An ideal gas is taken around the cycle $ \mathrm{ABCA} $ as shown in $ \mathrm{P}-\mathrm{V} $ diagram. The net work done during the cycle is equal to -

Question

An ideal gas is taken around the cycle $ \mathrm{ABCA} $ as shown in $ \mathrm{P}-\mathrm{V} $ diagram. The net work done during the cycle is equal to -, $ 12 \mathrm{P}_{1} \mathrm{~V}_{1} $, $ 6 \mathrm{P}_{1} \mathrm{~V}_{1} $, $ 5 \mathrm{P}_{1} \mathrm{~V}_{1} $, $ \mathrm{P}_{1} \mathrm{~V}_{1} $

MARKS App · Accepted Answer

From the graph, it is clear that the work done in the cyclic process is equal to the area of a triangle. Area of triangle = 1 2 × base × height = 1 2 × ( 3 V 1 - V 1 ) × ( 6 P 1 - P 1 ) = 5 P 1 V 1

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