The PV-graph for a monoatomic gas is shown in the figure. Find the energy absorbed by the gas during the process.
Since the process is cyclic the work done (W)=Heat or energy transfer or W=Q. …...(1)
Total Work done in the process is equal to the area of the cyclic process.
Therefore, W= ½ * (2Vo -Vo)(3Po -2Po) = ½ VoPo
Heat released during process B->C
QB->C = nCvT= 3/2 (2Po-3Po)(2Vo) = -3VoPo
Heat released during process C->A
QC->A = nCpT = 5/2 (2Po)(Vo-2Vo) = -5VoPo
Heat transfer during process A->C is
QA->C
From (1)
We have,
QA->C + QB->C + QC->A = W
=> QA->C = W -(QB->C + QC->A)
=> QA->C = ½ VoPo – (-3VoPo -5VoPo)
= 17/2 VoPo
= 8.5 VoPo.
Hope this helped you. Cheers..
Heat released during process B->C
QB->C = nCvT= 3/2 (2Po-3Po)(2Vo) = -3VoPo
Heat released during process C->A
QC->A = nCpT = 5/2 (2Po)(Vo-2Vo) = -5VoPo
Heat transfer during process A->C is
QA->C
From (1)
We have,
QA->C + QB->C + QC->A = W
=> QA->C = W -(QB->C + QC->A)
=> QA->C = ½ VoPo – (-3VoPo -5VoPo)
= 17/2 VoPo
= 8.5 VoPo.
Hope this helped you. Cheers..
