Answer to Question #104003 in Physical Chemistry for Bharti

Question

One mole of an ideal gas, cp = (7/2)R and Cv =(5/2)R is compressed adiabatically in a piston cylinder device from 2 bar and 25℃ to 7 bar. The process is irreversible and requires 35% more work than a reversible, adiabatic compression from the same initial state to the final pressure. What is entropy change of the gas?

Answer 1

$\Delta$S = n(C_pln(T₂/T₁)-R*ln(p₂/p₁))

p₂ = 7 bar

p₁ = 2 bar

T₁ = 25+273 = 298 K

If reversible:

W = $\alpha$*n*R*T₁*((p₂/p₁)^(($\gamma$-1))/$\gamma$ )-1)

$\gamma$ =C_p/C_v = 7/5 = 1.4

$\alpha$ =1/($\gamma$ -1) = 2.5

W = 2.5 * 8.314 * 298 * (3.5^(0.4/1.4)-1) = 2666 J

If irreversible:

W = 2666 *1.35 = 3599 J

C_v(T₂-T₁) = W

T₂ = W/C_v +T₁ = 3599/(5*8.314/2) + 298 = 471 K

ΔS = C_pln(T₂/T₁)-R*ln(p₂/p₁) = (7*8.314/2)*ln(471/298)-8.314*ln(7/2) = 2.91 J/K