Question

At

500 \mathrm{~K}

for an isobaric process.

\Delta \mathrm{S}_{\text {system }}=-10 \frac{\mathrm{kJ}}{\mathrm{mol} \mathrm{K}} \text { and } \Delta \mathrm{S}_{\text {surr }}=12 \frac{\mathrm{kJ}}{\mathrm{molK}}

Therefore,

\Delta \mathrm{G}

for the entire process is-

Accepted Answer

$-1000 \mathrm{~kJ} / \mathrm{mol}$

Answer

$-500 \mathrm{~kJ} / \mathrm{mol}$

Answer

$-600 \mathrm{~kJ} / \mathrm{mol}$

Answer

$-1100 \mathrm{~kJ} / \mathrm{mol}$

JEE PHYSICS MCQ: THERMODYNAMICS

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