An ideal mono-atomic gas follows the path ABCD. Then, the work done during the complete cycle is:

Zero

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- \frac{1}{2} p V

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- p V

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- 2 p V

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Solution

The correct option is B $- p V$
Net work done during the complete cycle is equal to the area under the rectangle and if the cycle is clockwise then negative, i.e. work done on the gas.

$W = - p \times V = - (2 p - p) \times (2 V - V) = - p V$

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An ideal mono-atomic gas follows the path ABCD. Then, the work done during the complete cycle is:

The correct option is B −pVNet work done during the complete cycle is equal to the area under the rectangle and if the cycle is clockwise then negative, i.e. work done on the gas. W=−p×V=−(2p−p)×(2V−V)=−pV

The correct option is B $- p V$
Net work done during the complete cycle is equal to the area under the rectangle and if the cycle is clockwise then negative, i.e. work done on the gas.

$W = - p \times V = - (2 p - p) \times (2 V - V) = - p V$