Question

Calculate the molar entropy of a constant-volume sample of argon at 250K, given that it is 154.84 JK−1mol−1 at 298K, and that CV = 12.47 JK−1mol−1

Answer 1

Since we know that entropy can be written as a function of T and V: S = S(T,V). So, considering the total differential of S, Now,

dS = (∂S/∂T)_V dT + (∂S/∂V)_T dV
It is known that, (∂S/∂T)_V = n*Cv/T, here Cv is the molar heat capacity at constant volume.

At constant volume, dV = 0, so then,

dS = n*Cv/T dT

If we assume that Cv is independent of temperature (i.e., if we assume Ar behaves as an ideal gas over this temperature range, which is a very good approximation), This can be integrate this expression to obtain:

S_final - S_initial = n*Cv*ln(T_final/T_initial)

S_final = S_initial + n*Cv*ln(T_final/T_initial)

From Kinetic gas theorywe know that Cv of a monoatomic ideal gas is 3R/2, and we have to calculate the molar entropy, so n = 1.

S(250K) = (154.54 J/(mol*K)) + (3*(8.314 J/(mol*K))/2)*ln((250/298)

S(250K) = 152.35 J/(mol*K)

Ans. 152.35 J/(mol*K)