Question

Calculate Δ S for the system when the state of 3 moles of ideal gas for which C p,m = 5/2R is changed from 25 0 C and 1 atm to 125 0 C and 5 atm. How do you rationalize the sign of Δ S?

Answer 1

we know that dU= dQrev- Pdv (1)

We also know that H= U+ PV delH= delU+ PdV+VdP substituting delU from Eq.1 gives

delH= dQrev- PdV+ PdV+ VdP =dqRev+ vdP

dqRev= delH- Vdp = TdelS= CpdT- (RT/P) dP

delS= CpdT/T- Rd(lnP)

on integration this gives delS= Cp ln (T2/T1)- R ln (P2/P1)

for n moles entropy change = n{Cp ln(T2/T1)-R ln (P2/P1)

Given

P2=5 atm and P1=1atm T2=125+273.15=398.15 k   and T1= 25+273.15 =298.15 K, R= 8.314 j/mol.K

delS= 3*{2.5R ln (398.15/298.15)- 8.314*ln (5/1)} 6.012-13.381=-22.107 joules/K

Increase in temperature increaases the entropy, But increase in pressure decreases the volume and decreases the entropy. Hence decrease in entropy due to volume is more than increase in entropy due to increase in temperature