In an isothermal reversible expansion, if the volume of 96 gm of oxygen at $27^{\circ} C$ is increased from 70 litres to 140 litres, then the work done by the gas will be

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Solution

The correct option is D

W = $μ R T l o g_{e} \frac{V_{2}}{V_{1}}$

= $\frac{m}{M} R T$ $l o g_{e} \frac{V_{2}}{V_{1}}$ = $2.3 \times \frac{m}{M} R T l o g_{10} \frac{V_{2}}{V_{1}}$

= 2.3 $\times$ $\frac{96}{32}$ R $(273 + 27) l o g_{10} \frac{140}{70}$ = $2.3 \times 900 R l o g_{10} 2$

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27^{\circ} C

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