Fig. shows the variation of internal energy `(U)` with the pressure `(P)` of 2.0 mole gas in cycle process `abcda`. The temperatures of gas at `c` and `d` are 300 and `500 K`, respectively. Calculate the heat absorbed by the gas during the process.

Explanation

Change in internal energy for cyclic process `(Delta U) = 0`
For process `a rarr b`, (P - constant)
`W_(a rarr b) = P Delta V = n R Delta T = - 400 R`
For process `b rarr c` (T - constant)
`W_(b rarr c) = - 2 R (300) 1n 2)`
For process `c rarr d` (P - constant)
`W_(c rarr d) = + 400 R`
For process `d rarr a` (T - constant)
`W_(d rarr a) = + 2 R (500) 1n 2`
Net work,
`(Delta W) = W_(a rarr b) + W_(b rarr c) + W_(c rarr d) + W_(d rarr a)`
`Delta W = 400 R 1n 2`
`dQ = dU + dW`, first law of thermodynamics
`dQ = 400 R 1n 2`