Question

Consider 2.00 mol of an ideal gas with a constant‑volume molar specific heat of 12.5 J/(mol·K), an initial temperature of 22.8 ∘C, and an initial pressure of 5.210×10^4 Pa. The state of the gas is then changed to a final temperature of 44.1 ∘C and a final pressure of 3.090×10^4 Pa.

What is the change in the entropy Δ of the gas during this process? The gas constant is 8.31 J/(mol·K).

I keep getting 10.7 J/K but it is wrong. Can someone explain?

Answer 1

We have that the  change in the entropy dS  is mathematically given as

dS=14.06J/K

Change in the Entropy dS

From the standard second law of thermodynamics we see that it is stated that entropy over time increases.

Generally the equation for the constant-pressure molar heat capacity   is mathematically given as

$Cp-Cv=nR\\\\Therefore\\\\Cp-26.0Jk^-1=2.0mol*8.31JK^-1\\\\Cp=42.62J/K$

Change in Entropy (dS) is given as

$dS=nCp*lnT2/T1+nR *lnp1/p2\\\\Therefore\\\\dS=nCp*lnT2/T1+nR *lnp1/p2\\\\dS=2*Cp*ln(321.6/298.3)+8.31*2 *ln(5.410*10^4/3.45*10^4)$

dS=14.06J/K

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