An ideal gaseous sample at initial state $i (P_{0}, V_{0}, T_{0})$ is allowed to expand to volume $2 V_{0}$ using two different process, in the first process, the equation of process is $2 P V^{2} = K_{1}$ and in the second process the equation of the process is $P V = K_{2}$ • Then :

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The order of value of work done cannot be compared unless we know the value of $K_{1}$ and $K_{2}$ .

I st process is not possible.

Value of work done (magnitude) in second process is greater in above expansion irrespective value of $K_{1}$ and $K_{2}$ .

Work done in first process will be greater than work in second process (magnitude wise).

answer is C.

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Detailed Solution

total gas at the outset $(P_{0}, V_{0}, T_{0})$

The adiabatic process involves: $w = \frac{K}{1 - γ} (V_{f}^{1 - γ} - V_{i}^{1 - γ})$

work done in $P V = k_{2}$ is more

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