ThermodynamicsMedium
Question
The standard molar entropy of an ideal gas $\left( \gamma = \frac{4}{3} \right)$is 2.5 cal/K-mol at 25°C and 1 bar. The standard molar entropy of the gas at 323°C and 1 bar is (ln 2 = 0.7)
Options
A.undefined
B.4R ln2
C.5.6 cal/K-mol
D.8.1 cal/K-mol
Solution
$S_{2} - S_{1} = n.C_{p,m} = \ln\frac{T_{2}}{T_{1}}$
Or, $S_{2} - 2.5 = 1 \times 4R.\ln\frac{596}{298} = 4R.\ln 2$
$\therefore S_{2} = 2.5 + 4 \times 2 \times 0.7 = 8.1\text{ cal/K.mol}$
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