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Exercise 1 Q.2 the Gas Is Cooled Such That It Loses 65 J of Heat

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Exercise 1 Q.2 the Gas Is Cooled Such That It Loses 65 J of Heat Chemistry | 4.43 Solved Examples JEE Main/Boards WAB= P A( V B −= V A ) nR( TB − T A ) −−11 Q.1 Calculate the increase in internal energy of 1 kg of = (2.0 mol )( 8.3 J K mol) (200K) = 3320 J. water at 100° C when it is converted into steam at the The work done by the gas during the process CA is zero same temperature and at 1 atm (100 kPa). The density as the volume remains constant. From (i), of water and steam are 1000 kg m-3 and 0.6 kg m-3 +=−=− respectively. 3320 J WBC 1200 J 4520 J The latent heat of vaporization of water = 2.25 × 106 J kg−1 Q.3 A mole of a monoatomic ideal gas at 1 atm and 273 Sol: Mass and density of water and steam is provided K is allowed to expand adiabatically against a constant so from the given data calculate the increase in volume. pressure of 0.395 bar until equilibrium is reached. Pressure term is given so by using pressure volume (a) What is the final temperature? relationship calculate the work done by the system. Change in internal energy can be calculated using heat (b) What is the final volume? and work relation. (c) How much Work is done by the gas? The volume of 1 kg of water (d) What is the change in internal energy? 11 = m33 and of 1 kg of steam= m 1000 0.6 Sol: Let the initial and final volumes of the gas be 1V 3 and V2m respectively. Given that the initial pressure (P1) The increase in volume 5 1 x 10 Pa, final temperature be 2T 1133 33 = = m − m =−≈(1.7 0.001m 1.7m) We have, P11V n 1RT 1 0.6 1000 1 x 8.314 x 273 V= =0.022697m3 The work done by the system is 1 1 x 1 0 5 35 p∆= V (100kPa)(1.7m )= 1.7x10 J. For an adiabatic expansion of 1 mole of monoatomic The heat given to convert 1 kg of water into steam ideal gas against a constant external pressure (P2), work = 2.25x106 J. done is given as 3R W PVV=−−=( ) CvTT TT( −=) ( −) The change in internal energy is 221 212 21 ∆=∆−∆=U Q W 2.25 × 1065 J −× 1.7 10 J = 2.08 × 10 6 3 × 8.314 Or− 0.395x105 (V −= 0.022697) (T − 273) ...... (i) (1) 222 Q.2 Consider the cyclic P C process ABCA on a sample Again, of 2.0 mol of an ideal gas P22 V= nRT 2; as shown in figure. The 5 0.395× 10 ×=× V 1 8.314 × T … (ii) temperatures of the gas at A B 22 V A and B are 300 K and 500 Solving eqns. (i) and (ii), we get, K respectively. A total of 1200 J heat is withdrawn from (a) The final temperature, T = 207 K the sample in the process. Find the work done by the 2 R = 8.3 JK−−11 mol 3 gas in part BC. Take R = 8.3 (b) The final volume 2V = 0.043578 m Sol: The change in internal energy during the cyclic Q.4 Metallic mercury is obtained by roasting mercury process is zero. Hence, the heat supplied to the gas (II) sulphide in a limited amount of air. Estimate the is equal to the work done by it. Work done can be temperature range in which the standard reaction is calculated by pressure volume relationship. product-favored. The change in internal energy during the cyclic process HgS( s) +→O( g) Hg( l) + S O( g) is zero. Hence, the heat supplied to the gas is equal to 22 the work done by it. Hence, ∆Ho= -238.6 kJ/ mole and ∆So= + 36.7 J / mole K 4.44 | Thermodynamics and Thermochemistry Sol: Assume that ∆H and ∆S values do not depend on P1 temperature. As ∆Ho is negative and ∆So is positive, using ∆=S 2.303nR log P2 the equation ∆Go = ∆Ho = T∆So ∆°G Will be negative P at all temperatures and so the reaction is product - 1 ∆=S 2.303nR log favored at all temperatures. In this problem, both the P2 o o factors ∆H and ∆S are favourable to spontaneity. w 64 1 n2= = = = 2.303 x 2 x 8.314 log m.wt 32 0.25 Q.5 An ideal gas has a molar heat capacity at constant = 23.053 J K−1 pressure Cp = 2.5 R. The gas is kept in a losed vessel of Volume 0.0083 m3, at a temperature of 300 K and a Q.7 An aluminium container of mass 100 g contains 62− 4 o pressure of 1.6× 10 Nm . An amount 2.49× 10 J of 200 g of ice at -20 C.Heat is added to the system at a rate -1 Heat energy is supplied to the gas. Calculate the final of 100 cal s . What is the temperature of the system after 4 minutes? Draw a rough sketch showing the variation temperature and pressure to the gas. in the temperature of the system as a function of time. Sol: First calculate the number of moles (amount Specific heat capacity of ice = 0.5 cal -1g oC-1, specific of gas) by using ideal gas equation, temperature heat capacity of aluminium = 0.2 cal g-1o C-1, specific heat can be calculated using internal energy and n. capacity of water = 1 cal g-1o C-1, and latent heat of fusion -1 after calculating temperature pressure (P2) can be of ice = 80 calg . calculated using following equation Sol: Total heat supplied to the system in 4 minutes is Q PV PV -1 4 11= 22 = 100 cal s x 240 s = 2.4 x 10 cal. The heat require to o o TT12 take the system from – 20 C to 0 C = (100g) x (0.2 cag-1 oC-1 x (20oC) + (200g) x (0.5 cal-1 oC-1 We have, C= C −= R 2.5R −= R 1.5R. vp x (20oC) PV The amount of the gas (in moles) is n = = 400 cal + 2000 cal = 2400 cal. RT 2400 The time taken in this process =S = 24 s. The 100 (1.6 x1 062 N m− ) x ( 0.0083 m 3) = = 5.33 mol heat required to melt the ice at 0oC =(200g) x (80 cal-1) −−11 (8.3 J K mol ) ( 300K) = 16000 cal. 1600 As the gas is kept in a closed vessel, its volume is constant, The time taken in this process =S = 160 s. 100 Thus, we have If the final temperature is θ , the heat required to take ∆Q the system to the final temperature is ∆∆Q= n Cv T or∆ T = nCv = (100g) × (0.2 cal-1 oC-1) θ + (200g) × -1 o -1 θ 2.49 x1 04 J (0.5 cal C ) . = = 377K 2.4 x 104 cal =2400 cal +16000 cal+(220 caloC-1) θ (5.3 mpl) 1.5 x 8.3 J K−−11 mol Thus, ( ) 5600 cal θ= =25.5 ° C. Or, −1 The final temperature is 300 K + 377 K = 677K. 200 cal° C PV PV The variation in the temperature as function of time is 11= 22 We have, sketched in figure TT12 Temp.(oC) Here V = V . Thus, 1 2 20 T2 677 62−− 62 P21= P = x 1.6 x1 0 N m= 3.6 x 10 N m . 10 T1 300 0 100 200 300 Time (s) Q.6 Oxygen gas weighting 64 is expanded from 1 atm to -10 0.25 atm at 30oC. Calculate entropy change, assuming -20 the gas to be ideal. Sol: First find out value of n and then ∆S can be determined by using following equation Chemistry | 4.45 Q.8 Calculate the enthalpy of the reaction Enthalpy of HO2= - 68.4 kcal H C= CH g + H g →− CH CH g 2 2( ) 2( ) 33( ) Enthalpy of O2= 0 (by convention) The bond energies of ∆H of eq. (i) = Enthalpies of products – Enthalpies of C−−= H.C C.C C and H − H are 99, reactants 83,147 and 104 kcal respectively. −207.9 = 2 x( − 94.48) + 2 ( −68.4) −∆ H CH3 COOH ( l) Sol: The reaction is: ∆=−+H 188.96 – 136.8 207.9 CH3 COOH ( l) HH HH =−+ 325.76 207.9 | | || = − 117.86 kcal C= C(g) +− H H(g) → H − C − C − H(g) | | || Q.10 100 cm3 of 0.5 N HCl solutions at 299.95 K were HH HH mixed with 100 cm3 0.5 N NaOH solution at 299.75 K in a thermos flask. The final temperature was found to ∆H = ? be 302.65 K. Calculate the enthalpy of neutralization of HCl. Water equivalent of thermos flask is 44 g. ∆H= Sum of bond energies of reactants – Sum of bond energies of products Sol: Here we are not provided with initial temperature. So by averaging the temperature of acid and base find = H ∆ + 4 ×∆ H +∆ H − ∆ H + 6 ×∆ H CC= CH −= HH CC − CH − out the initial temperature and thus rise in temperature =(147 +× 4 99 + 140) −( 83 +× 6 99) =− 30 Kcal and enthalpy of neutralization. The initial average temperature of the acid and the base. Q.9 Calculate the heat of formation of acetic acid from 299.95+ 299.75 = = 299.85 K the following data: 2 CH COOH(l)++ 2O(g) → 2CO (g) 2H O(l) … (i) 3 2 22 ..
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