Question

1 mole of an ideal gas at ${25}^{\circ }C$ is subjected to expand reversibly 10 times of its initial volume. Calculate the change in entropy of expansions.

• A. 25.15 $J{K}^{-1}mo{l}^{-1}$
• B. 30.15 $J{K}^{-1}mo{l}^{-1}$
• C. 19.15 $J{K}^{-1}mo{l}^{-1}$
• D. 29.15 $J{K}^{-1}mo{l}^{-1}$

Answer 1

The correct option is C 19.15 $J{K}^{-1}mo{l}^{-1}$

Here, we know that $\Delta T$ = 0
Therefore, it is an isothermal process.
We know for an isothermal process
$\Delta U$ = 0
Now,
Entropy change for an isothermal process
$\Delta U=q+w$
$\Delta U=0$ (Isothermal process)
$\Rightarrow q=-w$
But $w=-2.30nRTlog\frac{V_2}{V_1}$ $or-nRTln\frac{V_2}{V_1}$ [From first law of thermodynamics]
$\Rightarrow q$ $=2.303nRTlog\frac{V_2}{V_1}=nRTln\frac{V_2}{V_1}$
or for a reversible process
$q_{rev}$ $=2.303nRTlog\frac{V_2}{V_1}$ = $nRTln\frac{V_2}{V_1}$
$\therefore {\Delta }_{sys}S=\frac{q_{rev}}{T}=\frac{nRTln\frac{V_2}{V_1}}{T}=\frac{2.303nRTlog\frac{V_2}{V_1}}{T}$
Also $V\propto \frac{1}{P}$ (Boyle's law)
Substituting this relation in above equation, we get
$\Rightarrow \Delta S=2.303nRlog\frac{P_1}{P_2}=nRln\frac{P_1}{P_2}$
We have, $\Delta S$ =
Given, n = 1, R = 8.314 J, T = 298 K, $V_1$ = V, $V_2$ = 10 V
$\therefore \Delta S=2.303\times 1\times 8.314lo{g}_{10}\frac{10V}{V}$
= 19.15 $J{K}^{-1}mo{l}^{-1}$