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?
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Answer 1

Answer:

ΔS = 2.91 J/K

Explanation:

Change in Entropy is given as

= n(Cpln(T2/T1)-R*ln(p2/p1))

As per given information, p2 = 7 bar, p1 = 2 bar and T1  = 298 K

γ =Cp/Cv = 7/5 = 1.4

α =1/(γ - 1) = 2.5

Work done in reversible process;

W = α\alphaα*n*R*T1*((p2/p1)^((γ\gammaγ-1))/γ\gammaγ )-1)

Putting the values;

W = {2.5 * 8.314 * 298 * (3.5^(0.4/1.4)-1)} * 1.35

W =3599 J

Now we know that;

Cv(T2-T1) = W

Calculating final temperature = 471K

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