Consider the isothermal reversible expansion of an ideal gas an a real(van der Waals) gas, from the same initial volume $V_{1}$ to the same final volume $V_{2}$ , at the same temperature T. The absolute value of the work done.

By the real gas is always higher than the work done by the ideal gas, if the inter-particle interaction in the former is neglible.

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By the real gas is always higher than the work done by the ideal gas, if the volume of the particles in the former is neglible

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In both the cases will always be the same, since the initial and final volumes in both the cases are the same

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By the ideal gas is always higher than the work done by the real gas, since there is no hindrance due to interparticle interaction and the volume of the particles

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Solution

The correct option is B By the real gas is always higher than the work done by the ideal gas, if the volume of the particles in the former is neglible
Work done by ideal gas, isothermally and reversibly is:
$w_{i d e a l} = - n R T ln (\frac{V_{2}}{V_{1}})$
Vandewal gas equation is:
$(P + \frac{a n^{2}}{V^{2}}) (V - n b) = n R T P = \frac{n R T}{V - n b} - \frac{a n^{2}}{V^{2}} w_{r e a l} = - \int P d V = - \int (\frac{n R T}{V - n b} - \frac{a n^{2}}{V^{2}}) d V = - \int_{V_{1}}^{V_{2}} \frac{n R T}{V - n b} d V + \int_{V_{1}}^{V_{2}} \frac{a n^{2}}{V^{2}} d V = - n R T ln (\frac{V_{2} - n b}{V_{1} - n b}) - a n^{2} (\frac{1}{V_{2}} - \frac{1}{V_{1}}) i f n b << V,$
i.e. volume of particles is negligible in comparison to volume of gas occupied $w_{r e a l} = - n R T ln (\frac{V_{2}}{V_{1}}) - a n^{2} (\frac{1}{V_{2}} - \frac{1}{V_{1}}) w_{r e a l} - w_{i d e a l} = - n R T ln (\frac{V_{2}}{V_{1}}) - a n^{2} (\frac{1}{V_{2}} - \frac{1}{V_{1}}) + n R T ln (\frac{V_{2}}{V_{1}}) w_{r e a l} - w_{i d e a l} = a n^{2} (\frac{V_{2} - V_{1}}{V_{1} V_{2}}) > 0 a s V_{2} > V_{1} w_{r e a l} - w_{i d e a l} > 0 o r w_{r e a l} > w_{i d e a l}$
Therefore work done by real gas is always higher than work done by ideal gas, if the volume of the particles in real gas is negligible ( $b << V$ ).
Hence, option $B$ is correct.

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Consider the isothermal reversible expansion of an ideal gas an a real(van der Waals) gas, from the same initial volume V1 to the same final volume V2, at the same temperature T. The absolute value of the work done.

Consider the isothermal reversible expansion of an ideal gas an a real(van der Waals) gas, from the same initial volume $V_{1}$ to the same final volume $V_{2}$ , at the same temperature T. The absolute value of the work done.