Figure shows the variation of internal energy `(U)` with the pressure `(P)` of `2.0` mole gas in cyclic process abcda.The temperature of gas at c and d are `300 and 500 K` respectively. The heat absorbed by the gas during the process is `Xxx 10 R ln 2` . Find the value of x.

Answer

D

Explanation

Change in internal energy for cyclic process `(DeltaU)=0`
For process `ararrb`,(P-constant)
`W_(ararrb)=PDeltaV=nRDeltaT=-400R`
For process `brarrc` (T-constant)
`W_(brarrc)=-2R(300)1n2`
For process
`crarrd` (P-constant) `W_(crarrd)=+400R`
For process `drarra` (T-constant)
`W_(drarra)=+2R(500)1n2` Net work,
`(DeltaW)=W_(ararrb)+W_(brarrc)+W_(crarrd)+W_(drarra) DeltaW=400r 1n2`
`dQ=dQ+dW` , first law of thermodynamics
`dQ=400 R 1n 2`