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Pressure and volume of a gas changes fro...

Pressure and volume of a gas changes from `(p_0V_0)` to `(p_0/4, 2V_0)` in a process `pV^2=` constant. Find work done by the gas in the given process.

Text Solution

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The correct Answer is:
A, B
`pV^2=K=p_0V_0^2`
`:.` `p=(K)/(V^2)`
`W=intpdV=int_(V_0)^(2V_0)(K)/(V^2)dV`
`=[-(K)/(V)]_(V_0)^(2V_0)`
`=(K)/(V_0)-(K)/(2V_0)`
`(K)/(2V_0)`
Substituting, `K=p_0V_0^2`, we have
`W=1/2p_0V_0`
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