Question

One mole of an ideal gas at $300 \,K$ in thermal contact with surroundings expands isothermally from $1.0\, L$ to $2.0\, L$ against a constant pressure of $3.0\, atm$. In this process, the change in entropy of surroundings $\left(\Delta S _{ surr }\right)$ in $JK ^{-1}$ is- (1 L atm $=101.3 \,J )$

• A. 5.763
• B. 1.013
• C. -1.013
• D. -5.763

Answer 1

Answer: (C)

W.D. $=-P_{\text {ex }}\left(V_{2}-V_{1}\right)$
W.D. $=-3(2-1)=-3 \times 101.3=-303.9 \,J$
$\Delta E = q + W \,\,\{\Delta E =0$ for isothermal process $\}$
$q=-w$
We know $\Delta S=\frac{q}{T}$
$\Delta S_{\text {system }}=+\frac{303.9}{300}=+1.013\, J / K$
$\therefore \Delta S_{\text {Surrounding }}=-\Delta S_{\text {system }}$
$=-1.013\, J / K$