Figure shows a cycle `ABCDA` undergone by `2` moles of an ideal diatomic gas. The curve `AB` is a rectangular hyperbola and `T_(1)=300K` and `T_(2)=500K`. Determine the work done by the gas in the process `ArarrB`.

Options

A. `-3.320kJ`
B. `4.326kJ`
C. `2.326kJ`
D. `3.326kJ`

Answer

A

Explanation

Evidently, for the process `ArarrB`
`Vprop(1)/(T) ` or `VT`=constant (say,K)
or `TdV+VdT=0`or `dV=-(VdT)/(T)`
(a) Now, work done by the gas in a process is given by
`W=SigmaPdV` or `W_(AB)=Sigma(nRTdV)/(V)=Sigma(-nRdT)`
`W_(AB)=overset(500)underset(300)(int)-nRdT=-(2 mol) (8.314 J//mol-K)`
`[500-300]K=-3.326kJ`