100 moles of an ideal monatomic gas undergoes the thermodynamic process as shown in the figure
`ArarrB:` isothermal expansion `BrarrC:` adiabatic expansion
`CrarrD:` isobaric compression `DrarrA:` isochoric process The heat transfer along the process AB is `9xx10^4J`. The net work done by the gas during the cycle is [Take `R=8JK^-1mol^-1`]

Options

A. (a) `-0.5xx10^4J`
B. (b) `+0.5xx10^4J`
C. (c) `-5xx10^4J`
D. (d) `+5xx10^4J`

Answer

D

Explanation

`Q_(BC)=0`
`Q_(CD)=nC_pDeltaT`
`=n(5/2R)(T_D-T_C)`
`=5/2(p_DV_D-p_CV_C)`
`=5/2(10^5-2xx10^5)`
`=-25xx10^4J`
`Q_(DA)=nC_VDeltaT`
`=n(3/2R)(T_A-T_D)`
`=3/2(p_AV_A-p_DV_D)`
`=3/2(2.4xx10^5-10^5)`
`=21xx10^4J`
Now, `W_(n et)=Q_(n et)`
`=(9-25+21)xx10^4J=5xx10^4J`