Question

$5$ mole of an ideal gas expand reversibly from a volume of $8d{m}^3$ to $80d{m}^3$ at a temp of ${27}^0$C. The change in entropy is:

• A. $41.51J{K}^{-1}$
• B. $-95.73J{K}^{-1}$
• C. $95.73J{K}^{-1}$
• D. $-41.57J{K}^{-1}$

Answer 1

The correct option is B $-95.73J{K}^{-1}$

Solution:- (B) $-95.73J/K$

As we know that, for a reversible expansion,

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

Given:-

$n=5\text{moles}$

$V_f=80{dm}^3$

$V_i=8{dm}^3$

$R=8.314J/mol-K$

$\therefore \Delta S=-2.303\times 5\times 8.314\times \log \frac{80}{8}$

$\Rightarrow \Delta S=-95.73J/K$

Hence the change in entropy is $-95.73J/K$.