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Some gas (C(p)//C(V)=gamma=1.25) follows...

Some gas `(C_(p)//C_(V)=gamma=1.25)` follows the cycle ABCDA as shown in the figure. The ratio of the energy given out by the gas to its surrounding durning the isochoric section of the cycle to the expansion work done during the isobaric section of the cycle is

A

2

B

4

C

6

D

0.08

Text Solution

Verified by Experts

The correct Answer is:
B
Let n moles of gas follows the cycle ABCDA
`Q_(isochoric)=nC_(V)DeltaT=(nR)/(gamma-1)(T_(B)-T_(C))`
`W_(isobaric)=nRDeltaT=nR(T_(A)-T_(D))=nR(T_(B-T_(C)))`
`therefore` Required ratio =`(Q_(isochoric))/(W_(isobaric))=(1)/(gamma-1)=4`
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