Question

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

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

Answer 1

Answer: (C)

$95.73J{K}^{-1}$

Solution -

Given $V_1=8c{m}^3$

$V_2=80c{m}^3$

$n=5$

$T=27+273=300K$

$△S_T=2.303nRlog\frac{V_2}{V_1}$

On substituting the above value,

we get,

$△S_T=2.303\times 5\times 8.314\times log\frac{80}{8}$

$△S_T=2.303\times 5\times 8.314$

$△S_T=19.14\times 5$

$△S_T=95.74J{K}^{-1}$