One mole of an ideal gas, CP=3.5R, is compressed adiabatically in a piston/cylinder device from 2 bar and 25 oC to 7 bar. The process is irreversible and requires 35% more work than a reversible, adiabatic compression from the same initial state to the same final pressure. What is theentropy change of the gas?

the entropy change of the gas is ΔS = 2.913 J/K

Explanation:

starting from the first law o thermodynamics for an adiabatic reversible process

ΔU = Q - W

where

ΔU = change in internal energy

Q = heat flow = 0 (adiabatic)

W = work done by the gas

then

-W = ΔU

also we know that the ideal compression work Wcom = - W, then Wcom = ΔU. But also for an ideal gas

ΔU = n*cv * (T final - T initial)

where

n=moles of gas

cv = specific heat capacity at constant volume

T final = T₂ = final temperature of the gas

T initial = T₁ = initial temperature of the gas

and also from an ideal gas

cp - cv = R → cv = 7/2*R - R = 5/2*R

therefore

W com = ΔU = n*cv * (T final - T initial)

for an ideal gas under a reversible adiabatic process ΔS=0 and

ΔS = cp*ln (T₂/T₁) - R * ln (P₂/P₁) = 0

therefore

T₂ = T₁ * (P₂/P₁) ^ (R/cp) = T₁ * (P₂/P₁) ^ (R / (7/2R)) = T₁ * (P₂/P₁) ^ (2/7)

replacing values T₁=25°C = 298 K

T₂ = T₁ * (P₂/P₁) ^ (2/7) = 298 K * (7 bar/2 bar) ^ (2/7) = 426.25 K

then

W com = ΔU = n*cv * (T₂ - T₁)

and the real compression work is W real = 1.35*Wcom, then

W real = ΔU

W real = 1.35*Wcom = n*cv * (T₃ - T₁)

T₃ = 1.35*Wcom/n*cv + T₁ = 1.35 * (T₂ - T₁) + T₁ = 1.35*T₂ - 0.35*T₁ = 1.35*426.25 K - 0.35 * 298 K = 471.14 K

T₃ = 471.14 K

where

T real = T₃

then the entropy change will be

ΔS = cp*ln (T₃/T₁) - R * ln (P₂/P₁) = 7/2 * 8.314 J/mol K * ln (471.14 K / 298 K) - 8.314 J/mol K * ln (7 bar / 2 bar) = 2.913 J/K

ΔS = 2.913 J/K