In an isothermal expansion the work done in increasing the volume of one mole of a gas by $10$ % at a temperature of $100 K$ is :

79 J

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Zero

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8.31 J

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- 79 J

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Solution

The correct option is D $- 79 J$
Work done in increasing the volume
$W = μ R T {log}_{e} \frac{V_{2}}{V_{1}}$
$= - 1 \times 8.31 \times 100 \times {log}_{e} (\frac{110}{100})$
$= - 8.31 \times 100 \times {log}_{e} 1.1$
$W = - 79 J$ .

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In an isothermal expansion the work done in increasing the volume of one mole of a gas by 10% at a temperature of 100 K is :

The correct option is D −79 JWork done in increasing the volumeW=μRTlogeV2V1=−1×8.31×100×loge(110100)=−8.31×100×loge1.1W=−79 J.

In an isothermal expansion the work done in increasing the volume of one mole of a gas by $10$ % at a temperature of $100 K$ is :

The correct option is D $- 79 J$
Work done in increasing the volume
$W = μ R T {log}_{e} \frac{V_{2}}{V_{1}}$
$= - 1 \times 8.31 \times 100 \times {log}_{e} (\frac{110}{100})$
$= - 8.31 \times 100 \times {log}_{e} 1.1$
$W = - 79 J$ .