Fig. shows a process `ABCA` performed on 1 mole of an ideal gas. Find the net heat supplied to the gaseous system during the process.

Explanation

As we know in a cyclic process gas fully returns to its initial state and total change in initial energy of gas is zero, thus the total heat supplied to the gas is equal to work done by the gas. Now we find the work done by the gas in different paths of the cycle.
In process `AB`
As shown in graph, during process `AB`, volume of gas remains constant, thus work done by gas is zero.
`W_(AB) = 0`
In process `BC`
In process `BC`, volume of gas changes form `V_(1) = 2 m^(3)` to `V_(2) = 5 m^(3)` thus work done can be obtained as
` W_(BC) = int_(2)^(5) PdV`
An process `BC`, temperature of gas remains constant at `500 K`, we can write pressure of gas from gas law as
`P = (RT)/(V) = (500R)/(V)` `(as n = 1 mol)`
Now work done is
`W_(BC) = int_(2)^(5) (500 R)/(V) dV = 500 R 1n ((5)/(2))`
In process `CA`
As in this process, path is is a striaght line passing throught origin thus `V prop T` or pressure of gas remains constant and we know is constant work done is given as
`W_(CA) = nRT (T_(2) - T_(1)) = nR (300 - 500) = - 200 R`
This is negative as gas is being compressed from volume `5 m^(3)` to `2 m^(3)` or work is done on the gas.
Now we can find the total work done by the gas in the complete cycle `ABCA` as
`W_(ABCA) = W_(AB) + W_(BC) + W_(CA)`
`= - 0 + 500 R 1n ((5)/(2)) - 200 R - R (500 1n (5)/(2) - 200)`
`=2146.22 J =` heat supplied to the gas