1.

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

Answer»

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

Solution :
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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