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Two moles of an ideal monoatomic gas is taken through a cyclic process as shown in the P-T diagram. In the process BC, `PT^(-2)=` constant. Then the ratio of heat absorbed and heat released by the gas during the process AB and process BC respectively is.

A

A. 2

B

B. 3

C

C. 5

D

D. 6

Text Solution

Verified by Experts

The correct Answer is:
C

`DeltaQ_(AB)=nC_(PDelta)T=2xx(5R)/(2)(2T_(0)-T_(0))=5RT_(0)`
In the process `BC,PT^(-2)`= constant
`PV^(2)=`constant
`therefore` molar heat capacity `C=C_(V)+(R)/(1-x)=(3R)/(2)+(R)/(1-2)`
`C=(R)/(2)`
`thereforeDeltaQ_(BC)=nC_(Delta)T=2xx(R)/(2)(T_(0)-2T_(0))=-RT_(0)`
`therefore(|DeltaQ_(AB)|)/(|DeltaQ_(BC)|)=((5RT)_(0))/(RT_(0))=5`
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