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Work Done in Isothermal Reversible Process

A given mass ...

Question

A given mass of gas expands reversibly from state A to state B by three paths 1,2 and 3 as shown in the figure. If $w_{1}, w_{2}, a n d w_{3}$ respectively are the work done by the gas along three paths, then:

w_{1} < w_{2} > w_{3}

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w_{1} < w_{2} < w_{3}

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w_{1} > w_{2} > w_{3}

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w_{1} = w_{2} = w_{3}

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Solution

The correct option is B $w_{1} < w_{2} < w_{3}$
Work done for a process = Area under the PV curve.
More the area under the PV curve, more will be the work done for a process.
From the given graph, path 3 has the maximum area. So, maximum work is done in this case.
Similarly, path 1 has the least work done for a process since it has the least area under the curve. So, the order of work done is $w_{1} < w_{2} < w_{3}$ .

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