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 as.

Options

A. 2
B. 3
C. 5
D. 6

Answer

C

Explanation

`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`