One mole of an ideal gas at $300 \mathrm{~K}$ in thermal contact with surroundings expands isothermally from $1.0 \mathrm{~L}$ to $2.0 \mathrm{~L}$ against a constant pressure of $3.0 \mathrm{~atm}$. In this process, the change in entropy of surroundings $\left(\Delta S_{\text {surr }}\right)$ in $\mathrm{J} \mathrm{K}^{-1}$ is $(1 \mathrm{~L} \mathrm{~atm}=101.3 \mathrm{~J})$
IIT 2016, Easy
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The correct option is $C -1.013$
Since it is an isothermal process, $\Delta U=0$
$d q=-d W=P_{\text {ext }}\left(V_2-V_1\right)=3 L-\text { atm }=3 \times 101.3 J$
$\Delta S_{\text {surrounding }}-\frac{3 \times 101.3}{300} J K ^{-1}=-1.013 J K ^{-1}$
$\therefore \Delta S_{\text {surr }}=-1.013 J K ^{-1}$
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