Question

Calculate the entropy change when 2 mol of an ideal gas expands isothermally and reversibly from an initial volume of 10 $d{m}^3$ to 100 $d{m}^3$ at 300 K.

• A. 45 J K^-1
• B. 50 J K^-1
• C. 28.294 J K^-1
• D. 38.294 J K^-1

Answer 1

The correct option is D

38.294 J K^-1

Do you remember, how we can find the entropy change for isothermal process?
In an Isothermal process $\Delta U$ = 0
$\Delta U$ = q + w
$\Rightarrow q$ = -w
But w = -2.30 nRT log $\frac{V_2}{V_1}$ or -nRT ln $\frac{V_2}{V_1}$ (since this is a reversible process)
$\Rightarrow q$ = 2.303 nRT log $\frac{V_2}{V_1}$ = nRT ln $\frac{V_2}{V_1}$
or for a reversible process
$q_{rev}$ = 2.303 nRT log $\frac{V_2}{V_1}$ = nRT ln $\frac{V_2}{V_1}$
$\therefore {\Delta }_{sys}S=\frac{q_{rev}}{T}=\frac{nRTln\frac{V_2}{V_1}}{T}=\frac{2.303nRTlog\frac{V_2}{V_1}}{T}$
$\therefore \Delta S=nRln\frac{V_2}{V_1}=2.303nRlog\frac{V_2}{V_1}$
Also V $\propto \frac{1}{P}$ (Boyle's law)
Substituting this relation in above equation, we get
$\Rightarrow △S$ = 2.303 nR log $\frac{P_1}{P_2}$ = nR ln $\frac{P_1}{P_2}$
Now, we have
$V_1=10d{m}^3,V_2=100d{m}^3,n=2$ mol

For an isothermal process,
$\Delta S$ = $nRln\frac{V_2}{V_1}$
$V=10d{m}^3,V_2=100d{m}^3,n=2$ mol, $R=8.314J{K}^{-1}mo{l}^{-1}$
$\therefore \Delta S$=$\left(2mol\right)\times \left(8.314J{K}^{-1}mo{l}^{-1}\right)\times 2.303log\frac{100d{m}^3}{10d{m}^3}$
$=38.294J{K}^{-1}$