Statement - I : When μ amount of an ideal gas undergoes adiabatic change from state ( P _ 1 , V _ 1 , T _ 1 ) to state ( P _ 2 , V _ 2 , T _ 2 ) , then work done is W = ( T _ 2 - T _ 1 ) 1 - γ , where γ = C _ P C _ V

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Correct answer: A. Both statement-I and statement-II are true.

The work done in an adiabatic process is given by,

${W}_{adiabatic} = \dfrac{nR\left({T}_{f} - {T}_{i}\right)}{1 - \gamma} \to$ statement I is true.

For adiabatic process $∆ Q = 0$ . From the first law of thermodynamics,

$∆ Q = W + \DeltaU$

$0 = W + \DeltaU$

$\DeltaU = - W$

If work is done on the gas, i.e. work is negative

$∴ \DeltaU$ is positive.

$∴$ The temperature will increase. Hence, statement II is also true.

Statement - I : When μ amount of an ideal gas undergoes adiabatic...

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