Question

5 moles of an ideal gas expand reversibly from a volume of 8 $d{m}^3$ to 80 $d{m}^3$ at a temperature of ${27}^{\circ }C$. Calculate the change in entropy.

• A. $70.26J{K}^{-1}$
• B. $82.55J{K}^{-1}$
• C. $95.73J{K}^{-1}$
• D. $107.11J{K}^{-1}$

Answer 1

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

Given, number of mole, n=5
$V_2=80d{m}^3,V_1=8d{m}^3$
Entropy changes in reversible process in an ideal gas is given by
$△S=2.303n{C}_v\log \frac{T_2}{T_1}+2.303nR\log \frac{V_2}{V_1}$
for isothermal process
$△S=2.303nR\log \frac{V_2}{V_1}$
$△S=2.303\times 5\times 8.314\times \log \frac{80}{8}$
$△S=95.73J{K}^{-1}$