In figure, a cyclic process `ABCA` of `3` moles of an ideal gas is given. The temperature of the gas at `B` and `C` are `500k` and `1000K` respectively. If work done on the gas in process `CA` is `2500J` then find the net heat absorbed or released by an idela gas. Take `R = 25//3J//mol-K`.

Explanation

The change in internal energy during the cyclic process is zero. Hence , the heat supplied to the gas is equal to the work done by it. Hence,
`DeltaQ = W_(AB) +W_(BC) +W_(CA) ……….(i)`
The work done during the process `AB` is zero
`W_(BC) = P_(B) (V_(C)-V_(B))`
`= nR (T_(C)-T_(B))`
`= (3 mol) (25//3J//mol-K) (500K)` ltbr. `= 12500J`
As `W_(CA) = - 2500J` (given)
`:. DeltaQ = 0 + 12500 - 2500` [from ..........(i)]
`DeltaQ = 10 kJ`