5 moles of an ideal gas at $27^{\circ} C$ expands isothermally and reversibly from a volume of 6 L to 60 L. The work done in kJ is:

-28.72 kJ

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-32.60 kJ

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14.28 kJ

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20.32 kJ

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Solution

The correct option is A -28.72 kJ
$w = - 2.303 n R T log \frac{V_{2}}{V_{1}} = - 2.303 \times 5 \times 8.314 \times 300 log \frac{60}{6} = - 28.72 k J$

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5 moles of an ideal gas at 27∘C expands isothermally and reversibly from a volume of 6 L to 60 L. The work done in kJ is:

The correct option is A -28.72 kJ w=−2.303nRTlogV2V1=−2.303×5×8.314×300log606=−28.72 kJ

5 moles of an ideal gas at $27^{\circ} C$ expands isothermally and reversibly from a volume of 6 L to 60 L. The work done in kJ is:

The correct option is A -28.72 kJ
$w = - 2.303 n R T log \frac{V_{2}}{V_{1}} = - 2.303 \times 5 \times 8.314 \times 300 log \frac{60}{6} = - 28.72 k J$