Question

One mole of a perfect gas expands isothermally to ten times its original volume. The change in entropy is:

• A. 0.1 R
• B. 2.303 R
• C. 10.0 R
• D. 100.0 R

Answer 1

Answer: (B)

2.303 R

As we know that,

For reversible isothermal process,

$\Delta S=nR\ln \frac{V_f}{V_i}=2.303nR\log \frac{V_f}{V_i}$

Given:-

$n=1\text{mole}$

$V_i=V$

$V_f=10V$

$\therefore \Delta S=2.303\times 1\times R\times \log \frac{10V}{V}$

$\Rightarrow \Delta S=2.303R(\because {\log }_{10}10=1)$

Hence the change in entropy is $2.303R$.