Q23 / 90 · KVPY 2013 (SB/SX) · Chemistry
The entropy change in the isothermal reversible expansion of 2 moles of an ideal gas from 10 to $100 \mathrm{~L}$ at 300 $\mathrm{K}$ is
- A$42.3 \mathrm{JK}^{-1}$
- B$35.8 \mathrm{~J} \mathrm{~K}^{-1}$
- Correct answerC$38.3 \mathrm{~J} \mathrm{~K}^{-1}$
- D$32.3 \mathrm{~J} \mathrm{~K}^{-1}$
Solution
$\Delta \mathrm{S}=\frac{\mathrm{nRT}}{\mathrm{T}} \ln \frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}$
$=\mathrm{nR} \ln \frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}$
$=2.303 \mathrm{nR} \log \frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}$
$=2.303 \times 2 \times 8.314 \times 1$
$=38.3 \mathrm{~J} / \mathrm{k}$
Hence, the correct answer is option: 'C'.