.\'ll AT&T; 7:13 PM @ * 19% LO Done Homework due Monday 10/23/17. Will be checki
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.'ll AT&T; 7:13 PM @ * 19% LO Done Homework due Monday 10/23/17. Will be checking for the accuracy. Quiz on Monday on ideal gases. 1. A 1.5 mole sample of an ideal gas is expanded from 300K and an initial volume of 22.5 L to a final volume of 65.0L Calculate w, q, and Du for the following processes. C," is 52R (i). For the isothermal reversible expansion (ii). For the reversible adiabatic expansion (iii). For the expansion against a constant pressure of 0.498 atm (iv). For the reversible expansion at constant temperature. 2. Two moles of an ideal gas with C, -29.4 JK mol independent of temperature) are maintained at273K in a volume of 11.35 L is heated reversibly to 373 K at constant volume. Calculate w, q, Du and DH for this processExplanation / Answer
1. For isothermal expnasion of n moles of an ideal gas at 300K from initial volume of V1=22.5L to V2= 65L, change in internal energy=0, work done = nRT*ln(V1/V2)= 1.5*8.314*300*ln(22.5/65) =-3969 joules
From 1st law ofthermodynamics, deltaU(change in internal energy)= Q+W,
Q= -W = 3969 joules
2.For reversible adiabatic expansion, Q =0, for adiabatic process
(T2/T1) =(V1/V2)Y-1, where T1= 300K, T2= final temperature,Y= CP/CV, CV= CP-R
=5/2R-R= 1.5R, CP/CV= 5/3= 1.67
T2= 300*(22.5/65) (1.67-1 =147.4 K
Work done = R*(T2-T1)/(Y-1)=8.314*(147.4-300)/0.67=-2114.53 joules
From 1st law , deltaU= Q+W, deltaU=-2114.53 joules
3. For reversible expansion against external pressure of 0.498atm, work done= -Pext*dV= -0.498*(22.5-65)=
-21.2 L.atm, but 1L.atm= 101.3 Joules
Work done = -101.3*21.2= -2144 joules
Final temperature (T2) during constant external pressure can be calculated from ideal gas lawequation. First let us calculate the initial pressure (P1) from ideal gas law P1V1= nRT1, P1= nRT1/V1= 1.5*0.0821*300/22.5 atm =1.64 atm
From gas law, P1V1/T1= P2V2/T2
T2= P2V2T1/P1V1=0.498*65*300/(1.64*22.5)= 263K
Change in internal energy = nCV*(T2-T1)= 1.5*1.5*8.314*(263-300)=-692 joules
From 1st law, deltaU= Q+W
Q= -692+2144 =1452 joules
4.For reversuble isothermal expnasion of n moles of an ideal gas at 300K from initial volume of V1=22.5L to V2= 65L, change in internal energy=0, work done = nRT*ln(V1/V2)= 1.5*8.314*300*ln(22.5/65) =-3969 joules
From 1st law ofthermodynamics, deltaU(change in internal energy)= Q+W,
Q= -W = 3969 joules.
2. for constant volume, work done = 0
deltaU= Q=nCV*(T2-T1)= 2*(CP-R)*(373-273) = 2*(29.4-8.314)*100 =4217 joules
deltaH= nCP*(T2-T1)= 2*29.4*(373-273)= 5880 joules
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