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One mole of helium gas follows the cycle...

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

A

`8/9`

B

`9/8`

C

`12/13`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
`U_(1)=nCvT=(3R)/2TimpliesT_(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 four times the volume becomes double
`DeltaU_(31)=U_(0)=4U_(0)=-3U_(0)`
`[1/2xx3P_(0)xxV_(0)]=(-3)/2P_(0)V_(0)=(-3)/2 RT_(0)=(-3R)/2xx(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)/2xx(P_(0)V_(0))/2=(3RT_(0))/2=U_(0)`
`Q_(12)=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`
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