1. One mole of O2(g) undergoes the following change in state. O2(g, 353 K, 5.03

Question

1. One mole of O2(g) undergoes the following change in state.

O2(g, 353 K, 5.03 bar)   O2(g, 401 K, 2.21 bar)

What is S for the gas? Assume ideal gas behaviour.

The constant pressure molar heat capacity for O2(g) is Cp,m = 29.10 J K1 mol1.

Enter your answers in the specified units with three or more significant figures . Do not include units as part of your answer.

S = ____________ J K1

2. What are the values of  q, w, U, H, S, Ssurr, and Suniv for the following a constant pressure process for a system containing 0.300 moles of CH3OH ?

CH3OH(l, 22.0 ºC, 1.00 atm)    CH3OH(g, 115.0 ºC, 1.00 atm)

Assume that the volume of CH3OH(l) is much less than that of CH3OH(g) and that CH3OH(g) behaves as an ideal gas. Also, assume that the temperature of the surroundings is 115.0 ºC.

Molar heat capacity for CH3OH(l), Cp,m = 81.1 J K1 mol1

Molar heat capacity for CH3OH(g), Cp,m = 44.1 J K1 mol1

Enthalpy of vaporization, vapH = 35.2 kJ mol1 at 64.7 ºC and 1.00 atm

q = ____________ J

w = ____________ J

U = ____________ J

H = ____________ J

S = ____________ J K1

Ssurr = ____________ J K1

Suniv = ____________ J K1

Explanation / Answer

O2(g, 353 K, 5.03 bar)   O2(g, 401 K, 2.21 bar)

entropy change for one moles of gas , deltaS= CP*ln(T2/T1)- R*ln(P2/P1), where T1= 353K, T2= 401K, P1= 5.03 bar and P2= 2.21 bar

deltaS= 29.1* ln(401/353) - 8.314*ln(2.21/5.03)=10.55 J/K

2.

hat are the values of  q, w, U, H, S, Ssurr, and Suniv for the following a constant pressure process for a system containing 0.300 moles of CH3OH ?

CH3OH(l, 22.0 ºC, 1.00 atm)    CH3OH(g, 115.0 ºC, 1.00 atm)

first from 22 deg.c. the liquid need to be heated to liquid at 64.7 deg.c at constant pressure by supplying sensible heat of liquid and this heat = moles* specific heat of liquid* change in temperature = 0.3*81.1*(64.7-22)=1039 joules

at 64.7 deg.c, the liquid will be converted to vapor by supplying latent heat of vaporization and this heat =moles* latent heat= 0.3moles*35.2 KJ/mole= 10.56 KJ

up this point, deltaU= Q= 10.56 ( heat to convert vapor)+ sensible heat =10.56+1039/1000 kj=11.599 KJ

and since H= U+PV, deltaH= deltaU+P*dV= deltaU+0=10.56 KJ

there is no change in volume as long as CH3OH is in liquid form.

once the liquid gets vaproized, heat to be added at constant pressure, deltaH= Q=0.3*specific heat of vapor(CP)* temperature difference ( 64.7 to 115 deg.c)= 0.3*44.1*(115-64.7)= 665.5 Joules

work done =- moles*R*(T2-T1)=-0.3*8.314*(115-64.7)= -125 joules,

deltaU= Q+W=665.5-125=540.5 joules

total Q= 10.56+665.5/1000 Kj=11.225 KJ=11225 joules , and W=-125 joules, deltaU= 10.56+540.5/1000=11.10 KJ=11100joules

deltaH= 10.56*1000joules+665.5 joules =11225 joules

entropy change of liquid from 22 deg.c to 64.7 deg.c= moles*Cp of liquid*ln(T2/T1) ( Temperature has to be in K) =0.3*81.1*ln{(64.7+273)/(22+273)}=3.3 J/K, at 64.7 deg.c, deltaS= deltaH/T= 10.56*1000/(64.7+273)=31.27 J/K

from 64.7 deg.c vapor to 115 deg.c vapor at constant pressure, deltaS= 0.3*44.1*ln{(115+273)/(64.7+273)}=1.84 J/K

total entropy change of the system= 3.3+31.27+1.84= 36.41J/K

entropy change of surroundings =-Q/T= -Q/T=(-10.56*1000+0.3*CPvapor*(Tvap-Tsurr)/Tsurr)=-10.56*1000+(44.1*(115-64.7)/(22+273)= -35.7455 J/K

entropy change of universe= 36.41-35.7455= 0.6645 J/K

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