In a process, the pressure of an ideal g...

In a process, the pressure of an ideal gas is proportional to square of the volume of the gas. If the temperature of the gas increases in this process, then work done by this gas

is positive

is negative

is zero

may be positive

Text Solution

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The correct Answer is:

`P alpha 1/(V^(2))rArr P=k/(V^(2)) rArr P V^(2)=k`
`rArr PV.V=krArr nRTVrArr TV=k_(1)`
Since temperature increases therefore volume decreases

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