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One mole of a van der Waals PA gas obeyi...

One mole of a van der Waals PA gas obeying the equation P `(P + a/V^(2)) (V - b) = RT `

undergoes the quasi-static cyclic process which is shown in the P-V diagram. The net heat absorbed by the gas in this process is

A

`1/2 (P_(1) - P_(2)) (V_(1) - V_(2))`

B

`1/2 (P_(1) + P_(2))(V_(1) - V_(2))`

C

`1/2 (P_(1) + a/V_(1)^(2) - P_(2) - a/V_(2)^(2)) (V_(1) - V_(2))`

D

`1/2 (P_(1) + a/V_(1)^(2) + P_(2) + a/V_(2)^(2))(V_(1) - V_(2))`

Text Solution

Verified by Experts

The correct Answer is:
a

For a cyclic process , `Delta U = 0 `
` :. ` According to first law of thermodynamics, the net heat absorbed by the gas is
` Delta Q = " Work done , " Delta W = " Area of " Delta ABC `
` = 1/2 (P_(1) - P_(2)) (V_(1) - V_(2))`
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