Question

1 mole of an ideal gas at ${25}^{\circ }C$ is subjected to expand reversibly ten times of its initial volume. The change in entropy due to expansion is:

• A. $19.15J{K}^{-1}{\text{mol}}^{-1}|$
• B. $16.18J{K}^{-1}{\text{mol}}^{-1}|$
• C. $22.15J{K}^{-1}{\text{mol}}^{-1}|$
• D. $\text{none of these}$

Answer 1

Answer: (A)

The explanation for the correct option

Option(A) $19.15J{K}^{-1}{\text{mol}}^{-1}|$

1. The entropy change for the isothermal process is given as:
$\Delta S=2.303nRlog\frac{V_2}{V_1}$
as entropy change per mole is asked so
$\Delta S=2.303Rlog\frac{V_2}{V_1}$
putting the values, we get:

$\Delta S=2.303\times 8.314\times log\frac{V_2}{V_1}where,V_2=10V_1$

$\Delta S=2.303\times 8.314\times log\frac{10V_1}{V_1}$

$\Delta S=2.303\times 8.314\times log10$

$=19.15J{K}^{-1}mo{l}^{-1}$

Hence, Option(A) $19.15J{K}^{-1}mo{l}^{-1}$ is the change in entropy due to the expansion of 1 mole of an ideal gas at ${25}^{o}C$ is subjected to expand reversibly ten times of its initial volume.