Figure shows the variation in the internal energy U with the volume V of `2.0 mol` of an ideal gas in a cyclic process abcda. The temperatures of the gas at b and c are `500 K` and `300 K` respectively. Calculate the heat absorbed by the gas during the process.

Explanation

In the process a to b and c to d
as `Delta U = 0`, therefore `Delta T = 0` or `T = ` constant
`W = int_(V_(i))^(f_(i)) PdV`
We have
`PV = nRT implies P = (nRT)/(V)`
`:. W = int_(V_(i))^(V_(f)) (nRT) (RV)/(V)`
`= nRT |1n V|_(V_(i))^(V_(f)) = nRT 1n (V_(f))/(V_(i))`
`W_(ab) = nRT_(b) 1n (2 V_(0))/(V_(0)) = 2 R xx 500 1n 2 = 1000 R 1n 2`
and `W_(cd) = nRT_(c ) 1n (V_(0))/(2V_(0)) = 2 R xx 300 1n (1)/(2) = - 600 R 1n 2`
There is no volume change from `b` to `c` and from `d` to `a`, so
`W_(bc) = V_(da) = 0`
The work done in complete cycle
`W = W_(ab) + W_(bc) + W_(cd) + W_(da)`
`1000 R 1n 2 + 0 - 600 R 1n 2 + 0`
`= 400 R 1n 2`