100 mole of an ideal monoatomic gas undergoes a thermodynamic process as shown in the figure. The heat transfer along the process AB is `9 xx 10^(4)` J. The neet work done by the gas during the cycle is (Take R = 8 J. `K^(-1) "mole"^(-1)`)

Options

A. `-0.5 xx 10^(4)` J
B. `+0.5 xx 10^(4)` J
C. `-5 xx 10^(4)` J
D. `+5 xx 10^(4)` J

Answer

D

Explanation

`DeltaA_(AB) = 9 xx 10^(4) J`
`Delta Q_(BC) =0`
`DeltaQ_(CD) = n(5/2)R[(1 xx 10^(5) xx 1)/(nR) - (1 xx 10^(5) xx 2)/(nR)]`
`=-2.5 xx 10^(5)` J
`DeltaQ_(DA) = n(3/2)R[(2.4 xx 10^(5) xx 1)/(nR) - ( 1 xx 10^(5) xx 1)/(nR)]`
`=2.1 xx 10^(5)` J
`W = sumDeltaQ = 5 xx 10^(4)` J