One mole of helium gas follows the cycle 1-2-3-1 shown in the diagram. During process 3-1, the internal energy (U) of the gas depends on its volume (V) as `U=bV^(2)`, where b is a positive constant. If gas releases the amount of heat `Q_(1)` during process 3-1 and gas absorbs th amount of heat `Q_(2)` during process `1rarr2rarr3` them:

The value of `O_(1)//Q_(2)` is

Options

A. `(8)/(9)`
B. `(9)/(8)`
C. `(12)/(13)`
D. None of these.

Answer

C

Explanation

`U_(1)=nCvT=(3R)/(2)TimpliesT_(1)=(2U_(0))/(3R)`
Process `3-1impliesU=bV^(2)`
`implies(2bV_(1)^(2))/(3R)implies(PV_(1))/(R)=(2bV^(2))/(3R)impliesp_(1)=(2bV)/(3)`
When temperature becomes b four times, the volume becomes double.
`DeltaU_(31)=U_(0)-4U_(0)=-3U_(0)`
`W_(31)=`
`[(1)/(2)xx3P_(0)xxV_(0)]=(-3)/(2)P_(0)V_(0)=(-3)/(2)RT_(0)=(-3R)/(2)xx(2U_(0))/(3R)=-U_(0)`
`Q_(31)=-4U_(0)=-Q_(1)`
Process `1-2implies` isochoric process
`W_(12)=DeltaU_(12)=nCv(T_(2)-T_(1))=1xx(3R)/(2)((2P_(0)V_(0))/(R)(P_(0)V_(0))/(R))`
`=(3R)/(2)xx(P_(0)V_(0))/(2)=(3RT_(0))/(2)=U_(0)`
Process `2-3implies` isobaric process
`W_(23)=2P_(0)v_(0)=2xxRT_(0)=2Rxx(2U_(0))/(3R)=(4U_(0))/(3)`
`U_(23)=nCV(T_(3)-T_(2))=1xx(3R)/(2)`
`[(4P_(0)V_(0))/(R)-(2P_(0)V_(0))/(R)]=3P_(0)V_(0)=3RT_(0)=2U_(0)`
`Q_(23)=(10U_(0))/(3)`
`Q_(2)=Q_(12)+Q_(23)=(13U_(0))/(3)`