Question

For the process, 1 Ar (300 K, 1 bar) $ \rightarrow $ 1 Ar (200 K, 10 bar), assuming ideal gas behaviour, the change in molar entropy is

• A. -27.58 J/K/mol
• B. +27.58 J/K/mol
• C. -24.28 J/K/mol
• D. +24.28 J/K/mol

Answer 1

Answer: (A)

$\Delta S=nC_{p}ln \frac{T_{2}}{T_{1}}-nRln ⁡ \frac{p_{2}}{p_{1}}$

$=2.303C_{p}log \frac{T_{2}}{T_{1}}-2.303Rlog ⁡ \frac{p_{2}}{p_{1}}$

For monoatomic gas like Ar, $C_{p}=\frac{5}{2}R$

$=\frac{5 \times 8.314}{2}=20.8$

$\Delta S=2.303\times 20.8log \frac{200}{300}-2.303\times 8.314log ⁡ \frac{10}{1}$

$=2.303\times 20.8log \frac{2}{3}-2.303\times 8.314$

$=47.9\times \left(- 0.176\right)-19.15$

$=-8.43-19.15=-27.58$ J/K/mol