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Exercise 1 Q.5 State the Postulates of Kinetic Theory of Gases

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Exercise 1 Q.5 State the Postulates of Kinetic Theory of Gases 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.
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