Two moles of helium gas undergo a cyclic process as shown in figure. Assuming the gas to be ideal. The net work done by the gas is

Options

A. 200 Rln2
B. 100 Rln2
C. 300 Rln2
D. 400 Rln2

Answer

A

Explanation

At constant pressure,
`W = P(V_(f) - V_(i)) = nR(T_(f) - T_(i))`
At constant temperature,
`W = nRT l (V_(f))/(V_(i)) = nRT ln((P_(i))/(P_(f)))`
So work done for pths AB, BC, CD and DA respectively will be `W_(AB) = 2 xx R(400 - 300) = 200 R`
`W_(BC) = 2 xx R xx 400 ln ((2)/(1)) = 800 Rln2`
`W_(CD) = 2 xx R(300 - 400) = -200 R`
`W_(DA) = 2 xx R xx 300 ln((1)/(2)) =-600 Rln 2`
The net work done by the gas is `W = W_(AB) + W_(BC) + W_(CD) + W_(DA)`
`=200 R + 800 Rln2 - 200 R - 600 R ln2 = 200 R ln 2`