆ U +∆ W =µ CV ∆ T0 + 3 3 Example 4: One mole of monoatomic ideal gas is taken or ∆=Q 1x R (TC − 850) as CRV = through the cycle shown in figure. 2 2 Now along path BC, V = constant; P ∝ T A B PTCC (1 / 3)PAB T 850 i.e. = , TCB= xT = = = 425K ….(ii) PTBB (2 / 3)PA 2 2 P 3 D C So, ∆=Q 1x x8.31(425− 850) =− 5297.625J 2 V [Negative heat means heat is lost by the sys.] (c) DA→ process is isochoric AB→ Adiabatic expansion PT T BC→ DD D Cooling at constant volume = , i.e. PPDA= PTAA TA CD→ Adiabatic compression But C and D are on the same adiabatic DA→ Heating at constant volume 13.32 | Kinetic Theory of Gases and Thermodynamics
γ γ−11γ− η−1 TD P D PT AD = = = VV0 … (ii) η+1 TC P C PT CA
1 1− The work done by the agent is given by P γ 1/ γ A VV or (TDC )= T , 2P V V PT W=−= (P P )dV00 dV CA ∫∫21 22 2/5 00VV0 − TP i. e. T3/5 = BA C =−P V [ln(V2 − V 2Vv ) ] =− P V [ln(V2 −− V 2 ) lnV 2 ] 2 (1/ 3)PA 1000 0 0 0 0 00 0 0
2 2/3 2/5 η−1 12 3 2 22 3/5 =−−−P00 V ln V 0 V0 lnV 0 i.e. TD = x1000 η+1 2 3 1000 (η+ 1) 2 i.e. TD = 500K 2 =−P00 V [ln{4 η / ( η+ 1) }] = P0 V 0 ln . 4η Example 5: A piston can freely move inside a horizontal cylinder closed from both ends. Initially, the Example 6: An ideal gas has a density of 1.78 kg / m³ is piston separates the inside space of the cylinder into -3 contained in a volume of 44.8 x 10 m³. The temperature two equal part of volume V , in which an ideal gas is 0 of the gas is 273 K. The pressure of the gas is contained under the same pressure P and at the same 5 -1 -1 0 0.01 x 10 Pa. The gas constant R = 8.31 J-K mole . temperature. (a) What is the root mean square velocity of the air x molecules? (b) How many moles of gas are present? PA1 PA (c) What is the gas? F 2 agent (d) What is the internal energy of the gas? What work has to be performed in order to increase Sol: Use relation V and P given. m = ρ V = nM also. isothermally the volume of one of gas η times rms 1 2 compared to that of the other by slowly moving piston? (a) PV=ρ⇒ 3 Sol: Apply isothermal condition on both compartments. 1/2 1/2 3P (3)(1.01x1052 N/ m ) Then, proceed to find V (left)/V (right). V = = f f rms 3 ρ 1.78kg / m Let the agent move as shown. = 4.13 x 10² m/s In equilibrium position, PA1+= F agent P 2 A PV= n RT ⇒ Fagent= (P 2 − P 1 )A (b) m PV (1.01x105 Nm−− 2 )(44.8x1033 m ) Elementary work done by the agent nm = = RT (8.31JK−−11 mole )(273K) Fagent dx=−=− (P 2 P 1 )Axdx (P2 P 1 )dV … (i) = 2.0 moles Applying PV = constant for two parts, we have ρV (1.78kg / m3 )(44.8x10−33 m P10 (V+= Ax) P 00 V and P 2 (V 0 −= Ax) P 00 V Mmolar = = nm (2.0moles) PV PV 00 00 −3 P1 = and P2 = =340.0x10 kg / mole (V0 + Ax) (V0 − Ax) PV(2Ax) 2PVV (c) This ideal gas is Argon. ∴PP −=00 = 00 21 2 22 2 2 (d) Internal energy of monoatomic gas=3/2 nRT. VAxVV00−−
When the volume of the left end is η times the volume of right end, we have (V00+=η− V) (V V) Physics | 13.33
Example 7: Plot P-V, V-T and ρ -T graph corresponding P-V graph: As P is constant. Therefore, P-V graph is a VV> to the P-T graph for an ideal gas shown in the figure. straight line parallel to V-axis with CB (because V ∝ T in an isobaric process) P V-T graph: In an isobaric process V ∝ T, i.e., V-T graph BC is a straight line passing through the origin, with TC> TB
and VC> VB. 1 A D ρ-T graph: ρ∝ (when P = constant), i.e., ρ -T graph T T is a hyperbola with TTCB> and ρCB >ρ . Sol: Look for parameter which is constant in the each process. There is no need to discuss C-D and D-A processes as they are opposite to AB and BC respectively. The Process AB is an isothermal process with T = constant corresponding three graphs are shown above. and PB> PA. D V 1 P P P-V graph:P ∝ i.e., P-V graph is a hyperbola with V B A C B PPBA> and VVBA> . C C A D V-T graph: T= constant. Therefore, V-T graph is a B A straight line parallel to V-axis with VBA> V. V D T T PM D ρ−Tgraph : ρ= or ρ∝ P . V P RT P B A As T is constant. Therefore,C ρ−T graph is a straight line B parallel to ρ -axis with ρ >ρ as PP> . C C D BA BA A B Process BC is isobaric process with P = constant and A TT> V D CB. T T
JEE Main/Boards
Exercise 1 Q.5 State the postulates of Kinetic Theory of gases. Explain the pressure exerted by an ideal gas.
Q.1 Although the r.m.s. speed of gas molecules is of the Q.6 Find an expression for the pressure exerted by a order of the speed of sound in that gas yet on opening gas and establish its relation with kinetic energy of the a bottle of ammonia in one corner of a room, its smell gas. takes time in reaching the other corner. Explain Why?
Q.7 From Kinetic Theory of gases, explain kinetic Q.2 The pressure of a gas at – 173°C temperature is interpretation of temperature and absolute zero. 1 atmosphere, keeping the volume constant, to what temperature should the gas be heated so that its pressure becomes 2 atmosphere. Q.8 Explain the concept of mean free path.
Q.3 Explain (i) Boyle’s law (ii) Charle’s law. Why they are Q.9 Explain what is meant by Brownian Motion? not applicable to real gases at all states? Q.10 The density of water is 1000kg/m³. The density of Q.4 State and explain (i) Guy Isac’s law and (ii) Gas water vapour at 100°C and 2 atmospheric pressure is equation. Distinguish clearly between R and r for a gas. 0.6kg m³. The volume of a molecule multiplied by the 13.34 | Kinetic Theory of Gases and Thermodynamics total number gives what is called, molecular volume. Q.23 State the sign conventions used in all Estimate the ratio (or fraction) of the molecular volume thermodynamic processes. to the total volume occupied by the water vapour under the above conditions of temperature and pressure. Q.24 What do you learn by applying first law of thermodynamics to isothermal and adiabatic processes? Q.11 A 3000cm³ tank contains oxygen at 20°C and a 6 gauge pressure of 2.5 x 10 Pa. Find the mass of the Q.25 Explain what is meant by isothermal operations. 5 oxygen in the tank. Take 1 atm = 10 Pa. Give some examples.
Q.12 Calculate the r.m.s. velocity of air molecules at Q.26 What are adiabatic operations? Enumerate N.T.P. Given that 22400 c.c. of gas at N.T.P. weight 64 gm. some examples. State equations representing these operations. Q.13 How many collisions per second does each molecule of a gas make, when average speed of the Q.27 Obtain an expression for work done by a gas in -1 -7 molecule is 500ms and mean free path is 2.66 x 10 m? isothermal expansion.
Q.14 Calculate the mean free path of gas molecules, if Q.28 Derive an expression for work done in an adiabatic 19 number of molecules per cm³ is 3 x 10 and diameter process. of each molecule is 2Å.
Q.29 What are cyclic and non cyclic processes? Calculate 10 Q.15 The diameter of a gas molecules is 2.4 x 10- m. work done in such processes. Calculate the mean free path at N.T.P. Given Boltzmann constant k = 1.38 x 10-23 J molecule-1 K-1. Q.30 What are reversible and irreversible processes? Give some examples of each. Q.16 Which molecules, ice at 0°C or water 0°C have greater potential energy and why? Q.31 What is a heat engine? Obtain an expression for its efficiency. Q.17 An ideal gas is compressed at a constant temperature, will its internal energy increases of decrease? Q.32 A tyre pumped to a pressure of 3 atmosphere suddenly bursts. Calculate the fall in temperature due to adiabatic expansion. The temperature of air before Q.18 Which type of motion of the molecules is expansion is 27°C and value of γ = 1.4. responsible for internal energy of a monoatomic gas?
Q.33 A quantity of air at 27°C and atmospheric pressure Q.19 The volume of an ideal gas is V at a pressure P. On is suddenly compressed to half its original volume. Find ∆ increasing the pressure by P, the change in volume of the final (i) pressure and (ii) temperature. Given γ for air the gas is ( ∆ V ) under isothermal conditions and ( ∆ V ) 1 2 = 1.42. under adiabatic conditions, Is ∆VV12 >∆ or vice-versa and why? Q.34 A Cylinder containing one gram mole of gas was put on boiling water bath and compressed adiabatically Q.20 200 joule of work is done on a gas to reduce till its temperature rose by 70°C. Calculate the work its volume by coming it. If this change is done under done and increase in energy of the gas, γ = 1.5, R = 2 adiabatic conditions, find out the change in internal cal. mole-1 K-1. energy of the gas and also the amount of heat absorbed by the gas? Q.35 One gram mole of an ideal gas at S.T.P. is subjected to reversible adiabatic expansion to double its volume. Q.21 Give briefly the concept of internal energy. Find the change in internal energy in the process. Take γ =1.4. Q.22 Define the four thermodynamic processes. What is meant by indicator diagram? Physics | 13.35
Q.36 If 1 gram oxygen at 760mm pressure and 0°C has Q.7 A barometer tube, containing mercury, is lowered it volume doubled in an adiabatic change, calculate in a vessel containing mercury until only 50 cm of the the change in internal energy. Take R=2 cal. mole-1 K-1, tube is above the level of mercury in the vessel. If the J=4.2 J cal-1 and γ =1.4. atmospheric pressure is 75 cm of mercury, what is the pressure at the top of the tube? Exercise 2 (A) 33.3 kPa (B) 66.7 kPa (C) 3.33 MPa (D) 6.67 MPa Single Correct Choice Type Q.8 A vessel contains 1 mole of O gas (molar mass Q.1 Find the approx. number of molecules contained in 2 32) at a temperature T. The pressure of the gas is P. An a vessel of volume 7 litres at 0°C at 1.3 x 105Pascal identical vessel containing one mole of He gas (molar (A) 2.4 x 1023 (B) 3 x 1023 mass 4) at a temperature 3T has a pressure of (C) 6 x 1023 (D) 4.8 x 1023 (A) P/8 (B) P (C) 2P (D) 8P
Q.2 An ideal gas mixture filled inside a balloon Q.9 The ratio of average translational kinetic energy expands according to the relation PV2/3 = constant. The to rotational kinetic energy of a diatomic molecule temperature inside the balloon is temperature T is (A) Increasing (B) Decreasing (A) 3 (B) 7/5 (C) 5/3 (D) 3/2 (C) Constant (D) Can’t be said Q.10 One mole of an ideal gas at STP is heated in an Q.3 A rigid tank contains 35 kg of nitrogen at 6 atm. insulated closed container until the average speed of Sufficient quality of oxygen is supplied to increase its molecules is doubled. Its pressure would therefore the pressure to 9 atm, while the temperature remains increase by factor. constant. Amount of oxygen supplied to the tank is: (A) 1.5 (B) 2 (C) 2 (D) 4 (A) 5 kg (B) 10 kg (C) 20 kg (D) 40 kg
Q.11 One mole of an ideal gas is contained within Q.4 At temperature T K, the pressure of 4.0g argon in a cylinder by a frictionless piston and is initially at bulb is p. The bulb is put in a bath having temperature temperature T. The pressure of the gas is kept constant higher by 50K than the first one. 0.8g of argon gas had while it is heated and its volume doubles. If R is molar to be removed to maintained original pressure. The gas constant, the work done by the gas in increasing its temperature T is equal to volume is? (A) 510 K (B) 200 K (C) 100 K (D) 73 K (A) RT ln2 (B) 1/2RT (C) RT (D) 3/2 RT
Q.5 When 2 gms of a gas are introduced into an Q.12 A polyatomic gas with six degrees of freedom evacuated flask kept at 25°C the pressure is found to be does 25J work when it is expanded at constant pressure. one atmosphere. If 3 gms of another gas added to the The heat given to the gas is? same flask the pressure becomes 1.5 atmosphere. The ratio of the molecular weights of these gases will be (A) 100J (B) 150J (C) 200J (D) 250J (A) 1: 3 (B) 3: 1 (C) 2: 3 (D) 3: 2 Q.13 In thermodynamic process pressure of a fixed mass of gas is changed in such a manner that the gas Q.6 During an experiment, an ideal gas obeys an release 30 joule of heat and 18 joule of work was done addition equation of state P2V = constant. The on the gas. It the initial internal energy of the gas was initial temperature and pressure of gas are T and V 60 joule, then, the final internal energy will be? respectively. When it expands to volume 2 V, then its temperature will be? (A) 32 joule (B) 48 joule (C) 72 joule (D) 96 joule (A) T (B) 2T (C) 2 T (D) 2 2T 13.36 | Kinetic Theory of Gases and Thermodynamics
Q.14 An ideal gas undergoes an adiabatic process Q.3 70 cal of heat are required to raise the temperature obeying the relation PV4/3 = constant. If its initial of 2 moles of an ideal diatomic gas at constant pressure temperature is 300 K and then its pressure is increased from 30°C to 35°C. The amount of heat required upto four times its initial value, then the final (in calorie) to raise the temperature of the same gas temperature is (in Kelvin)? through the same range (30°C to 35°C) at constant 3 volume is (1985) (A) 300 2 (B) 300 2 (A) 30 (B) 50 (C) 70 (D) 90 (C) 600 (D) 1200 Q.15 1 kg of a gas does 20kJ of work and receives 16kJ Q.4 If one mole of a monatomic gas (γ=5/3) is mixed of heat when it its expanded between two states. A with one mole of a diatomic gas ( γ =7/5), the value of second kind of expansion can be found between the γ for the mixture is (1988) initial and final state which requires a heat input of 9 kj. The work done by the gas in the second expansion is: (A) 1.40 (B) 150 (C) 1.53 (D) 3.07 (A) 32 kJ (B) 5 kJ Q.5 The temperature of an ideal gas is increased from (C) -4 kJ (D) 13 kJ 120 K to 480 K. If at 120 K the root mean square velocity of the gas molecules is v, at 480 K it becomes (1996) Q.16 A mixture of ideal gases 7 kg of nitrogen and 11 (A) 4 v (B) 2 v (C) v/2 (D) v/4 γ kg of CO2. Then (Take for nitrogen and CO2 as 1.4 and 1.3 respectively) Q.6 The average translational energy and the rms (A) Equivalent molecular weight of the mixture is 36. speeds of molecules in a sample of oxygen gas at (B) Equivalent molecular weight of the mixture is 18. 300 K are 6.21 x 10-21J and 484 m/s respectively. The corresponding values at 600 K are nearly (assuming (C) γ for the mixture is 5/2 ideal gas behaviour) (1997) (D) γ for the mixture is 4/3 (A) 12.42x10-21 J, 968 m/s (B) 8.78x10-21 J, 684 m/s
-21 Previous Years’ Questions (C) 6.21x10 J, 968 m/s (D) 12.42x10-21 J, 684 m/s
Q.1 An ideal mono-atomic gas is taken round the cycle ABCD as sown in the P-V diagram (see figure). The work Q.7 A vessel contains 1 mole of O2 gas (molar mass done during the cycle is (1983) 32) at a temperature T. The pressure of the gas is p. An identical vessel containing one mole of the gas (molar P 2p, V 2p, 2V mass 4) at a temperature 2T has a pressure of (1997) B C (A) p/8 (B) p (C) 2 p (D) 8 p
AD Q.8 Two identical containers A and B with frictionless p, V p, 2V pistons contain the same ideal gas at the same temperature and the same volume V. The mass of the gas in A is m and that in B is m . The gas in each (A) PV (B) 2 PV A B cylinder is now allowed to expand isothermally to the 1 (C) pV (D) Zero same final volume 2V. The changes in the pressure in 2 A and B are found to be ∆ P and 1.5 ∆ p respectively. Then (1998) Q.2 At room temperature, the rms speed of the (A) 4mAB= 9m (B) 2mAB= 3m molecules of a certain diatomic gas is found to be 3m= 2m 9m= 4m 1930 m/s. The gas is (1984) (C) AB (D) AB
(A) H2 (B) F2 (C) O2 (D) Cl2 Physics | 13.37
Q.9 A mixture consists of 2 moles of oxygen and 4 moles P of argon at temperature T. Neglecting all vibrational B A modes, the total internal energy of the system is 3P (1999) (A) 4 RT (B) 15 RT (C) 9 RT (D) 11RT 1P C D Q.10 A monoatomic ideal gas, initially at temperature 0 1V 3V 9V V T1, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to Column I Column II a temperature T2 by releasing the piston suddenly.
If L1 and L2 are the lengths of the gas column before (A) Process AB→ (p) Internal energy decreases and after expansion respectively, thenT1/T2 is given by (2000) (B) Process BC→ (q) Internal energy increases 2/3 (A) (L12 / L ) (B) (L12 / L ) (C) Process CD→ (r) Heat is lost 2/3 (C)L21 /L (D) (L21 / L ) (D) Process DA→ (s) Heat is gained
Q.11 An ideal gas is expanding such that pT² = (t) Work is done on the gas constant. The coefficient of volume expansion of the gas is (2008) Q.14 For an ideal gas 1 2 3 4 (A) (B) (C) (D) T T T T (A) The change in internal energy in a constant pressure process from temperature T1 to T2 is equal to nCV2 (T− T 1 ) , where CV is the molar heat capacity at Match the following for the given process (2006) Q.12 constant volume and n the number of moles of the gas P (B) The change in internal energy of the gas and the J 30 work done by the gas are equal in magnitude in an adiabatic process 20 M (C) The internal energy does not change in an isothermal 10 KL process
3 10 20 V (m ) (D) No heat is added or removed in an adiabatic process. (1989)
Column I Column II Q.15 One mole an ideal gas in initial state A undergoes (p) Q>0 (A) Process JK→ a cyclic process ABCA, as shown in figure. Its pressure at A is P . Choose the correct option(s) from the following (q) W<0 0 (B) Process KL→ (2010) V (r) W>0 B (C) Process LK→ 4V0 (s) Q<0 (D) Process MJ→ V A 0 C
T T Q.13 One mole of a monatomic ideal gas is taken 0 through a cycle ABCDA as shown in the P-V diagram. (A) Internal energies at A and B are the same column II gives the characteristics involved in the cycle. Match them with each of the processes given in (B) Work done by the gas in process AB is P0V0 In 4 column I. (2011) P (C) Pressure at C is 0 4 T (D) Temperature at C is 0 4 13.38 | Kinetic Theory of Gases and Thermodynamics
Q.16 The speed of sound in oxygen (O2) at a certain Q.23 The potential energy function for the force temperature is 460 ms−1. The speed of sound in helium between two atoms in a diatomic molecule is (He) at the same temperature will be (assumed both ab U(x) = − (2008) approximately given by 12 6 , where a and b gases to be ideal) xx (A) 460 ms−1 (B) 500 ms−1 are constants and x is the distance between the atoms. If the dissociation energy of the molecule is D = [U(x = (C) 650 ms−1 (D) 1420 ms−1 ∞) – Uat equilibrium], D is (2010)
2 2 2 2 Q.17 An insulated container of gas has two chambers b b b b (A) (B) (C) (D) separated by an insulating partition. One of the chambers 2a 12a 4a 6a has volume V1 and contains ideal gas at pressure P1 and temperature T1. The other chamber has volume V2 and Q.24 Carnot engine operating between temperatures contains ideal gas at pressure P2 and temperature T2. If 1 the partition is removed without doing any work on the T and T has efficiency . When T is lowered by 62 1 2 6 2 gas, the final equilibrium temperature of the gas in the 1 K, its efficiency increases to . Then T and T are, container will be (2008) 3 1 2 T T (P V+ P V ) PVT+ P V T respectively: (2011) (A) 12 11 2 2 (B) 1 11 2 2 2 PVT+ P V T PV+ P V 112 2 21 11 2 2 (A) 372 K and 330 K (B) 330 K and 268 K PVT+ P V T T T (P V+ P V ) (C) 310 K and 248 K (D) 372 K and 310 K (C) 112 2 21 (D) 12 11 2 2 PV11+ P 2 V 2 PVT1 11+ P 2 V 2 T 2 Q.25 Helium gas goes through a cycle ABCDA Q.18 One kg of a diatomic gas is at a pressure of (consisting of two isochoric and two isobaric lines) 8 × 104 N/m2. The density of the gas is 4 kg m-3. What as shown in figure. Efficiency of this cycle is nearly: is the energy of the gas due to its thermal motion? (Assume the gas to be close to ideal gas) (2012) (2009) BC (A) 3 × 104 J (B) 5 × 104 J 2P0 (C) 6 × 104 J (D) 7 × 104 J P D 0 A Q.19 Assuming the gas to be ideal the work done on the gas in taking it from A to B is (2009) V0 2V0 (A) 200 R (B) 300 R (C) 400 R (D) 500 R (A) 15.4% (B) 9.1% (C) 10.5% (D) 12.5% Q.20 The work done on the gas in taking it from D to A is (2009) Q.26 A Carnot engine, whose efficiency is 40%, takes (A) – 414 R (B) + 414 R in heat from a source maintained at a temperature of 500 K It is desired to have an engine of efficiency 60%. (C) – 690 R (D) + 690 R Then, the intake temperature for the same exhaust (sink) temperature must be (2012) Q.21 The net work done on the gas in the cycle ABCDA is (2009) (A) Efficiency of Carnot engine cannot be made larger than 50% (A) Zero (B) 276 R (B) 1200 K (C) 1076 R (D) 1904 R (C) 750 K
Q.22 A diatomic ideal gas is used in a Car engine as the (D) 600 K working substance. If during the adiabatic expansion part of the cycle, volume of the gas increases from V to Q.27 An ideal gas enclosed in a vertical cylindrical 32V the efficiency of the engine is (2010) container supports a freely moving piston of mass M. (A) 0.5 (B) 0.75 (C) 0.99 (D) 0.25 The piston and the cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of Physics | 13.39
the gas is V0 and its pressure is P0. The piston is slightly (i) Sequentially keeping in contact with 2 reservoirs displaced from the equilibrium position and released. such that each reservoir supplies same amount of heat. Assuming that the system is completely isolated from (ii) Sequentially keeping in contact with 8 reservoirs its surrounding, the piston executes a simple harmonic such that each reservoir supplies same amount of heat. motion with frequency: (2013) 1 APγ 1 V MP In both the cases body is brought from initial (A) 0 (B) 00 temperature 100°C to final temperature 200°C. Entropy 2π VM 2π 2 0 A γ changes of the body in the two cases respectively is 2 (2015) 1 APγ 0 1 MV0 (C) (D) (A) ln 2, 4 ln2 (B) ln 2, ln 2 2π MV0 2πγ AP0 (C) ln 2, 2 ln 2 (D) 2 ln 2, 8 ln 2
Q.28 One mole of diatomic ideal gas undergoes a Q.31 Consider an ideal gas confined in an isolated cyclic process ABC as shown in figure. The process closed chamber. As the gas undegoes an adiabatic BC is adiabatic. The temperatures at A, B and C are expansion, the average time of collision between 400 K, 800 K and 600 K respectively. Choose the correct molecules increases as Vq, where V is the volume of the statement (2014) B CP 800 K gas. The value of q is : γ= (2015) Cv
P 35γ+ 35γ− 600 K (A) (B) A C 6 6 400 K γ+1 γ−1 V (C) (D) 2 2 (A) The change in internal energy in whole cyclic process is 250R Q.32 ‘n’ moles of an ideal gas undergoes a process (B) The change in internal energy in the process CA is A →B as shown in the figure. The maximum temperature (2016) 700R of the gas during the process will be: (C) The change in internal energy in the process AB is P –350R A 2P0 (D) The change in internal energy in the process BC is – 500R P0 B
Q.29 Consider a spherical shell of radius R at V temperature T. The black body radiation inside it can V0 2V0 be considered as an ideal gas of photons with internal U 4 3P V 9P V 9P V 9P V energy per unit volume uT= ∝ and pressure (A) 00 (B) 00 (C) 00 (D) 00 V 2nR 2nR nR 4nR 1U P = . If the shell now undergoes an adiabatic 3V Q.33 An ideal gas undergoes a quasi static, reversible expansion the relation between T and R is (2015) process in which its molar heat capacity C remains constant. If during this process the relation of pressure −R −3R n (A) Te∝ (B) Te∝ P and volume V is given by PV = constant, then n is given by (Here C and C are molar specific heat at 1 1 p V (C) T ∝ (D) T ∝ constant pressure and constant volume, respectively): R 3 R (2016) CC− p CCp − (A) n = (B) n = Q.30 A solid body of constant heat capacity 1 J/°C is CC− CC− being heated by keeping it in contact with reservoirs v v in two ways: CC− Cp (C) n = v (D) n = CC− p Cv 13.40 | Kinetic Theory of Gases and Thermodynamics
JEE Advanced/Boards
Exercise 1 Q.9 An ideal gas has a molar heat capacity CV at constant volume. Find the molar heat capacity of this Q.1 A closed vessel of volume V contains oxygen at gas as a function of volume, if the gas undergoes the 0 αV process: T = T0e . a pressureP0 and temperatureT0. Another closed vessel of the same volume V0 contains helium at a pressure V
P0 and temperatureT0/2. Find ratio of the masses of oxygen to the helium.
Q.2 V-T curve for 2 moles of a gas is straight line as 1 shown in the graph here. Find the pressure of gas at A. T
V(lit.) B Q.10 One mole of an ideal monoatomic gas undergoes a process as shown in the figure. Find the molar specific A heat of the gas in the process. 53o
T(K) Q.11 The pressure of an ideal gas changes with volumes as P = aV where ‘a’ is a constant. One mole of this gas is Q.3 A gas is undergoing an adiabatic process. At a expanded to 3 times its original volume V . Find certain stage A, the values of volume and temperature 0 = (V00 ,T ) and the magnitude of the slope of V-T curve (i) The heat transferred in the process. is m. Find the value of C and C . P V (ii) The heat capacity of the gas
Q.4 Find the molecular mass of a gas if the specific Q.12 In a cycle ABCA consisting of isothermal expansion heats of the gas are C= 0.2 cal/gm°C and C =0.15 P V AB, isobaric compression BC and adiabatic compression cal/gm°C. [Take R = 2 cal/mole°C] CA, find the efficiency of cycle. A A Q.5 The average degrees of freedom per molecules P for a gas is 6. The gas performs 25 J of work when it B expands at constant pressure. Find the heat absorbed C by the gas. >V V0 2V0
Q.6 A mixture of 4gm helium and 28 gm of nitrogen in (Given: T= T = 400K, γ= 1.5) enclosed in a vessel of constant volume 300 K. Find the AB quantity of heat absorbed by the mixture to double the room mean velocity of its molecules. (R = Universal gas Q.13 A highly conduction solid cylinder of radius a constant) and length is surrounded by co-axial layer of a material having thermal conductivity K and negligible heat capacity. Temperature of surrounding space (out Q.7 One mole of an ideal gas is compressed from side the layer) is T , which is higher than temperature 0.5 lit to 0.25 lit. During the compression, 23.04 x 10² J 0 of the cylinder. If heat capacity per unit volume of of work is done on the gas and heat is removed to keep cylinder material is s and outer radius of the layer is b, the temperature of the gas constant at all times. Find calculate time required to increase temperature of the the temperature of the gas. (Take universal gas constant cylinder from T to T . Assume and faces to be thermally R = 8.31 J mol-1K-1) 1 2 insulated.
Q.8 Ideal diatomic gas is taken through a process Q.14 A vertical brick duct (tube) is filled with cast iron. ∆=∆Q 2U. Find the molar heat capacity for the process The lower end of the duct is maintained at a temperature (where ∆ Q is the heat supplied and ∆ U is change in T which is greater than the melting point T of cast iron internal energy) 1 m Physics | 13.41
T and the upper end at a temperature T2 which is less than Q.19 At a temperature of 0 = 273°K, two moles of the temperature of the melting point of cast iron. It is an ideal gas undergoes a process as shown. The total given that the conductivity of liquid cast iron is equal to amount of heat imparted to the gas equals Q = 27.7 kJ. k times the conductivity solid cast iron. Determine the Determine the ratio of molar specific heat capacities. fraction of the duct filled with molten metal. T C
k A 273 K B
V4V h
Q.20 A cylinder containing a gas is closed by a movable piston. The cylinder is submerged in an ice- water mixture. The piston is quickly pushed down from Q.15 A lagged stick of cross section area 1cm² and position 1 to position 2. The piston is held at position length 1 m is initially at a temperature of 10°C. It is then 2 until the gas is again at 0°C and then slowly raised kept between 2 reservoirs of temperature 100°C and back to position 1. Represent the whole process on P-V 0°C. Specific heat capacity is 10 J/kg°C and linear mass diagram. If m = 100 gm of ice are melted during the density is 2 kg/m. Find cycle, how much work is done on the gas, Latent heat of ice = 80 cal/gm.
100oC 0Co x 1
(a) Temperature gradient along the rod in steady state. 2 (b) Total heat absorbed by the rod to reach steady state.
Q.16 A cylindrical block of length 0.4 m and area of cross-section 0.04m² is placed coaxially on a thin metal Q.21 A parallel beam of particles of mass m moving with disc of mass 0.4 kg and of the same cross-section. The velocities v impinges on a wall at an angle θ to its normal. upper face of the cylinder is maintained at a constant The number of particles per unit volume in the beam is temperature of 400K and the initial temperature of the n. If the collision of particles with the wall is elastic, then disc is 300K. If the thermal conductivity of the material of find the pressure exerted by this beam on the wall. the cylinder is 10 watt/m-K and the specific heat of the material of the disc in 600 J/kg-K, how long will it take for Q.22 For the thermodynamic process shown in the figure. the temperature of the disc to increase to 350K? Assume, P= 1x1055 Pa;P = 0.3x10 Pa for purposes of calculation, the thermal conductivity of AB 5 the disc to be very high and the system to be thermally PDA= 0.6x10 Pa;V= 0.20litre VD = 1.30litre insulated expect for the upper face of the cylinder. P
A Q.17 A liquid takes 5 minutes to cool from 80°C to PA 50°C. How much time will it take to cool from 60°C P to 30°C? The temperature of surrounding is 20°C. Use D D B C exact method. PD V V V V Q.18 An ideal gas at NTP is enclosed in an adiabatic A C D vertical cylinder having area of cross section A = 27 cm², (a) Find the work performed by the system along path AD. between two light movable pistons as shown in the figure. Spring with force constant k = 3700 N/m is in (b) If the total work done by the system along the path a relaxed state initially. Now the lower piston is moved ADC is 85J find the volume at point C. upwards a height h/2, h being the initial length of gas (c) How much work is performed by the system along column. It is observed that the upper piston moves up the path CDA? by a distance h/16. Find h taking γ for the gas to be 1.5. Also find the final temperature of the gas. 13.42 | Kinetic Theory of Gases and Thermodynamics
2 Exercise 2 mean square velocity of molecules of gas B is v and mean square of x component of velocity of molecules w2 w22 /v Single Correct Choice Type of gas A is . The ratio of is? (A) 1 (B) 2 (C) 1/3 (D) 2/3 Q.1 A perfect gas of a given mass is heated first in small vessel and then in a large vessel, such that their Q.7 A reversible adiabatic path on a P-V diagram for an volumes remain unchanged. The P-T curves are ideal gas passes through state A where P=0.7x105 N/m-2 (A) Parabolic with same curvature and v = 0.0049m³. The ratio of specific heat of the gas is 1.4. The slope of path at A is? (B) Parabolic with different curvature 75− 75− (A) 2.0 x10 Nm (B) 1.0 x10 Nm (C) Linear with same slopes 75− 75− (C) −2.0 x10 Nm (D) −1.0 x10 Nm (D) Linear with different slopes Q.8 A cylinder made of perfectly non conducting Q.2 An open and wide glass tube is immersed vertically material closed at both ends is divided into two equal in mercury in such a way that length 0.05 m extends parts by a heat proof piston. Both parts of the cylinder above mercury level. The open end of the tube is closed contain the same masses of a gas at a temperature and the tube is raised further by 0.43 m. The length of t0 = 27°C and pressure P0 =1 atm. Now if the gas in air column above mercury level in the tube will be? Take one of the parts is slowly heated to t = 57°C while P= 76cm atm of mercury. the temperature of first part is maintained at t0 the (A) 0.215 m (B) 0.2 m (C) 0.1 m (D) 0.4 m distance moved by the piston from the middle of the cylinder will be (length of the cylinder = 84cm)
Q.3 A container X has volume double that of container (A) 3cm (B) 5cm (C) 2cm (D) 1cm Y and both are connected by a thin tube. Both contains same ideal gas. The temperature of X is 200K and that Q.9 A vessel contains an ideal monoatomic gas which of Y is 400 K. If mass of gas in X is m then Y it will be: expands at constant pressure, when heat Q is given to (A) m/8 (B) m/6 (C) m/4 (D) m/2 it. Then the work done in expansion is: 3 2 2 (A) Q (B) Q (C) Q (D) Q Q.4 An ideal gas of Molar mass M is contained in a 5 5 3 vertical tube of height H, closed at both ends. The tube is accelerating vertically upwards with acceleration g. Multiple Correct Choice Type Then, the ratio of pressure at the bottom and the mid point of the tube will be Q.10 Two bodies A and B have thermal emissivities of 0.81 respectively. The outer surface areas of the two (A) exp[2MgH/RT] (B) exp[-2MgH/RT] bodies are the same. The two bodies radiate energy at (C) exp[MgH/RT] (D) MgH/RT the same rate. The wavelength λB , corresponding to the maximum spectral radiancy in the radiation from B, is shifted from the wavelength corresponding to the Q.5 Two monoatomic ideal gas at temperature T1 and maximum spectral radiancy in the radiation from A by T2 are mixed. There is no loss of energy. If the masses 1.00 µm. If the temperature of A is 5802 K, of molecules of the two gases are m1 and m2 and number of their molecules are n1 and n2 respectively. (A) The temperature of B is 1934K The temperature of the mixture will be? (B) λ=B 1.5 µ m TT+ TT 12 12+ (A) nn+ (B) s (C) The temperature of B is 11604 K 12 nn12 (D) The temperature of B is 2901 K n T+ nT nT+ n T (C) 21 12 (D) 11 2 2 nn+ nn+ 12 12 Q.11 During an experiment, an ideal gas is found to obey a condition VP² = constant. The gas is initially Q.6 At temperature T, N molecules of gas A each having at a temperature T, pressure P and volume V. The gas mass m and at the same temperature 2N molecules of expands to volume 4V. gas B each having mass 2m are filled in a container. The Physics | 13.43
P Q.15 The figure shows a radiant energy spectrum graph (A) The pressure of gas changes to 2 for a black body at a temperature T. Choose the correct statement(s) (B) The temperature of gas changes to 4T dEl T (C) The graph of above process on the P-T diagram is dl parabola (D) The graph of above process on the P-T diagram is O l l hyperbola. m (A) The radiant energy is not equally distributed among Q.12 The total kinetic energy of translatory motion all the possible wavelengths of all the molecules of 5 litres of nitrogen exerting a (B) For a particular wavelength the spectral intensity is pressure P is 3000 J. maximum (A) The total K.E. of 10 litres of N at a pressure of 2P is 2 (C) The area under the curve is equal to the total rate at 3000 J which heat is radiated by the body at that temperature (B) The total K.E. of 10 litres of He at a pressure of 2P (D) None of these is 3000 J (C) The total K.E. of 10 litres of Q at a pressure of 2P is 2 Q.16 Two metallic sphere A and B are mode of same 20000 J material and have got identical surface finish. The mass (D) The total K.E. of 10 litres of Ne at a pressure of 2P of sphere A is four times that of B. Both the spheres are is 12000 J heated to the same temperature and placed in a room having lower temperature but thermally insulated from each other. Q.13 A container holds 1026 molecules/m³, each of mass 3 x 10-27 kg. Assume that 1/6 of the molecules move (A) The ratio of heat loss of A to that of B is 24/3. with velocity 2000 m/s directly towards one wall of the (B) The ratio of heat loss of A to that of B 22/3. container while the remaining 5/6 of the molecules move either away from the wall or in perpendicular (C) The ratio of the initial rate of cooling of A to that of direction, and all collisions of the molecules with the B is 2-2/3. wall are elastic (D) The ratio of the initial rate of cooling of A to that of (A) Number of molecules hitting 1 m² of the wall every B is 2-4/3. second is 3.33 x 1028. (B) Number of molecules hitting 1 m² of the wall every Q.17 50 gm ice at = 10°C is mixed with 20 gm steam at second is 2 x 1029. 100°C. When the mixture finally reaches its steady state inside a calorimeter of water equivalent 1.5 gm then: (C) Pressure exerted on the wall by molecules is [Assume calorimeter was initially at 0°C, take latent 5 24 x 10 Pa. heat of vaporization of water = 1 cal/gm-°C, specific (D) Pressure exerted on the wall by molecules is heat capacity of ice = 0.5 cal/gm°C] 5 4 x 10 Pa. (A) Mass of water remaining is: 67.4 gm (B) Mass of water remaining is: 67.87 gm Q.14 Two gases have the same initial pressure, volume and temperature. They expand to the same final volume, (C) Mass of steam remaining is: 2.6 gm one adiabatically and the other isothermally (D) Mass of steam remaining is: 2.13 gm (A) The final temperature is greater for the isothermal process Q.18 A gas expands such that it is initial and final (B) The final pressure is greater for the isothermal temperature are equal. Also, the process followed by process the gas traces a straight line on the P-V diagram: (C) The work done by the gas is greater for the (A) The temperature of the gas remains constant isothermal process. throughout. (D) All the above options are incorrect (B) The temperature of the gas first increases and then decreases. 13.44 | Kinetic Theory of Gases and Thermodynamics
(C) The temperature of the gas first decreases and then Q.21 If the temperature of the body is raised to higher increases. temperature T’, then choose the correct statement(s) (D) The straight line has a negative slopes (A) The intensity of radiation for every wavelength increases Q.19 A cyclic process ABCD is shown in the P-V (B) The maximum intensity occurs at a shorter diagram. Which of the following curves represents the wavelength same process if BC & DA are isothermal processes. (C) The area under the graph increases A B (D) The area under the graph is proportional to the P fourth power of temperature C
D Paragraph 2: V Two rods A and B of same cross-sectional area A and length / connected in series between a source AB D C (T1 =100°C) and a sink (T2 = 0°C) as shown in figure. The rod is laterally insulated (A) (B) P V C B T1 T2 D A 3K K o 100oC 0C T T l l
A B B Q.22 The ratio of the thermal resistance of the rod is A R 1 R (C ) (D) (A) A = (B) A = 3 R3 R P C V B B D D C R 3 4 (C) A = (D) T T R4B 3
Q.23 If T and T are the temperature drops across the Comprehension Type A B rod A and B, then Paragraph 1: T 3 T 1 (A) A = (B) A = T1B T3B dEl T dl TA 3 TA 4 (C) = (D) = T4B T3B O l m l Q.24 If GA and GB are the temperature gradients across the rod A and B, then Q.20 The figure shows a radiant energy spectrum graph for a black body at a temperature T. G 3 G 1 (A) A = (B) A = G1 G3 Choose the correct statement (s) B B (A) The radiant energy is not equally distributed among G 3 G 4 (C) A = (D) A = all the possible wave lengths G4B G3B (B) For a particular wavelength the spectral intensity is maximum Paragraph 3: (C) The area under the curve is equal to the total rate at In fluids heat transfer tales place and molecules of which heat is radiated by the body at that temperature. the medium takes very active part. The molecules (D) None of these take energy from high temperature zone and move towards low temperature zone. This method is known Physics | 13.45 as convection, when we require heat transfer with fast Q.29 The work done in the complete cycle ABCA is phase, we use some mechanism to make the flow of fluid on the body fast. The rate of loss of heat is proportional (A) 90 J (B) 60 J (C) 0 J (D) 30 J to velocity of fluid (v), and temperature difference ( ∆ T) between the body any fluid, of course more the surface area of body more the rate of loss of heat. We can write Paragraph 5: dQ Five moles of helium are mixed with two moles of the rate of loss of heat as =KAv ∆ T dt hydrogen to form a mixture. Take molar mass of helium M= 4g and that of hydrogen M= 2g Where K is positive constant. 1 2 Now answer the following questions Q.30 The equivalent molar mass of the mixture is 13g Q.25 A body is being cooled with fluid. When we (A) 6g (B) 7 increase the velocity of fluid 4 times and decrease the 18g temperature difference 1/2 time, the rate of loss of heat (C) (D) None increases. 7 (A) Four times (B) Two times Q.31 The equivalent degree of freedom f of the mixture (C) Six times (D) No change is (A) 3.57 (B) 1.14 (C) 4.4 (D) None Q.26 In the above question if mass of the body increased two times, without change in any of the other γ parameters, the rate of cooling Q.32 The equivalent value of is (A) Decreases (A) 1.59 (B) 1.53 (C) 1.56 (D) None (B) Increases Q.33 If the internal energy of He sample of 100J and (C) No effect of change of mass that of the hydrogen sample is 200J, then the internal (D) None of these energy of the mixture is (A) 900J (B) 128.5J Paragraph 4: (C) 171.4J (D) 300J When a thermo-dynamic process is shown on P-V diagram, area under curve represents work done Match the Column during process. During cyclic process work done is area enclosed. The P-V graph for a thermodynamically Q.34 An ideal gas at pressure P and volume V is system is shown in figure. expanded to volume 2V. Column I represents the thermodynamic processes used during expansion.
) 20 A column II represents the work during these processes
-2 in the random order.
10
e P (in Nm B Column I Column II C −γ essur PV(1− 21 ) Pr (p) isobaric (x) 12 γ−1 6 3 Volume V (in m) (q) isothermal (y) PV Q.27 The work done by the system in the process A to B is (r) adiabatic (z) PV/n 2
(A) 90 J (B) 60 J (C) 0 J (D) 30J The correct matching of column I and column II is given by: Q.28 The work done in the process B to C is (A) p-y, q-z, r-x (B) p-y, q-x, r-z (A) -90 J (B) -60 J (C) 0 J (D) -30J (C) p-x, q-y, r-z (D) p-z, q-y, r-x 13.46 | Kinetic Theory of Gases and Thermodynamics
Previous Years’ Questions d d d d Q.1 When the ideal diatomic gas is heated at constant pressure the fraction of the heat energy supplied which increases the internal energy of the gas is (1990) T T 2 3 3 5 D D (A) (B) (C) (D) T T+ T T T+ T 5 5 7 7 T T d T T+DT d T T+DT
Q.2 A vessel contains a mixture of one mole of oxygen d d and two moles of nitrogen at 300 K. The ratio of the average rotational kinetic energy per O2 molecule to per N2 molecule is (1998) T T D D (A) 1: 1 T T+ T T T+ T T T (B) 1: 2 T T+DT T T+DT (C) 2: 1 (D) Depends on the moment of inertia of the two Q.7 P-V plots for two gases during adiabatic processes molecules are shown in the figure. Plots 1 and 2 should correspond respectively to (2001) Q.3 Two cylinders A and B fitted with pistons contain P equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each 1 cylinder. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of the gas in B is (1998) 2 V (A) 30 K (B) 18 K (C) 50 K (D) 42 K
(A) He and O2 (B) O2and He Q.4 The ratio of the speed of sound in nitrogen gas to (C) He and Ar (D) O and N that in helium gas, at 300 K is (1999) 2 2
(A) (2 / 7) (B) (1 / 7) Q.8 An ideal gas is taken through the cycle A→→→ B C A, as shown in the figure. If the net heat (C) ( 3) /5 (D) ( 6)/5 supplied to the gas in the cycle is 5J, the work done by the gas in the process CA→ is (2002) 3 Q.5 Starting with the same initial conditions, and ideal V(m ) gas expands from volume V12 to V in three different 2 CB ways. The work done by the gas is W1 if the process is purely isothermal, W2 if purely isobaric and W3 if purely adiabatic, then (2000) 1 A (A) W213>> WW; (B) WWW231>>
(C) WW123>> W; (D) WWW132>> 2 10 p(N/m ) Q. 6 An ideal gas is initially at temperature T and volume V. Its volume is increased by ∆V due to an increase (A) -5J (B) -10 J (C) -15 J (D) -20 J in temperature ∆T , pressure remaining constant. The quantity δ=V/V ∆ T varies with temperature as (2000) Q.9 Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which of the following graphs represent the variation of temperature with time? (2004) Physics | 13.47
Temp Temp Temp Temp Q.12 The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then be (2007) (A) (B) Time Time P0 Time Time (A) P (B) 0 2
P0 Mg P0 Mg Temp Temp (C) + (D) − Temp Temp 2 πR2 2 πR2
(C) (D) Q.13 Whilethe piston is at a distance 2L from the top, Time Time the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is (2007) Q.10 An ideal gas expands isothermally from a volume 2Pπ R 2 Pπ− R2 Mg V to V V (A) 0 (2L) (B) 0 (2L) 12 and then compressed to original volume 1 2 2 π+R P0 Mg πRP0 adiabatically. Initial pressure is P1 and final pressure is P3 . The total work done is W. Then, (2004) 2 2 P0π+ R Mg PR0π (A)P31>> P ,W 0 (B) P31= P ,W > 0 (C) (2L) (D) (2L) 2 2 πRP0 π−R P0 Mg (C) P31>< P ,W 0 (D) P31= P ,W = 0
Q.14 The piston is take completely out of the cylinder. Q.11 Statement-I The total translational kinetic energy The hole at the top is sealed. A water tank is brought of all the molecules of a given mass of an ideal gas is below the cylinder and put in a position so that the 1.5 times the product of its pressure and its volume water surface in the tank is at the same level as the top Statement-II The molecules of a gas collide with each of the cylinder as shown in the figure. The density of other and the velocities of the molecules change due to the water is ρ . In equilibrium, the height H of the water the collision. (2007) column in the cylinder satisfies (2007)
(A) If statement-I is true, statement-II is true; statement-II is the correct explanation for statement-I (B) If statement-I is true, statement-II is true; statement-II is not a correct explanation for statement-I (C) If Statement-I is true; statement true is false L0 (D) If Statement-I is false; statement-II is true H
Paragraph 1: A fixed thermally conducting cylinder has a radius R ρ −2 + −+ = and height L0 . The cylinder is open at its bottom and (A) g(L0 H) p 0 (L 0 H) L 00 p 0 has a small hole at its top. A piston of mass M is held at 2 (B) ρg(L − H) − p (L −− H) L p = 0 a distance L from the top surface as shown in the figure. 0 0 0 00 2 The atmospheric pressure is P0 . (C) ρg(L0 − H) + p 0 (L 0 −− H) L 00 p = 0 2R 2 (D) ρg(L0 − H) − p 0 (L 0 −+ H) L 00 p = 0 L
Q.15 Cv and Cp denote the molar specific heat L0 capacities of a gas at constant volume and constant pressure, respectively. Then, (2009)
(A) CCPV− is larger for a diatomic ideal gas than for a Piston monoatomic ideal gas 13.48 | Kinetic Theory of Gases and Thermodynamics
(B) C + C is larger for a diatomic ideal gas than for a P V Column I Column II monoatomic ideal gas C (B) An ideal monoatomic gas (p) The temperature ( P C) is larger for a diatomic ideal gas than for a expands to twice its original of the gas increases CV volume such that its pressure or remains constant monoatomic ideal gas 1 (D) C *C is larger for a diatomic ideal gas than for a p,∝ where V is the volume PV V2 monoatomic ideal gas of the gas. (C) An ideal monoatomic gas (q) The gas loses Q.16 The figure shows the P-V plot an ideal gas taken expands to twice its original heat through a cycle ABCDA. The part ABC is a semi-circle volume such that its pressure and CDA is half of an ellipse. Then, (2009) 1 p,∝ where V is its volume. P V4/3 A 3 (D) An ideal monoatomic gas (r) The gas gains 2 D B expands such that its pressure heat p and volume V follows the 1 C behaviour shown in the graph. 0 1 23 V P
(A) The process during the path AB→ is isothermal (B) Heat flows out of the gas during the path BC→→ D (C) Work done during the path AB→→ Cis zero V V 2V (D) Positive work is done by the gas in the cycle ABCDA 1 1
Match the Columns Q.18 A real gas behaves like an ideal gas if its (2010) Q.17 Column I contains a list of processes involving (A) Pressure and temperature are both high expansion of an ideal gas. Match this with column II (B) Pressure and temperature are both low describing the thermodynamic change during this process. Indicate your answer by darkening the (C) Pressure is high and temperature is low appropriate bubbles of the 4 x 4 matrix given in the (D) Pressure is low and temperature is high ORS. (2008)
Q.19 One mole of an ideal gas in initial state A Column I Column II undergoes a cyclic process ABCA, as shown in the figure. (A) An insulated container has two (p) The temperature Its pressure at A is P0. Choose the correct option(s) from chambers separated by a valve. of the gas decreases the following: (2010) Chamber I contains an ideal gas V and the Chamber II has vacuum. 4V B The valve is opened. 0
I II A Ideal gas Vacuum V0 C T T0
(A) Internal energies at Aand B are the same
(B) Work done by the gas in process AB is P0V0 ln 4 P (C) Pressure at C is 0 4 T (D) Temperature at C is 0 4 Physics | 13.49
Q.20 5.6 liter of helium gas at STP is adiabatically Q.25 A water cooler of storage capacity 120 litres can compressed to 0.7 liter. Taking the initial temperature cool water at a constant rate of P watts. In a closed ( ) to be T1, the work done in the process is 2011 circulation system (as shown schematically in the figure), the water from the cooler is used to cool an 9 3 15 9 (A) RT1 (B) RT1 (C) RT1 (D) RT1 external device that generates constantly 3 kW of heat 8 2 8 2 (thermal load). The temperature of water fed into the device cannot exceed 30°C and the entire stored 120 Q.21 A mixture of 2 moles of helium gas (atomic litres of water is initially cooled to 10°C. The entire mass = 4 amu) and 1 mole of argon gas (atomic mass system is thermally insulated. The minimum value of = 40 amu) is kept at 300 K in a container. The ratio of P (in watts) for which the device can be operated for 3 hours is Vrms (helium) the rms speeds is (2012) Cooler Device Vrms (argon) Hot
(A) 0.32 (B) 0.45 (C) 2.24 (D) 3.16
Q.22 Two non-reactive monoatomic ideal gases have their Cold atomic masses in the ratio 2 : 3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant (Specific heat of water is 4.2 kJ kg-1 K-1 and the density temperature, is 4 : 3. The ratio of their densities is (2013) of water is 1000 kg m-3) (2016) (A) 1 : 4 (B) 1 : 2 (C) 6 : 9 (D) 8 : 9 (A) 1600 (B) 2067 (C) 2533 (D) 3933
Q.23 A thermodynamic system is taken from an initial Q.26 A cylindrical vessel of height 500 mm has an orifice state i with internal energy U = 100 J to the final state (small hole) at its bottom. The orifice is initially closed i and water is filled in it upto height H. Now the topis f along two different paths iaf and ibf, as schematically completely sealed with a cap and the orifice at the bottom shown in the figure. The work done by the system along is opened. Some water comes out from the orifice and the paths af, ib and bf are Waf = 200 J, Wib = 50 J and Wbf = the water level in the vessel becomes steady with height 100 J respectively. The heat supplied to the system along of water column being 200 mm. Find the fall in height (in mm) of water level due to opening of the orifice. the path iaf, ib and bf are Qiaf, Qib and Qbf respectively. If the internal energy of the system in the state b is [Take atmospheric pressure = 1.0 × 105 N/m2, density of water = 1000 kg/m3 and g = 10 m/s2. Neglect any effect Ub = 200 J and Qiaf = 500 J, the ratio Qbf / Qib is (2014) of surface tension.] (2009) a f Q.27 A diatomic ideal gas is compressed adiabatically P 1 to of its initial volume. If the initial temperature of 32 i b the gas is Ti (in Kelvin) and the final temperature is aTi, v the value of a is (2010)
A container of fixed volume has a mixture of Q.24 Q.28 One mole of a monatomic gas is taken through one mole of hydrogen and one mole of helium in a cycle ABCDA as shown in the P-V diagram. column equilibrium at temperature T. Assuming the gases are II give the characteristics involved in the cycle. Match (2015) ideal, the correct statement(s) is(are) them with each of the processes given in column I. (2011) (A) The average energy per mole of the gas mixture is 2RT. P B A (B) The ratio of speed of sound in the gas mixture to 3P that in helium gas is. (C) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/2. 1P C D (D) The ratio of the rms speed of helium atoms to that of hydrogen molecules is. 0 1V 3V 9V V 13.50 | Kinetic Theory of Gases and Thermodynamics
Q.30 If the piston is pushed at a speed of 5 mms–1, the Column I Column II air comes out of the nozzle with a speed of (p) Internal energy decreases (A) Process A → B (A) 0.1 ms–1 (B) 1 ms–1 (C) 2 ms–1 (D) 8 ms–1
(B) Process B → C (q) Internal energy increases. Q.31 If the density of air is ρa and that of the liquid ρ , (r) Heat is lost (C) Process C → D then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional (D) Process D → A (s) Heat is gained to (t) Work is done on the gas ρa ρ (A) (B) ρρ (C) (D) ρ ρ a ρ a Q.29 The figure shows the variation of specific heat capacity (C) of a solid as a function of temperature Q.32 An ideal monoatomic gas is confined in a horizontal (T). The temperature is increased continuously from cylinder by a spring loaded piston (as shown in the
0 to 500 K at a constant rate. Ignoring any volume figure). Initially the gas is at temperature 1T , pressure P1 change, the following statement(s) is (are) correct to a and volume V1 and the spring is in its relaxed state. reasonable approximation. (2013)
C The gas is then heated very slowly to temperature
T2, pressure P2 and volume V2. During this process the piston moves out by a distance x. Ignoring the friction between the piston and the cylinder, the correct 100 200 300 400 500 statement(s) is(are) (2015) T(K)
(A) The rate at which heat is absorbed in the range (A) If V2 = 2V1 and T2 = 3T1, then the energy stored in 0-100 K varies linearly with temperature T. 1 the spring is P1 V1 (B) Heat absorbed in increasing the temperature from 4
0-100 K is less than the heat required for increasing the (B) If V2 = 2V1 and T2 = 3T1, then the change in internal temperature from 400 – 500 K. energy is 3P1V1 (C) There is no change in the rate of heat absorption in (C) If V = 3V and T = 4T , then the work done by the range 400 – 500 K. 2 1 2 1 7 gas is P V (D) The rate of heat absorption increases in the range 3 1 1 200 – 300 K. (D) If V2 = 3V1 and T2 = 4T1, then the heat supplied to 17 the gas is P V Paragraph 1: 6 1 1 A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross Q.33 A gas is enclosed in a cylinder with a movable section is connected to the nozzle. The other end of the frictionless piston. Its initial thermodynamic state at 5 -3 3 tube is in a small liquid container. As the piston pushes pressure Pi = 10 Pa and volume Vi = 10 m changes × 5 × -3 air through the nozzle, the liquid from the container to a final state at Pf = (1/32) 10 Pa and Vf = 8 10 3 3 5 rises into the nozzle and is sprayed out. For the spray m in an adiabatic quasi-static process, such that P V gun shown, the radii of the piston and the nozzle are = constant. Consider another thermodynamic process 20 mm and 1 mm respectively. The upper end of the that brings the system from the same initial state to the container is open to the atmosphere. (2014) same final state in two steps: an isobaric expansion at Pi followed by an isochoric (isovolumetric) process at
volume Vf. The amount of heat supplied to the system in the two-step process is approximately (2016) (A) 112 J (B) 294 J (C) 588 J (D) 813 J Physics | 13.51
MASTERJEE Essential Questions
JEE Main/Boards JEE Advanced/Boards
Exercise 1 Exercise 1 Q. 11 Q.20 Q.32 Q.12 Q.15 Q.18 Q.33 Q.34 Q.35 Q.19 Q.22 Q.36
Exercise 2 Exercise 2 Q.4 Q.5 Q.15 Q.8 Q.10 Q.11 Q.16 Q.17 Q.20 Q.28 Q.29 Q.30
Answer Key
JEE Main/Boards
Exercise 1
2x(273− 173) Q.2 P /T= P /T orT = P T /P = =200K =−° 73 C 112 2 2211 1
Q.10 6 x 10-4 Q.11 0.103 kg Q.13 1.88 x 109 s-1
Q.14 1.87 x 10-7 m Q.15 1.47 x 10-7 m
Q.17 No, because internal energy of an ideal gas depends only on temperature of the gas
∆V1 Q.18 Translational motion of molecules. Q.19 =γ.As1, γ> ∴∆ (V)(V)12 > ∆ ∆V2 Q.20 Internal energy increasing by 200 J. Heat absorbed is zero.
Q.32 80.8°C Q.33 (i) 2.675 atm (ii) 128.3°C
Q.34 -1176 joule, -280 cal Q.35 – 1374 J Q.36 – 43.38 J 13.52 | Kinetic Theory of Gases and Thermodynamics
Exercise 2
Single Correct Choice Type Q.1 A Q.2 A Q.3 C Q.4 B Q.5 A Q.6 B Q.7 A Q.8 C Q.9 D Q.10 D Q.11 C Q.12 A Q.13 B Q.14 A Q.15 D Q.16 A
Previous Years’ Questions
Q.1 A Q.2 A Q.3 B Q4 B Q.5 B Q.6 D
Q.7 C Q.8 C Q.9 D Q.10 D Q.11 C
Q.12 A → s; B → r, p; C → p; D → s Q.13 A → p; r, t; B → p, r; C → q, s; D → r, t Q.14 A, B, C, D
Q.15 A, B Q.16 D Q.17 A Q.18 B Q.19 C Q.20 A
Q.21 B Q.22 B Q. 23 C Q.24 D Q.25 A Q.26 C
Q.27 C Q.28 D Q.29 C Q.30 B Q.31 C Q.32 D
Q.33 A
JEE Advanced/Boards
Exercise 1
Q.1 4:1 Q.2 1.25 × 104 Pa mRT T m 00+ Q.3 1R Q.4 The molar mass of the gas is 40 gm, VV00
Q.5 100J Q.6 3600 R Q.7 400 K Q.8 5R R R Q.9 C + Q.10 V α V 2 P 1
atic 31− 1/3 Adiabatic γ+1 2 γ+1R 2 (i) 4aV , (ii) 1− Q.11 Isob 0 Q.12 γ−1 γ−12 ln2 Isotherm V
as2 b TT− l k(T− T ) log log 01 1= 1m Q.13 ee Q.14 2K a T02− T l k(T1m− T ) +− (T m T 2 )
Q.15 (a) -100°C/m, (b) 1000 J Q.16 166.3 sec
Q.17 10 minutes Q.18 1.6 m, 364 K
Q.19 1.63 Q.20 8000 cal.
2 2 Q.21 2mv cos θ Q.22 (a)WAD = 88 J, (b)VC = 1.23litre,
(c)WCDA = − 85J Physics | 13.53
Exercise 2
Single Correct Choice Type
Q.1 D Q.2 C Q.3 C Q.4 C Q.5 D Q.6 D Q.7 C Q.8 C Q.9 C
Multiple Correct Choice Type
Q.10 A, B Q.11 A, D Q.12 C, D Q.13 A, D Q.14 A, B, C Q.15 A, B Q.16 A, C Q.17 A, C Q.18 B, D Q.19 A, B
Comprehension Type
Paragraph 1: Q.20 A, B Q.21 A, B, C, D Paragraph 2: Q.22 A Q.23 B Q.24 B
Paragraph 3: Q.25 B Q.26 A Paragraph 4: Q.27 A Q.28 B Q.29 D
Paragraph 5: Q.30 D Q.31 A Q.32 C Q.33 D
Match the Column
Q.34 A
Previous Years’ Questions
Q.1 D Q.2 A Q.3 D Q.4 C Q.5 A Q.6 C Q.7 B Q.8 A Q.9 C Q.10 C Q.11 B Q.12 A Q.13 D Q.14 C Q.15 B, D Q.16 B, D Q.17 A → q; B → p, r; C → p, s; D → q, s Q.18 D Q.19 A, B Q.20 A Q.21 D Q.22 D Q.23 2 Q.24 A, B, D Q.25 B Q.26 6 Q.27 4 Q.28 A → p, r, t; B → p, r; C → q, s; D → r, t Q.29 A, B, C, D Q.30 C Q.31 A Q.32 B or A, B, C Q.33 C
Solutions
JEE Main/Boards aren’t. So, the molecules are not going in straight line. Thus, it takes time for the smell to spread in the room. Exercise 1 Sol 2: Initial Sol 1: The speed of molecules follows directly from Pi = 1 atm measuring the pressure and density-you don’t need to know the size of molecules. In standard kinetic theory, T0 = -173°C = 100 k collisions with other molecules have always been Vi = V ignored, because the molecules were tiny. Though, they 13.54 | Kinetic Theory of Gases and Thermodynamics
Finally 3. Attraction or repulsion forces between any two particles are negligible Pf = 2 atm 4. The collisions between gas molecules or molecule T = ? f and wall of container are completely elastic meaning
Vf = V = const. no energy is gained or lost from collisions. By ideal gas law 5. The time it takes to collide is negligible PV = nRT 6. All gases at a given temperature have same Kinetic T Energy. ⇒ V = = constant P 7. Motion of particles is random
Ti Tf 8. Effect of gravity on gas molecule is neglected ∴ = P P i f 9. Average momentum of gas molecule is zero.
TPif× 100× 2 ⇒ Expression for the pressure and vrms of a gas- Tf = = Pi 1 2 222 v=++ vvvxyz = 200 K = -73°C By postulates of KTG we know
(ii) Ideal gas equation: - PV = nRT [Assuming n molecules in the vessel] R → Gas constant v22++ v ...... v 2 → mn x1 x2 xn M 2 r Radius of gas molecule = = < v1 > n Sol 5: Postulates of Kinetic Theory Gases (KTG): M = mn : - Total mass 1. Gases consist of particles in random motion. M ⇒ Total force =
Force Sol 8: Mean free path is the average distance travelled Pressure (P) = Area by the molecules between successive collisions.
Assume vessel to be a cube. Sol 9: Brownian motion is the random motion of 1M <>v2 particles in a fluid. Resulting from their collision with 1 M ∴ 3 2 the other atoms or molecules in the fluid. P = = 3
6 5 6 [Mm: molecular weight] P = 2.5 × 10 Pa + 10 Pa = 2.6 × 10 Pa Using gas equation we get PV 2.6×× 1066 3000 × 10− n = = 1 RT 8.31× 293 PV = n M V 2 3 m rms _ 3.20 c/moles 1 3 ⇒ M v2 = PV ; V is molar volume ∴ 2 m rms 2 m m Mass of O2 in tank = 3.204 × 32 _ 102.5 gm = _ 0.103 kg L.H.S. is the expression for kinetic energy per mole of gas. 3RT 3×× 8.314 273 Sol 12: Vrms = = −3 Mm 64× 10 Sol 7: We know that molar kinetic energy _ 3 3 326.18 m/s = PV = RT [From previous question] 2 m 2 1 3 Sol 13: Average time per collision ∴ Kinetic energy per molecule = [ RT] −7 NA 2 2.66× 10 m = = 5.32 × 10-10 sec 1 3 3 500m sec−1 = × KNA T = KT NA 2 2 ∴ No. of collision per second {N : Avogadro’s number A 1 = _ 1.88 × 109 sec K: Boltzmann constant} 5.32× 10−10 1 3 ⇒ m v2 = kT [m: molecular mass] 1 2 rms 2 Sol 14: Mean free path estimate = 2π dn2 mv 2 2 v ∴ T = rms = × kinetic energy per molecule. 3k 3k d: diameter of molecule, nv molecules per unit volume ∴ 19 -3 25 -3 The average energy of the molecule is proportional to nV = 3 × 10 cm = 3 × 10 m absolute temperature. Absolute zero is the temperature d = 2 × 10-10 m at which the kinetic energy of the molecules becomes zero. i.e. they stop. 13.56 | Kinetic Theory of Gases and Thermodynamics
∴ Mean free path ∆V2 1 −V ⇒ = + …….(ii) 1 ∆P γ P = −10 2 25 1.414× 3.14 × (2 × 10 ) ×× 3 10 By (i) and (ii) we get
-5 ∆ = 0.0187 × 10 m V2 1 ∆V1 = _ -7 ∆P γ ∆P 1.87 × 10 m ∆V 1 ⇒ 1 = γ (x) ∆V Sol 15: Mean free path = 2 2 2π dnv nNA As r > 1 nv = V ∴ D D V1 > V2 n: number of moles in volume V 23 Sol 20: ∆W = -200 J NA = Avogadro’s number = 6.02 × 10 n P DQ = 0 [adiabatic conditions] By gas equation: - = V RT By 1st law of thermodynamics:- PN ∴ n = A DQ = DU + DW v RT ⇒ D RT U = + 200 J ∴ x = 2π d2 PNA 8.314× 273 Sol 21: The internal energy is the total energy contained = by a thermodynamics system. It is the energy needed 1.414× 3.14 × (2.4 × 10−10 ) 2 ×× 10 5 6.02 × 1023 to create the system but excludes the energy due to _ -7 1.47 × 10 m external force fields. E.g. Kinetic, energy if the sample is moving or potential energy if the sample is at a height from the ground. Internal energy has two major Sol 16: Water has higher potential energy. As we already components i.e. kinetic and internal potential energy. know that kinetic energy depends on temperature. So, as we cool water to ice its K.E. remains constant at 0°C. Therefore, the heat energy removed from water, is Sol 22: Four thermodynamics process are:- accounted as decrease in potential energy of ice. (i) Isothermal process: Temperature remains constant. i.e. PV = constant [∵ DT = 0] Sol 17: We know that internal energy is a state function ∴ D D U = nCV T = 0 and depends on temperature. (ii) Adiabatic process:- No flow of heat either into or out ∴ D → U = 0 No change from the system. i.e. DQ = 0 = DU + W Sol 18: Translational motion of molecules. ∴ DU = -W Sol 19: For isothermal (iii) Isochoric process:- Volume of the system remains constant. PV = constant ⇒ D D i.e. V = constant P V1 + V P = 0 ⇒ DV = 0 ∆V1 −V ⇒ = – …...(i) ∴ W = PdV = 0 ∆P P ∫ Thus, Dθ = DU For adiabatic process (iv) Isobaric process:- Pressure remains constant γ PV = constant i.e. P = constant γ γ−1 ⇒ V∆+γ P PV ∆ V= 0 2 DP = 0
W = P (Vf – Vi) Physics | 13.57
Indicator diagram: are graphical representation of cyclic nRT We know, PV = nRT ⇒ P = variations of pressure and volume within a heat engine. V
Vf nRTdV Vf Vf Sol 23: Sign Convention ∴ DW = = nRT ln V = nRT ln ∫ Vi V V Vi DQ → +ve for heat supplied to gas i Vf Pi → -ve for heat rejected by gas ⇒ DW = nRT ln = nRT ln Vi Pf DW→+ve for work done by gas → -ve for work done on gas Sol 28: For adiabatic process DQ = 0 Sol 24: Isothermal process ∴ DU = -DW DT = 0 ∴ nCVdT = -PdV …(i) ∴ D D U = nCv T = 0 By equation of gas:- PV = nRT st By 1 law of thermodynamics:- ⇒ PdV + VdP = nRdT ∴ DQ = DU + DW ⇒ DQ = DW PdV+ VdP ∴ ndT = No change in internal energy R Work done by system = heat supplied to the system. ∴ Substituting in (i) we get :- Adiabatic Process:- PdV+ VdP CV = -PdV DQ = 0 R So, DQ = DU + DW ⇒ (CV+R) PdV = -CV VdP ⇒ D D U = - W C dV −dP ⇒ P = [∵ CV + R = CP] No heat is supplied to the system CVV P Change in internal energy = Work done on gas dV −dP ⇒ γ = V P Sol 25: Isothermal Operation: Operations/ process in On integrating both sides, we get which temperature stays constant. V P f dV f dP V P Example:- System immersed in a large constant γ ∫ = −∫ ⇒ γ ln f = ln i temperature ice-water bath. V P V P Vi Pi i f γ P V Sol 26: Adiabatic operations are those operations in ⇒ i f ⇒ PVγγ = P V = ii f f = k (say) which neither heat enters or leaves a system. Pf Vi
Example:- A system having a perfectly non-conducting γ k ∴ PV = const. = k ⇒ P = boundary. Vγ D V V Q = U + W f f k ∴ PdV W = ∫ = ∫ γ dV ∵ Q = 0 V Vi Vi ∴ DU + W = 0 ⇒ DU = -W Vf V−γ+1 kV11−γ− KV −γ ⇒ W = k = fi Sol 27: DT = 0 −γ +1 1−γ Vi D D U = nCV T = 0 (P Vγ )V11 −γ− (P V γ )V −γ ∴ DW = DQ - DU ⇒ DW = DQ = ff f ii i 1−γ Vf D PdV W = ∫ Pff V− PV ii nR ⇒ W = = (T – T ) Vi 1−γ 1−γ f i 13.58 | Kinetic Theory of Gases and Thermodynamics
Sol 29: Cyclic processes:- In cyclic process, thermo- w dynamic process start from the same state and end at T T the initial state. 1 2
Hot Q1 Q2 Cold A Working P (source) substance (sink)
Here, Q1 is the heat given to the substance whereas Q2 B is the heat released. ∴ V Qgiven = Q1 W QQ− Q Since, initial and final states are same, ∴ η = = 12 = 1 - 2 Q Q Q Thus DU = 0, 1 ⊥ 1 ∴ D D Heat released Q = W ∴ Efficiency ( η ) = 1 - Heat absorbed ∴ |DQ| = |DW| = Area enclosed by the curve. Sign of DW is ‘+’ if process is clockwise and ‘–‘ if process Two types of heat engines:- in anti-clockwise. (i) Internal combustion engines Non-cyclic process: If initial and final states are not (ii) External combustion engines same. A P Sol 32: For adiabatic process γ PV = constant B nRT We know V = P γ V nRT B ∴ P = const. DW = ∫PdV = Area enclosed by P-V curve P A 1−γ γ ⇒PT = const. Sol 30: Reversible process: - The process in which the 1 Pi = 3 atm; Pf = atm system and surroundings can be restored to the initial state from the final state without producing any change Ti = 300 K; Tf = ? in the thermodynamic properties of the universe. PT1−γ γ ∴ Tγ ii Ex: - An infinitesimal compression of a gas in a cylinder. f = 1−γ Pf
Irreversible process:- In irreversible process system is 1− 1.4 not in equilibrium throughout the process. Initial state γ 3 1.4 ⇒ Tf = (300) can’t be obtained from final state without producing 1 changes in the universe. −0.4 Example:- (i) Processes having friction. ⇒ T = 3 1.4 × 300 _ 219.2 K f (ii) Heat transfer through a finite temperature difference. ∴ Fall in temperature = Ti – Tf = 80.8° C Sol 31: Heat engine is a device which convert heat energy into mechanical energy by using a cyclic process. Sol 33: Ti = 27°C = 300 K; Tf = ?
Pi = 1 atm; Pf = ? W = Q1 - Q2 W V = V; V = V/2 η i f Efficiency: = γ Qgiven (i) PV = constant γ Vi ∴ P = P = (2)1.42 × 1 _ 2.675 atm f i Vf Physics | 13.59
γ 1.4− 1 (ii) PV = constant 1 = × 273 = 206.89 K nRT γ 2 ⇒ V = constant V γ−1 nR ⇒ TV = constant ∴ DU = (T – T ) 1−γ f i γ−1 Vi ∴ T = T 1 f V i − ××(2 4.2) f 32 = (206.89 - 273) = (2)0.42 × 300 1− 1.4 ⇒ T ≈ 128.38 °C −43.38 Joule
Sol 34: n = 1 T = T; T = (T + 70) K i f Exercise 2 nR 12× W = [T – T ] = [70] = -280 Cal 1−γ f i 1− 1.5 Sol 1: (A) V = 7L = 7 × 10-3 m3 ∴ Work done = -1176 joules T = 273 K DQ = DU + DW ⇒ DU = 280 Cal P = 1.3 × 105 Pa Work done on gas can be seen by increase in PV 1.3× 1053 ×× 7 10− temperature of the gas, which accounts for 280 calories ∴ m = = _ 0.4 moles of energy. RT 8.314× 273 ∴ No. of molecules = 0.4 × 6.02 × 1023 Sol 35: n = 1 = 2.4 × 1023 molecules D Vi = V; Q = 0 −nR 2/3 D D Sol 2: (A) PV =const. Vf = 2Vi; U = - W = [Tf – Ti] 1−γ nRT ⇒ V2/3 = const. Ti = 273K V γ−1 TV = constant ⇒ TV-1/3 = const. γ−1 ⇒ T = conts. × V1/3 Vi ∴ T = × T f i ∴ Vf On increasing volume, temperature increase
⇒ T = (0.5)0.4 × 273 = 206.89 K f Sol 3: (C) T = constant, V = constant, 1× 8.314 ∴ DU = - [206.89-273] 35 1− 1.4 n = = 1.25 Kmoles i 28 ⇒ DU −1374 Joule P RT ∴ = = const. n V Sol 36: Vi = V; Vf = 2V P = 760 mm of Hg= 1 atm _ 105 Pascal P P i ∴ f = i 1 nf ni T = 273 K, n = i 32 P 9 35 ⇒ f ≈ nf = × ni = × 1.875 kmoles ∵ DQ = 0 Pi 6 28 ∴ For adiabatic process:- Moles of O2 supplied γ−1 TV = constant = 1.875 – 1.25 = 0.625 kmoles γ−1 V ∴ Amount of oxygen = 20 kg ∴ i Tf = Ti Vf 13.60 | Kinetic Theory of Gases and Thermodynamics
4 ∴ PV = 1×RT \P’V=1×R×(2T) Sol 4: (B) PV = RT .....(i) M ⇒ PV = RT ⇒ P’ = 2P (4− 0.8) 3.2R PV = R (T + 50) = (T + 50) .....(ii) M M Sol 9: (D) For diatomic molecule By (i) and (ii) we get, Translational degree of freedoms = 3
4 3.2R Rotational degree of freedoms = 2 RT = (T + 50) ⇒ 4T = 3.2T + 3.2 × 50 M M 3KT 3 Ratio = = 2KT 2 ⇒ T = 200 K Sol 10: (D) We know Sol 5: (A) For P = 1 atm, m = 2gms, T=298 K 1 v = 1.085 v 2R rms avg ∴ 1× V= × 298 ….(i) M 3P 1 ⇒ = 1.085 v ; ρ is density ρ avg For P = 1.5 atm, m1=2 gm, m2=3 gm, T=298 K. In this case ρ is constant 23 ∴ +×R 298 1.5× V= ….(ii) ∴ P ∝ (v )2 MM12 avg 2 P (vavg ) f Using (ii)/(i) we get, ∴ f = = 4 P (v ) i avg i 22 + MM12 3M1 M 1 1.5 = ⇒ 3 = 2 + ⇒ 1 = Sol 11: (C) Frictionless piston means, no generation of 2 M M 3 2 2 heat due to the motion of piston. M1 n = 1
nRT Ti = T Sol 6: (B) P = V P = const. 2 2 nRT T V = V V = 2V ∴ V = constant ⇒ = constant i f V V Initially:- PV = RT 1 V 2 Work done = PDV = P(2V – V)=PV=RT ∴ f 2T Tf = × Ti = Vi Sol 12: (A) f = 6 D ⇒ D ⇒ D Sol 7: (A) Pat top of tube + Pdue to 50 cm of Hg= Patmospheric W = 25 P V = 25 nR T = 25 J
⇒ [By using gas equation] Pat top of tube = (75 – 50) cm of Hg ∴ P = constant = 25 cm of Hg DQ = DU + DW [By 1st law of thermodynamics] 105 = 25 × Pa = 33.3 k Pa nf 6 75 and DU = RDT = × 25 = 75 J 2 2 ∴ D Sol 8: (C) Q = 100 J
For O2 For H Alternative Method n = 1 n = 1 PV∆ At constant pressure, DQ = nC DT = C P P R T 2T [⇒ PV = nRT P P’ PDV = nRDT at const. pressure] V V Physics | 13.61
DW = PDV = 25 J 11kg nCO = = 0.25 k moles f 2 44 1+ f = 6; CP = R = 4R 2 2 20 2 f = = = CO2 1.3− 1 0.3 3 25 ∴ DQ = (4R) = 100 J R ∴ Equivalent molecular weight
nM11+ n 2 M 2 mm12+ D = = Sol 13: (B) Q = -30 J nn12+ nn12+ D W = -18 J 7+ 11 = = 36 gm By 1st law of thermodynamics 0.25+ 0.25 DQ = DU + DW nγ+ n γ + ..... γ ≠ 11 2 2 mix ⇒ DU = -30 – (-18) = -12 J n12++ n ..... ⇒ U – U = -12 n f++ n f ..... B i f = 11 2 2 ∴ mix nn+ UB = -12 + 60 = 48 Joules 12 20 0.25×+ 5 0.25 × Sol 14: (A) T = 300 K T = ? 3 35 i f = _ 0.25+ 0.25 6 Pi = P Pf = 4P 2 2 47 nRT ∴ γ = 1 + = 1 + = = 1.34 By gas equation we know: - V = mix f 35 / 6 35 P mix ∴ PV4/3 = constant 4 4 nRT 3 T 3 ⇒ P =constant ⇒ = constant Previous Years’ Questions P 1 P3 14 ÷ 1 Sol 1: (A) Work done in a cyclic process = area between 33 4 Pf 4P the cycle ∴ T = × T = × 300 K f P i P i = AB × BC = (2P – P) × (2V – V) = PV ⇒ 2 Tf = 300 K Note if cycle is clockwise (p on y-axis and V on x-axis) work done is positive and if it is anticlockwise work Sol 15: (D) For first kind of expansion:- done is negative. DW = 20 KJ 3RT D Sol 2: (A) v = Q = 16 KJ rms M ∴ DU = DQ – DW = -4 KJ Room temperature T = 300 K Since, U is a state function. Therefore, value of DU in both expansions remain same. 3× 8.31 ×× 103 300 ∴ 1930 = Thus, for second expansion:- M DU = -4KJ, DQ = 9 KJ ∴ M = 2.0 g/mol or the gas is H2. ∴ By first law of the thermodynamics:- Q C 1 D D D D D 2 V W = Q – U =13 KJ Sol 3: (B) Q1 = nCP T, Q2 = nCv T, = = Q1 CP γ 7kg Q1 70 Sol 16: (A) n = = 0.25 k moles or Q = = = 50 cal N2 28 2 γ 1.4 2 2 f = = = 5 N2 γ−1 1.4− 1 13.62 | Kinetic Theory of Gases and Thermodynamics
5 3 nA RT nA RT Sol 4: (B) g = means gas is monatomic or C = R – DP = (p ) – (p ) = – 1 3 V1 2 A i A f V 2V n RT 7 5 = A .…. (i) g = means gas is diatomic or C = R 2V 2 5 V2 2 In chamber B → CV (of the mixture) n RT n RT – 1.5DP = (p ) - (p ) = B – B 35 B i B f V 2V nC+ n C (1) R + (1) R 1V12 2V 22 n RT = = = 2R = B ……. (ii) nn12+ 11+ 2V From Equation (i) and (ii) CP (of the mixture) = CV + R = 3R n 1 2 m /M 2 m 2 C 3R A = = or A = or A = or 3m ∴ g P A mixture = = = 1.5 nB 1.5 3 mB /M 3 mB 3 CV 2R
= 2mB 3RT Sol 5: (B) v = rms M Sol 9: (D) Internal energy of n moles of an ideal gas at temperature T is given by ∝ T i.e., Vrms f U = n RT 2 When temperature is increased from 120 K to 480 K (i.e., four times), the root mean square speed will become where, f = degrees of freedom
4 or 2 times i.e., 2v. = 5 for O2 and 3 for Ar Hence, U = U + U 3 O2 Ar Sol 6: (D) The average translational KE = kT which is 2 5 3 = 2 RT + 4 RT = 11 RT directly proportional to T, while rms speed of molecules 2 2 is given by
3RT Sol 10: (D) During adiabatic expansion, we know v = i.e., v ∝ T rms M rms TVγ–1 = constant γ When temperature of gas is increased from 300 K to γ–1 TV –1 or T1 V = 22 600 K (i.e., 2 times), the average translational KE will 1 5 increase to 2 times and rms speed to 2 or 1.414 times. For a monoatomic gas, γ = 3 ∴ –21 Average translational KE = 2 × 6.21 × 10 J 5 γ–1 –1 –21 3 = 12.42 ×10 J T1 V2 AL2 ∴ = = V ≈ T2 1 AL1 And vrms = (1.414)(484)m/s 684 m/s (A = Area of cross-section of piston) nRT 2/3 Sol 7: (C) PV = nRT or P = or P ∝ T V L2 = L If V and n are same. Therefore, if T is doubled pressure 1 also becomes two times i.e., 2p. Sol 11: (C) pT2 = constant Sol 8: (C) Process is isothermal. Therefore nRT 1 ∴ T2 = constant or T3V–1= constant T = constant. p ∝ V V Volume is increasing, therefore, pressure will decrease. Differentiating the equation, we get 2 3 → 3T T In chamber A .dT – dV = 0 V V2 Physics | 13.63
T Sol 14: (A) DU = nC DT = nC (T – T ) in all processes. or 3dT = .dV ……. (i) V V 2 1 V (B) In adiabatic process DQ = 0 From the equation, dV = Vγ dT ∴ DU = – DW or |DU| = |DW| dV γ = coefficient of volume expansion of gas = V.dT (C) In isothermal process DT = 0 dV 3 ∴ D D D From Equation (i) γ = = U = 0 (as U = nCV T) V.dT T (D) In adiabatic process DQ = 0 Sol 12: A → s; B → r, p; C → p; D → s Sol 15: (A, B) T = T ∴ U = U In process J → K V is constant whereas p is decreasing. A B A B Therefore, T should also decrease. Vf 4V0 W = (1)(R)T ln = RT ln = p V ln(4) \ D AB 0 0 0 0 W = 0, U = – ve and Q < 0 Vi V0 → In process K L p is constant while V is increasing. Information regarding p and T at C cannot be obtained Therefore, temperature should also increase. from the given graph. Unless it is mentioned that line W > 0, DU > 0 and Q > 0 BC passes through origin or not. In process L → M This is inverse of process J → K. Sol 16: (D) No option is correct ∴ W = 0, DU > 0 and Q > 0 γRT In process M → J v = M V is decreasing. Therefore, W < 0 7 (PV) < (PV) vMγ × 4 J M 1= 12 = 5 \ D vMγ 5 Tf < TM or U < 0 2 21 × 32 3 Therefore, Q < 0 460 21 460×× 5 2 2 = ⇒=v2 =14201420 msm/s v 25× 8 21 Sol 13: A → p, r, t; B → p, r; C → q, s; D → r, t 2 Internal energy ∝ T ∝ PV
nf f Sol 17: (A) U=U1 + U2 This is because U = RT = (PV) 2 2 (P V+ P V )T T T = 11 2 2 12 Here, n = number of moles, f = degree of freedom. (P112 V T+ P 2 V 21 T ) ∴ If the product PV increases then internal energy will increase and if product decreases the internal energy Sol 18: (B) Thermal energy corresponds to internal will decrease. energy Further, work is done on the gas, if volume of gas decreases. For heat exchange, Mass = 1 kg 3 Q = W + DU Density = 8 kg/m
mass 1 3 Work done is area under p–V graph. If volume increases ⇒ Volume = = m work done by gas is positive and if volume decrease density 8 work done by gas is negative. Further DU is positive 4 2 if product of PV is increasing and DU is negative if Pressure = 8 × 10 N/m product of PV is decreasing. 5 4 ∴ Internal Energy =P ×=× V 5 10 J If heat is taken by the gas, Q is positive and if heat is 2 lost by the gas, Q is negative.
Keeping the above points in mind the answer to this Sol 19: (C) WAB = ΔQ - ΔU = nCpdT – nCvdT (at constant question is as under. pressure) (A) → (p, r, t); (B) → (p, r); (C) → (q, s); (D) → (r, t) 13.64 | Kinetic Theory of Gases and Thermodynamics
P ab 5 AB Sol 23: (C) U(x) = − 2 x 10 Pa xx12 6
n=2, g=1.67 U(x = ∞) =0
5 dU 12a 6b 1 x 10 Pa as, F =−=−+ D C dx xx13 7
At equilibrium, F = 0 300 K 500 KT 6 2a = (C – C )dt ∴=x n p v b = nRdT = 2 × R × (500 – 300) = 400 R a bb− 2 ∴U = −= at equilibrium 2 2a 4a 2a b Sol 20: (A) At constant temperature (isothermal process) b P 105 b2 W= nRTln1 = 2.303 ×× 2R 300log ∴D = [(U x = ∞ ) – U ]= DA 5 at equilibrium P2 2× 10 4a 1 =2.303 × 600log TT− 1 2 TTη=− 121 = Sol 24:η= (D)121 = 1 T61 =×=−0.693 600lR 414R T61 T−− (T 62) 1 Tη=−− (T12 62) 1 =TT− 62 1 η=122 = ⇒12 += and 2 T31 T31 T11 T3 Sol 21: (B) Net work done in a cycle = WAB + WBC + WCB 1 62 1 62 1 + WBA +=⇒ = 6T 3 TTT6− 1 = 400 R + 2 × 2.303 × 500 R ln 2 – 400R – 414 R 1 121 = T61TT12− 1 = 1000R x ln 2 – 600R x ln 2 = 400R x ln 2 = 276R TT12− 1 = ∴T1 = 62×= 6 = 372K T2 T61 1 T61 1−= TT− 1 T6T 1 The efficiency of cycle is 12T 1 11−=2 Sol 22: (B) 1−=2= ⇒ T61 T2 T651 T T61 = η=1 − 2 T 1 372T 6 5 T2 5 2 = T1 1−=2 ⇒= ⇒ T6= T3722 310K 6 3721 6 ⇒=T 310K for adiabatic process T ⇒=T5 310K 2 2 = 2 TVγ−1 = constant 372 6 Sol⇒=T 25: 310K(A) Work done in complete cycle = Area under 7 2 For diatomic gas γ= P–V graph = P0V0 5 From A to B, heat given to the gas TVγ−11= TV γ− 11 22 33 3 = ∆= ∆= ∆= γ−1 nC v T n R T V0 P PV 00 V 22 2 TT= 2 12 V1 From B to C, heat given to the system
7 −1 55 5 5 2/5 T= T (32) = T (2 ) = T × 4 =nC v ∆= T n R ∆= T( 2P0) ∆= V 5P 00 V 12 2 2 22 T= 4T 12 From C to D and D to A, heat is rejected. 13 η=1 − = =0.75 Work done by gas 44 Efficiency, η= ×100 Heat given to the gas PV η= 00 =15.4% 3 PV+ 5PV 2 00 00 Physics | 13.65
40 500− T 1 = S = τ= Sol 26: (C) ,TS 300K Sol 31: (C) Since 2 100 500 nπ 2vrms d 600 T− 300 1 = ⇒=T 750K n ∝ and vTrms ∝ 100 T V V ⇒τ∝ Sol 27: (C) T
Mg γγ n= CV−1 < v >= C T1/2 = P0 P00 V= PV 12 A γ+1 = γγ= − TVγ−1 = ⇒τ∝V 2 Mg P0 A...... (1)(i) P00 AX PA(x 0 x) Since constant γ P00 Ax Let piston is displaced by x P = P P0 γ Sol 32: (D)0 P−=− P (V − 2V ) (x0 − x) P−=− P00P0 (V −00 2V ) P−=− PV (V − 2V ) V0 Pxγ 00V0 Pxγ Mg = 00 AF= 0 P 00 γ restoring P0 0 Mg = AF= P= 3P − P P V= 3P0 − V ....(1) ....(1) γ restoring(x0 − x) 0 0 …(i) (x− γ x) P= 3P − V V V0 ....(1) Px000γ 0 0 = = γ V0 Mg Px00γ AFrestoring x nRT P Mg = x γ AF= 0 nRT P0 0 (x− 0 x) P0 Arestoring 1− =Frestoring [x0−≈ x x 0 ]= − =3P − V P A 1−0 γ =F [xγ −≈ x x ] nRT 3P0 P0 V 0 0 (x0 − x) γ restoring(x− x) 0 0 VV=3P − VV V 0 (x−γ x) 0 0 0 0 x0γ VV0 − = γP Ax −≈ P0 2 P0 A 1 x0 Frestoring0 [x0 x x 0 ] P0 2 γP Ax γ F = = −nRT= 3P0 V − V P0 A 10−(x− x) =Frestoring [x0−≈ x x 0 ] nRT 3P0 VP0 V 2 F = 0 γ x nRT= 3P V − V V V0 x (x0 − x) 0 0 0 γP0 Ax V0 = 0 γPA F γP0 Ax 1 0 F = x1 γPA0 ∴=f ∴=f 0 2π xM x0 0 Differentiate w.r.t. Volume 2π xM0 1 γPA0 ∴=f 1 γPA 1 γPA2 ∴=f1 2πγPA xM20 = 0 2P 3V = 2π 0 xM0 3P−00 V =⇒= 0 V 0 2π MV0 0 2π MV02 V20 1 γPA0 = 1 γPA2 = 2π MV0 2π MV0 5R Sol 28: (D)0 ∆=U C ∆=× T1 ∆ T Putting in equ (i) nV 2 P 3V 3P P=−= 3P 00 0 For BC, ΔT = –200 K 0 V20 2 ⇒ ΔU = –500R 9P V Now, PV = xRT 00= nRT 4 1U PV p = 9P00 V 9 00 Sol 29: (C) 3V = nRT ⇒ T = 4 4 xR nRT 1 4 9 PV ∝ T T = 00 V3 4 xR R Sol 33: (A)R CC= + CC=V + V VT3 = const 1n− 1n− −−CCPv−− CC Pv CCPvC− C = CC Pv ;1 −= n 4 33 C− CV = ;1V −= n π=R T const 1n−−1n−− C C C Cv 3 v C−− C CCCPv−− C CC P =−=Pvn1=−= P 1 n1 CC−− CC TR= const ⇒∝ T CC−−vv CC vv R
Sol 30: (B) Since entropy is a state function, therefore change in entropy in both the processes should be same. 13.66 | Kinetic Theory of Gases and Thermodynamics
JEE Advanced/Boards Sol 4: CP’ and Cv’ be molar specific heats. ∴ CP’ = CP × M = 0.2 M cal/mole °C Exercise 1 CV’ = CV × M = 0.15 M cal/mole °C
and CP’ – CV’ = R Sol 1: For O2 For He ⇒ 0.2 M - 0.15 M = 2 P = P P = P 0 0 2 ⇒ M = = 40 gms V = V0 V = V0 0.05
T = T0 T = T0/2 M = 32 gm M = 4gm Sol 5: f = 6 PV = n RT At constant pressure D D m PV Q = nCP T ⇒ T = = constant M R PV∆ = CP × mO m R ∴ 2 T = He T M O2 M He [PV = nRT ⇒ P DV = nR DT at constant pressure] O2 He C P D mO m mO = × W ⇒ 2 He ⇒ 2 R × T0 = × T0/2 = 4 32 4 mHe f C = 1+ R = 4R nR P 2 Sol 2: We have, V = T and we know that V – T P And DW = 25 J curve is a straight line. 4R DQ = × 25 = 100 J nR R ∴ Pressure is constant and its slope = = tan 53o P 3 2×× 8.314 10 Pa L / moleK 4 Sol 6: He N ⇒ = 2 P 3 m = 4 gm m = 28 gm N2 ⇒ P ≈ 1.25 × 104 Pa 4 28 ∴ nHe= =1 mole nN2= =1 mole γ 4 28 Sol 3: PV = constant γ−1 ⇒ TV = constant f = 3 f = 5 γ−1 γ−2 3 ⇒ V dT + (γ-1) V T dV = 0 ∴ C = ∴ C = 5/2 R V 2 V dV −V dV V nC+ nC ⇒ = ⇒ = He VHe N 2 V N dt (γ− 1) T dT (γ− 1) T ∴ C = 2 V mix nn+ He N2 +V0 1 mT0 ⇒ m = ⇒ = + 35 (γ− 1) T0 γ−1 V0 RR+ 22 ∴ C = = 2R V mix 11+ R mT R ∴ 0 CV = = + γ−1 V0 3RT We know v = rms M
And Cp = CV + R ∴ To double the vrms temperature must be made 4 (V+ mT )R mRT00 T m times the original temperature. = 00 = 1R+ V VV ∴ 0 00 Tf = 4 Ti = 1200 K ∴ DT = 900 K Physics | 13.67
∴ D D Q = n CV T [At constant volume] Sol 10: From graph, it is clear that V and 1/T have a linear relationship = (nHe+∆ n N )C V T 2 mix m = 2 × 2R × 900 = 3600 R ∴ V = T m is the slope of straight line Sol 7: T = constant ⇒ VT = constant ∴ DU = 0 ⇒ PV2 = constant = k (say) Thus DW = –23.04 × 102 J Vf Vf V −2 ⇒ nRT ln f = –23.04 × 102 DW = ∫ PdV = ∫ kV dV Vi Vi Vi
0.25 Vf ⇒ 2 −kV−1 -1 -1 8.31 × T ln = –23.04 × 10 = = – [k Vf – kVi ] 0.50 Vi
2 2 -1 2 -1 +×23.04 10 = – [(PfVf ) vf – (PiVi ) Vi ] = – [PfVf – PiVi] ∴ T = ≈ 400 k 8.31× ln2 = – nRDT. Thus DQ = DU + DW Sol 8: Let C be the molar heat capacity of this gas. ⇒ D D D nC T = nCV T – nR T ∵ DQ = nCDT ⇒ C = CV – R And we know DU = nC DT V 3 R ⇒ C = R – R = ∴ ∆=∆Q 2U 2 2 ∴ D D nC T = 2nCV T Vf Vf 5 ⇒ R Sol 11: (i) DW = PdV = aVdV C = 2CV = 2 = 5R ∫ ∫ 2 Vi Vi
PPfi 22 Sol 9: C → Molar heat capacity of the process V VV− 2 f 22VV fi aV aVfi− aV fi st = = = By 1 law of thermodynamics:- 2 2 2 V dQ = dU + dW i P P ⇒ f i nC dT = nCv dT + dW …..(i) ( a = = ) Vf Vi dW = PdV αV Pff V− PV ii nR∆ T nRT Te0 ⇒ D = dV = nR dV W = = V V 2 2 a nR T = T e V DU = nC DT = DT 0 V γ−1 ∴ a aV dT = T0 e dV DQ = DW + DU dT ⇒ dV = αTeαV nR∆ T nR 11 0 = + DT = nRDT + 2 γ−1 21γ− nR ⇒ dW = dT αV nR∆ T γ+1 Substituting value of dW in (i) we get = ....(i) 2 γ−1 nR nCdT = nC dT + dT V αV We know P = aV 2 R nRT aV ∴ C = C + ⇒ = aV ⇒ T = V αV V nR 2 2 aV0 9aV0 ∴ T = ad T = i R f R 13.68 | Kinetic Theory of Gases and Thermodynamics
2 D -1/3] 8aV0 W = - 400 n R [1-2 ∴ DT = .....(ii) R For process AC Substituting (ii) in (i) DQ = 0 R 8aV 2 γ+1 γ+1 DU = nC DT = 2 ×400 nR [1 – 2-1/3] D 0 2 V Thus Q = × = 4aV0 2 R γ−1 γ−1 DW= -DU = -2 × 400 nR [1-2-1/3]
(ii) C → Heat capacity of gas Q Efficiency (e) =1 - released So DQ = nCDT Qabsorbed Using (i) we get 3×− 400nR[1 2−1/3 ] 3(1− 2−1/3 ) nR∆ T γ+1 = 1 – = 1 – nCDT = 400nR ln2 ln2 2 γ−1 γ+1 R ⇒ C = dQ −kAdT −k[2 π−× (b x) ]dT γ−1 2 Sol 13: = = dt dx dx Sol 12: At A x V = V , T = 400 K d A 0 A b 400nR ∴ PA = V0 At B 400nR VB = 2V0 , TB = 400 K, PB = T0 2V0 At C 400nR VC = ?, TC = ?, PC = 2V0 ∵ AC is adiabatic process γ ∴ PV = constant 1 On integrating with respect to dx we get γ 1 P 2/3 A 1.5 ba− ba− ⇒ V = V = (2) V ⇒=VC0 2V dQ 1 dT C P A 0 ⇒ × dx= −k × dx C ∫ π− ∫ 0 dt 2 (b x) a dx PV ⇒ CC -1/3 ba− TC = =400 × (2) dQ −ln nR ⇒ × (b− x) =- kDT dt 2π 0 For process AB dQ b DU = 0 .ln=−π 2 kl(T − T ) dt a 0 2V0 DQ = DW = nR × (400) × ln dT b V π2 =−π − 0 ( a l) 3 ln 2 kl(T T0 ) dx a = 400 nR ln 2 For process BC ⇒ On integrating w.r.t. t we get nR -1/3 b DU = nC DT = × [400 (2) – 400] −πa2 ls ln V γ− T2 t 1 a dT ⇒ = dt = -2 × 400 n R [1-2-1/3] 2π kl ∫ TT− ∫ T⊥ 0 0 nRγ DQ = nC DT = - × 400 [1-2-1/3] P γ−1 as2 b TT− ⇒ 01 -1/3 ln ln = t = -3 × 400 n R [1-2 ] 2K a TT02− Physics | 13.69
→ 2 Sol 14: kS Conductivity of solid cast iron. At the Area of cross-setion(A)=0.04m point of junction of solid cast iron and liquid cast iron, temperature would be equal to melting point of iron. T1=400k
T2 k=10/m-k l=0.4cm Solid cast
iron (ks) T l m Liquid cast This metalic x T =300K iron (kks) 0 disk s=60 J/kRk
T1 m=0.1kg
dQ dQ On integrating we get = dt solid cast iron dt liquid cast iron 350 t ms dT ⇒ - = dt ∫ − ∫ kA 300 T 400 0 −−ks2m A[T T ] −−kksm [T T 1 ] ⇒ = − x x −×0.4 600 × 0.4 350− 400 ⇒ t = ln x k[T− T ] 10× 0.04 300− 400 ⇒ = m1 − x TT2m− ≈ 166.3 sec
x k[Tm1− T ] k(T1m− T ) ⇒ = = -kt (T−+ T ) k(T − T ) k(T−+− T ) (T T ) Sol 17: (T – T0) = (T1 – T0) e 2m m1 1m m 2 1 TT10− ⇒ t = ln dT k TT− Sol 15: (a) Temperature gradient = 0 dx T : Surrounding temp (0− 100)o C 0 = = -100°C/m (1− 0)m T1 : Initial temp of object T : Final temp of object dT T− 100 (b) = = -100°C/m dx x0− 1 80− 20 1 ∴ 5 = ln = ln (2) k 50− 20 R ⇒ T = 100 (1 - x) °C
1 60− 20 1 2 1 Total heat absorbed and t= ln= ln(2) = 2× ln2 k 30− 20 k k = mass×× specific heat change of temp ∫ ∴ t = 10 min 1 1 = ∫(2dx)(10)(T− 0) = 20 ∫100(1− x)dx Sol 18: A = 27 cm2 = 27 × 10-4 m2 0 0 k = 3700 N/m 1 2 x 5 = 2000 x − = 1000 J Pi = 1 atm = 10 Pa 2 0 Ti = 273 K
Vi = Ah. dQ −−kA(T 400) Sol 16: = Initial contraction in spring dt × 54− dT −−kA(T 400) PAi 10×× 27 10 27 ⇒ ms = = = = m dt k 3700 370
ms dT Pf = ? ⇒ = -dt kA T− 400 13.70 | Kinetic Theory of Gases and Thermodynamics
⇒ Tf = ? TC = 4TA 9hA D D D QAB = U + W Vf = 16 l = nRTA n 4 Since, spring contract by h/16 length DQ = DU + DW h 27 BC ∴ Force exerted by spring = k + = nC [T – T ] = 3n C T 16 370 V C A V A ∴D D 3 QAB + QBC = 27.7× 10 J Avg. pressure due to spring (P0) ⇒ l k h 27 nRTA n 4 + 3 nCV TA = 27700 = + A 16 370 27700 ⇒ R ln 4 + 3C = ≈ 50.73 V 2× 273 Since process is adiabatic γ ∴ C ≈ 13.068 J/mol-K V V ∴ i Pf = Pi + V CP CRV R f = = 1 + ≈ 1.63 CV CV CV 3/2 16 4× 16 105 = × 105 09 27 Sol 20: Initially 0 Ti = 0 C= 273K = T1 Pf must be equal to P0 for equilibrium
Pi = P1 k h 27 4× 16 Thus, + = × 105 V = V A 16 370 27 i 1 From (1) to (2) h 27 4×× 16 1054 ×× 27 10− ⇒ + = Process is adiabatic 16 370 27× 3700 DQ = 0 4× (16)2 27× 16 ⇒ h = - DU = -DW 370 370 P = P ⇒ h = 1.6 m 2 γ−1 T = T In adiabatic process TV = constant 2 γ−1 V = V2 < V1 V ⇒ i At (2) Tf = Ti Vf Process is isochoric 1/2 16 ∴ V = constant = × 273 9 ∴ V = V2 ⇒ T = 364 K ∴ f P = P3 < P1
T =273 K = T1 < T2 Sol 19: Process AC is isobaric From (2) to (1)
d Process is isothermal
T = 273 K = T1 273k V = V A B 1 ∴ P = P1
44 V
V 4V ∴ C TC = TA = × TA VA V Physics | 13.71
Exercise 2 P P 2 Single Correct Choice Type
aric Adiabatic Sol 1: (D)
Isob P3
P1 Isothemal n moles n moles V1 V1 V Small vessel D Qprocess = 100 × 80 = 8000 cal P D Large vessel Uprocess =0 cal ∴ DW + DU = DQ T ∴ D W =8000 cal V = constant and PV = nRT Sol 21: Number of molecules hitting 1m2 of wall per nR molecules ⇒ P = T second (N)== nv cos θ V m2 sec ∴ P = mT where m is a constant ∴ P-T curve for both vessels will be linear but with q different slopes, since the constant volumes have q different values in both cases.
Sol 2: (C) Initially P = 76 cm of Hg Change in momentum of 1 molecule i V = 5A = m (2v cos θ) =2mv cos θ i ∴ Pressure exerted = N ×change of momentum of 1 5cm molecule = nv cos θ × 2 mv cos θ = 2mv2 cos2 θ
Sol 22: (a) Work along path A-D Finally = Area under curve AD P = (76) – (48 – x) = (28 + x) cm of Hg 1 f = [PA + PD] × [VD - VA] 2 Vf = x A 1 = × 1.6 × 105 × 1.1 × 10-3 = 88 J 2 x D D D ⇒ D (b) WADC = WAD + WDC WDC = -3J 1 ∴ - [ [P + P ] × [V – V ]] = -3 2 D B D C 48-x ⇒ 5 -3 0.9 × 10 (1.3 – Vc) × 10 = 6 ⇒ ≈ VC 1.23 L
(c) WCDA = –WADC = –85 J 13.72 | Kinetic Theory of Gases and Thermodynamics
P Since, temperature remains constant B dP H/2 2Mg = dh ∴ P V = P V ∫ P ∫ RT i i f f Pm 0 ⇒ 76 × 5A = (28 + x) x A PB MgH ⇒ x2 + 28 x – 380 = 0 ln = Pm RT ⇒ (x + 38) (x – 10) = 0 PB ⇒ x = 10 cm or x = -38 cm = exp (MgH/RT) Pm x = -38 rejected since x can’t be -ve Sol 5: (D) Since there is no loss of energy Sol 3: (C) ∴ Sum of change of internal energies must be zero. Container X Container Y D D i.e. U1 + U2=0 2V 2V V ⇒ n1 Cv [T-T1]+ n2 Cv [T – T2] = 0 xY ⇒ (n1 + n2) T = n1 T1 + n2 T2 nT+ n T Process (P) in both vessels will be same. ⇒ T = 11 2 2 nn12+ Thus V ∝ n T V 3RT 1 ∴ n ∝ Sol 6: (D) v = ⇒ v 2 ∝ T rms M rms M
2V 2 ∴ n = k [k is some constant] (vrms ) B MA mN× A 1 X 200 ∴ = = = ....(i) (v2 ) M 2m× N 2 kV rms A B A n = Y 400 Since by the postulate of KTG that, molecules move in n 4 random motions. ∴ X = n 1 ∴ 2 2 2 2 2 2 2 Y
mX nX 2 2 2 2 ∴ = = 4 Thus (vrms )A = 3w & (vrms )B = v m n Y Y v2 1 w2 2 ⇒ m m Thus, by (i) we get: - 2 = 2 = ∴ m = X = 3w 2 v 3 Y 4 4
Sol 4: (C) Pressure gradient will develop due to the Sol 7: (C) P = 0.7 × 105 N/m2 upward acceleration so V = 0.0049 m3 dP = 2ρg C dh γ = P = 1.4 = ratio of specific heats CV dP PM = 2g CV dh RT ≠ 1.4 since, we know CP > CV CP
g In reversible adiabatic process γ PV = constant H/2 γ γ−1 ∴ V DP + Pγ V DV = 0
P 5 m ∆P −γP 1.4×× 0.7 10 h ⇒ = = – ∆V V 0.0049 dP ⇒ = -2 × 107 Nm-5 PB dV Physics | 13.73
Sol 8: (C) Equal mass of same gas ⇒ Equal moles Multiple Correct Choice Type Initially
Sol 10: (A, B) eA = 0.01 and eB = 0.81 X n moles n moles Y AA = AB T = 300 k T = 300 k P = 1 atm P = 1 atm EA = EB V = 42 A V = 42 A ⇒ σ 4 σ 4 eA AA TA = eB AB TB 42cm 42cm ⇒ 4 4 0.01 TA = 0.81 TB Area of 1 cross-sextion = A ⇒ T = × T B 3 A 1 Using gas equation we get ⇒ T = × 5802 = 1934 K B 3 PV = nRT By Wien’s displacement law ⇒ 42 A = nR 300 l T = constant = 2.93 × 10-3 mK nR 42 m ⇒ = ……(i) ∴ λ = 0.5 mm A 300 mA Finally Since, it is given in the question that λ = 1 mm + λ X n moles n moles mB mA T = 300 k T = 300 k ∴ λ = 1.5 mm mB P=f P P = 1 atm V = 42 A V = (42+x) A Sol 11: (A, D) P2 V = const. (42+x)cm 42-x 1/2 1/2 V V ∴ i × Using gas equation we get Pf = Pi = P = P/2 Vf 4V P (42 + x) A = n R 330 …..(ii) nRT P (42 – x) A = n R 300 …..(iii) P2 = constant P (ii)/(iii) gives ⇒ PT = constant 42+ x 330 = 42− x 300 ∴ P – T curve is hyperbola
Pi 2x 30 and T = T = 2T ⇒ = f P i 42− x 300 f ⇒ 20x = 42 – x Sol 12: (C, D) P = 3000 J ⇒ x = 2 cm V = 5
3RT Sol 9: (C) At constant pressure v = rms M W = PDV m 1 3 3 5 ⇒ 2 For monoatomic: - C = R (nMm) vrms = nRT = PV P 2 2 2 2 D 3 and Q = nCP T Translational K.E. of all molecules = PV 2 5R PV∆ 5 ⇒ Q = n × × = W 3 2 nR 2 ∴ P × 5 = 3000 J 2 2 ∴ W = Q 5 ∴ P = 400 J/l 13.74 | Kinetic Theory of Gases and Thermodynamics
Total K.E. of 10 of gas at Pressure 2P for a monoatomic Alternative method: 3 gas = × 800 × 10 = 12000 J 2 Adiabatic process 5 Isothermal process Total K.E. of 10 , 2P pressure for a diatomic gas = PV P 2 Initial stage 5 Final stage = × 800 ×10 = 2000 Joules 2 Pf isothermal P adiabatic Note: In options its asked total K.E. not translational K.E. f V V Sol 13: (A, D) No. of molecules moving towards wall f 1 molecules per unit volume = × 1026 From graph, it is very clear, that 6 m3 P > P ∴ No. of molecules hitting 1m2 of the wall every sec (n) f isothermal f adiabatic and for same volume and moles of gas 1 26 = ×10 × (2000) ∝ PV= nRT 6 T P [ ] molecules Thus, T >T = 3.33 × 1028 f isothermal f adiabatic m2 sec Vf Pressure exerted = (change in momentum) × no. of Work done = ∫ PdV = area under P-V diag molecules putting wall per unit area per sec Vi
-27 = (3 × 10 kg/molecules) × (2 × 2000 m/s) × 3.33 × Thus, (work done)isothermal > (work done)adiabatic 1028 molecules/m2 sec) ≈ 4 × 105 Pa Sol 15: (A, B) Area under the curve gives the rate at which heat per unit surface is radiated by the body. Sol 14: (A, B, C) Isothermal process Adiabatic process i.e. total rate of heat radiation = (Area under the curve) γ × (Surface area of the body) PV = constant PV = constant T m ⇒ = constant ⇒ TVr-1 =constant Sol 16: (A, C) ρ = V V γ−1 4 V V ⇒ ρ p 3 ∴ f ⇒ i × r = m Tf = × Ti Tf = Ti 3 Vi Vf ⇒ r ∝ (m)1/3 (Since there is expansion) 2 V V And area of sphere (A) ∝ r ∴ f > 1 ⇒ i < 1 ∴ ∝ 2/3 Vi Vf A (m) γ−1 Vi A Thus < 1 ∴ A 2/3 4/3 V = (4) = (2) f AB eAσ− (T T )4 Since γ-1 is +ve ∴ Ratio of heat loss = A0 eAσ− (T T )4 (∴ B0 Isothermal temp. > adiabatic temp.) A A 4/3 γ = = (2) V V AB i i Pf = × Pi Pf = × Pi Vf Vf By Newton’s law of cooling: V dQ γ i = ms(COOH) = -k (T – T ) (Since > 1 and < 1) dt 2 0 Vf γ dT −k V V ⇒ i i = (T – T0) ∴ > dt ms Vf Vf
∴ Isothermal pressure > Adiabatic pressure Physics | 13.75
Where k = 4e A σ T 3 nR 0 ∴ V is a linear function of T with slope and passing dT A P ∴ ∝ through origin in V-T curve. dt m Secondly for process CD: P-T will be a linear curve
dT AA passing through origin. 4/3 dt mA (2) ∴ A = = = 2-2/3 dT AB 4 Comprehension Type dt m B B Paragraph 1:
Sol 17: (A, C) Sol 20: (A, B) Area under the curve gives the rate at
o o o o which heat per unit surface is radiated by the body. -10 C 0C 0C 100 C 50 gm 250cal 50 gm 4000cal 50 gm 5000cal 50 gm i.e. total rate of heat radiation = (Area under the curve) ice ice water water × (Surface area of the body) 0Co 100oC Calorimeter l 150cal 1.5 gm Sol 21: (A, B, C, D) m T = constant [By Wien’s 1.5 gm water Displacement Law] water T1 dEl T >>T T ∴ Heat absorbed by ice and calorimeter to reach 100° T 1 2 3 dl 2 C water T = 250 + 4000 + 5000 + 150 = 9400 cal 3
∴ Amount of steam converted into water l l l m1 m2 m3 x 9400 = = 17.4 gm 540 4 4 Area under graph = Eλ = e σ T ∝ T ∴ Amount of water remaining = 50 + 17.40 = 67.4 gm ∴ Amount of steam remaining = 20 – 17.4 = 2.6 gm Paragraph 2:
D Sol 18: (B, D) T = 0 Sol 22: (A) AA = A AB = A ∝ The slope of straight line can’t be +ve. Since, T PV A = l B = l and if slope is +ve, then both P and V are increasing. k = 3k k = k Therefore, temperature will always increase. A B
A B T1 < T2 < T3 < T4 ∴ RA = = RB = = kAAA 3kA kABB kA
P T1T2T3 T4 is thermal curve R 1 ∴ A = i.e. PV = const. RB 3
St. line Sol 23: (B) Rate at which heat flows from A = Rate at which heat flows from B Thus, from graph it can be seen, that temperature first increases and then decrease. dQ dQ ⇒ = dt A dt B Sol 19: (A, B) Process AB is isochoric: - i.e. P = constant T T T R 1 ⇒ A B ⇒ A A PV = nRT = = = RA RB TB RB 3 nR ⇒ V = T P 13.76 | Kinetic Theory of Gases and Thermodynamics
TA TA TB TB Sol 31: (A) fH = 5 and f = 3 Sol 24: (B) G = = and G = = 2 He A B A B nf+ nf H22 H He He G TA 1 ∴ f = ∴ = = mix nnH+ He GB TB 3 2 2553×+× 25 = = = 3.57 Paragraph 3: 25+ 7
dQ 2 2 D ∴ 1+ 1+ Sol 25: (B) = kAV T Sol 32: (C) rmix = = = 1.56 dt initially fmix 25 / 7
dQ ∆T dQ Sol 33: (D) Since internal energy is an extensive =kA(4V) = 2 dt finally 2 dt initially property ∴ U = U + U = 200+100=300 J mix He H2 Sol 26: (A) If all the parameters are kept constant then dQ dT = ms = kA v DT Match the Columns dt dt dT kAv∆ T ∴ = Sol 34: (A) Vi = V Vf = 2V dt ms Pi = P (P) → (y) isobaric process Paragraph 4: W = PDV = PV (q) → (z) isothermal Sol 27: (A) wab= Area below curve AB V W = nRT ln f = PV ln 2 1 V = [10 + 20] × (12 – 6) i 2 1 (r) → (x) Adiabatic = × 30 × 6 = 90 J γ 2 V i 2γ Pf = Pi = P Vf Sol 28: (B) wBC = -Area below curve BC −γ = – 10 [12 – 6] = – 60 J Pff V− PV ii (2 P)(2V)− PV W = = 1−γ 1−γ Note:- wBC is –ve, since volume is decreasing PV(1− 21−γ ) ⇒ W = Sol 29: (D) wCA = 0 since volume is constant γ−1 ∴ wABCA = wAB + wBC + wCA = (90) + (-60) + 0 = 30 J
Paragraph 5: Previous Years’ Questions n = 5 moles n = 2 moles He H2 The desired fraction is M = 4g M = 2 g Sol 1: (D) He H2
∆U nCV∆ T CV 1 + f = = = = nMHe He nM H H ∆Q nC∆ T C γ Sol 30: (D) Equivalent molar mass = 22 p p nn+ He H2 5 7 54×+× 22 24 or f = as γ= = = gm 7 5 + 52 7 Physics | 13.77
Sol 2: (A) Average kinetic energy per molecule per Area under the graph gives the work done by the gas 1 degree of freedom = kT. Since, both the gases are 2 P diatomic and at same temperature (300 K), both will A 2 have the same number of rotational degree of freedom i.e., two. Therefore, both the gases will have the same 1 average rotational kinetic energy per molecule 1 3 = 2 × kT or kT 2 V Thus, ratio will be 1: 1. V1 V2
Sol 3: (D) A is free to move, therefore, heat will be (Area)2 > (Area) > (Area)3 supplied at constant pressure ∴ W2 > W1 > W3 ∴ dQA = nCP dTA …… (i) B is held fixed, therefore, heat will be supplied at Sol 6: (C) For an ideal gas : PV = nRT constant volume. For P = constant ∴ dQ = nC dT …… (ii) B V B PDV = nRDT But dQ = dQ (given) A B ∆V nR nR V ∴ = = = Cp ∆T p nRT T nC DT = nC DT ∴ dT = dT p A V B B C A V V ∆V 1 1 ∴ = or δ = γ γ VT∆ T T = (dTA) [ = 1.4 (diatomic)]
(dTA = 30 K) d = (1.4)(30 K) ∴ dTB = 42 K
Sol 4: (C) Speed of sound in an ideal gas is given by
γRT T v = M Therefore, δ is inversely proportional to temperature T. γ δ ∴ v ∝ (T is same for both the gases) i.e., when T increases, decreases and vice-versa. M Hence, δ-T graph will be a rectangular hyperbola as v γ shown in the above figure. N2 N2 MHe 7/5 4 ∴ = . = = 3/5 v γ M 5 / 3 28 He He N2 Sol 7: (B) In adiabatic process 7 dP P γ = (Diatomic) Slope of P-V graph, = – γ N2 5 dV V 5 Slope ∝ γ (with negative sign) γ He = (Monoatomic) 3 From the given graph, (slope) > (slope) Sol 5: (A) The corresponding p–V graphs (also called 2 1 ∴ g g indicator diagram) in three different processes will be 2 > 1 as shown: γ Therefore, 1 should correspond to O2 ( = 1.4) and 2 should correspond to He (γ = 1.67). 13.78 | Kinetic Theory of Gases and Thermodynamics
D D Sol 8: (A) WAB = P V = (10)(2 – 1) = 10 J Sol 12: (A) Since it is open from top. Pressure will be P0. D WBC = 0 (as V = constant) Sol 13: (D) Let p be the pressure in equilibrium. From first law of thermodynamics PA DQ = DW + DU DU = 0 (process ABCA is cyclic) ∴ D D D D Mg PA Q = WAB + WBC + WCA 0 ∴ D D D D WCA = Q – WAB – WBC Then, PA = P0A – Mg = 5 – 10 – 0 = – 5 J Mg Mg P = P0 – = p0 – 2 Sol 9: (C) Temperature of liquid oxygen will first increase A πR in the same phase. Then, phase change (liquid to gas) Applying, P V = P V will take place. During which temperature will remain 1 1 2 2 constant. After that temperature of oxygen in gaseous ∴ P (2AL) = (P)(AL’) state will further increase. 0 2P0 L P Sol 10: (C) Slope of adiabatic process at a given state ∴ L’ = = 0 (2L) (P, V, T) is more than the slope of isothermal process. P Mg P0 − The corresponding P-V graph for the two processes is πR2 as shown in figure. PRπ 2 = 0 (2L) 2 P π−R P0 Mg C P3
P1 Sol 14: (C) B
p V V1 V2 1 2
In the graph, AB is isothermal and BC is adiabatic.
WAB = positive (as volume is increasing) p1 = p2 and WBC = negative (as volume is decreasing) plus, r P0 + g (L0 – H) = P ……..(i)
|WBC| > |WAB|, as area under p–V graph gives the work Now, applying P1V1 = P2V2 for the air inside the cylinder, done. we have
Hence, WAB + WBC = W < 0 P0(L0) = P(L0 – H)
From the graph itself, it is clear that P3 > P1 PL ∴ p = 00 Note: At point B, slope of adiabatic (process BC) is LH0 − greater than the slope of isothermal (process AB) Substituting in Equation (i), we have PL Sol 11: (B) Total translational kinetic energy P + ρg(L – H) = 00 0 0 LH− 3 3 0 = nRT = PV = 1.5 PV 2 2 ρ 2 or g(L0 – H) + P0(L0 – H) – P0L0 = 0 Physics | 13.79
Sol 15: (B, D) In case of free expansion under adiabatic Sol 16: (B, D) For monoatomic gas conditions, change in internal energy DU = 0 35 Cvp= R, C = R ∴ Internal energy and temperature will remain constant. 22 57 1 For diatomic gas, C= R, C = R (B) P ∝ vp25 V2 ∴ 2 PV = constant ……… (i) Sol 17: A → q; B → p, r; C → p, s; D → q, s nRT or V2 = constant (A) P-V graph is not rectangular hyperbola. Therefore, V process A – B is not isothermal. 1 ∴ T ∝ ……… (ii) V (B) In process BCD, product of PV (therefore temperature and internal energy) is decreasing. Further, volume is If volume is doubled, temperature will decrease as per decreasing. Hence, work done is also negative. Hence, equation (ii). Q will be negative or heat will flow out of the gas. Further, molar heat capacity in process PVx = constant is (C) WABC = positive R (D) For clockwise cycle on p- v diagram with P on y-axis, C = Cv + 1– x net work done is positive. From equation (i) , x = 2 3 R R Sol 18: (D) At low pressure and high temperature inter ∴ C = R + = + molecular forces become ineffective. So a real gas 2 1– 2 2 behaves like an ideal gas. Since, molar heat capacity is positive, according to Q = nCDT, Q will be negative if DT is negative. Or gas Sol 19: (A, B) V loses heat if temperature is decreasing. B 4V0 1 ∝ (C) P 4/3 V V A 0 C PV4/3 = constant T T nRT 0 ∴ 4/3 f V = constant U= nRT V 2 1 ∴ T ∝ Where f, n, R are constants. Also temperature T is same V1/3 at A and B. Further, with increase in volume temperature will ∴ U =U decrease. A B 4 Here, x = Also, 3 V 4V Also,∆= W nRT lnf = nRT ln0 = nRT ln4 = P V ln4 AB 0 0 0 0 0 3 R VVi0 ∴ C = R + = – 1.5 R 2 4 1– 3 Sol 20: (A) TV λ -1 = C As molar heat capacity is negative, Q will be positive 2/3 2/3 ⇒ 2/3 T1(5.6) = T2 (0.7) T2 = T1(8) = 4T1 if DT is negative. Or gas gains heat with decrease in nR∆ T 9 temperature. ∴∆w(work done on the system) = = RT γ−181 (D) T ∝ PV 3RT In expansion from V to 2V , product of PV is increasing. Sol 21: (D) V = 1 1 rms M Therefore, temperature will increase. Or DU = +ve. Further, in expansion work done is also positive. MAr 40 Hence, Q = W + DU = +ve or, gas gains heat. Required ratio = = =10 = 3.16 M4He 13.80 | Kinetic Theory of Gases and Thermodynamics
m Sol 22: (D) PVm= nRT = RT (3kW− P) ×× 3 3600 PV= nRT = RT M 30−= 10 M 120×× 4.2 103 ⇒=ρPM RT⇒=ρPM RT 20×× 120 42 2800 ρ PM P M 42 8 3kW−= P = ρ PM1 P= 11 M = 142 × 1 8 =×= 1= 11 = 1 × 1 =×= 3× 36 3 ρ2 PM 22 P 2 M 2 3 3 9 ρ2PM 22 P 2 M 2 3 3 9 P = 3000 - 933 = 2067 W Here ρ1 and ρ2 are the densities of gases in the vessel containing the mixture. 3 2 Sol 26: (6) P = P0 − ρgh = 98 × 10 N/m
Sol 23: (2) Ub = 200 J, Ui =100 J Process iaf 500 cm
Process W (in Joule) ∆U (in Joule) Q (in Joule) 500 mm ia 0 H af 200 200 mm Net 300 200 500
⇒ Uf =400 Joule P0V0 = PV Process ibf 105[A(500 − H)] = 98 × 103[A(500 − 200)] Process W (in Joule) ∆U (in Joule) Q (in Joule) H = 206 mm ia 100 50 150 af 200 100 300 Level fall = 206 − 200 = 6 mm Net 300 150 450 γ−1 SolTVγ− 27:1 = (4) constant TV= constant Qbf 300 ⇒==2 7/5− 1 7/5− 1 v7/5− 1 v Qib 150 7/5− 1 TV= aT TV= aT 32 32 ∴=a4 ∴=a4 Sol 24: (A, B, D) U= CT + C T U= CT + C T nV12 nV nV12 nV 53 53 → → → → =×1 RT +× 1 RT=× =1 4RT RT +× 1 RT = 4RT Sol 28: A p, r, t; B p, r; C q, s; D r, t 2222 ⇒=2C T 4RT Process A → B → Isobaric compression ⇒=2C T 4RT Vmix Vmix Process B → C → Isochoric process Average energy per mole =⇒=2RT C 2R Vmix Process C → D → Isobaric expansion Process D → A → Polytropic with T = T C λ M 334 6 A D mix= mix He = ×× = C λ M 253 5 He He mix Sol 29: (A, B, C, D) Option (A) is correct because the M VrmsHe H 1 graph between (0 – 100 K) appears to be a straight line =2 = VM upto a reasonable approximation. rmsH2 he 2 Option (B) is correct because area under the curve in dQ the temperature range (0 - 100 K) is less than in range Sol 25: (B) Rate of heat generated ==3KW 3kW dt (400 - 500 K.) Let at any time ‘t’, temperature of cooler = T P AC Rate of cooling: dT ms= 3kW − P dt B
30 (3kW− P) 3 V ∫∫dT = dt 10 ms 0 Physics | 13.81
Option (C) is correct because the graph of C versus T is For process CD constant in the temperature range (400 - 500 K) 33 ∆Q2 = nv C ∆= T n R ∆= T V( ∆ P) Option (D) is correct because in the temperature range 22 (200 – 300 K) specific heat capacity increases with 31−35 temperature. ∆Q2 = × 8 × 10 ×−+ 1 × 10 2 32
−93 22 Sol 30: (C) By A V = A V ∆=Q2 × 10 =− 11.625 × 10 1 1 2 2 8 22 2 ∆=Q 1750 − 1162 = 588 J ⇒π(20 ) × 5 =π (1) V22 ⇒ V = 2m / s net
1122 Sol 31: (A) ρ=ρVV 22aa
For given Va ρ V ∝ a ρ
kx Sol 32: (B or A, B, C) P (pressure of gas) =P + 1 A
kx2 (P−− P )(V V ) W= PdV = P (V −+ V ) = P (V −+ V ) 2 12 1 ∫ 12 122 12 1 3 ∆=U nC ∆= T (P V − P V ) V2 2 2 11 QW= +∆ U
5PV 17PV PV Case I: ∆=U 3P V , W =11 , Q = 11 , U =11 1 1 44spring 4
9PV 7PV 41PV PV Case II: ∆=U11 ,W = 11 ,Q = 11 ,U =11 236spring 3
Note: A and C will be true after assuming pressure to the right of piston has constant value P1.
Sol 33: (C) For adiabatic process P3V5 = constant 5 PV3 = constant 5 γ= gas is monoatomic 3 For process AC 55 ∆Q1 = np C ∆= T n R ∆= T P. ∆ V 22 5 ∆Q = × 1053 × (8 −× 1) 10− 1 2 2 ∆=Q1 17.5 × 10 J = 1750J