A system undergoes three quasi-static process sequentially as indicated in Fig. 1-2 is an isobaric process, 2-3 is a polytropic process with `gamma = 4//3` and 3-1 is a process in which `PV` = constant. `P_(2) = P_(1) = 4 xx 10^(5) N//m^(2), P_(3) = 1 xx 10^(5) N//m^(2)` and `V_(1) = 1 m^(3)`.
The heat transfer for the cycle is `Delta Q`, the change in internal energy is `Delta U` and the work done is `Delta W`. Then

Options

A. `Delta W = 0`
B. `Delta Q = 1.08 xx 10^(5) J`
C. `Delta U = 0`
D. `Delta Q gt Delta W`

Answer

A, B, C

Explanation

For the process `3 rarr 1`
`P_(1)V_(1)=P_(3)V_(3)implies V_(3)=(P_(1)V_(1))/(P_(3))=(4xx10^(5)xx1)/( 1 xx 10^(5))=4 m ^(3)`
For the process ` 2 rarr 3`
`P_(2)V_(2)^(gamma)=P_(3)V_(3)^(gamma)implies V_(2)=((P_(3))/(P_(2)))^((1)/(gamma))V_(3)=4((1)/(4))^((1)/(4))`
`=4^((1)/(4))m^(3)=sqrt(2)m^(3)`
For the process `1 rarr 2 `,
`W_(12)=P_(1)(V_(2)-V_(1))`
`= ( sqrt(2)-1)4 xx 10^(5)J`
For the process `2 rarr 3`,
`W_(23)=((P_(2)V_(2)-P_(3)V_(3)))/((gamma-1))`
`=((4 sqrt(2)-1 xx ) xx 10^(5))/(((4)/(3)-1))=12 (sqrt(2)-1)xx10^(5)J`
For the process `3 rarr 1 `,
`W_(31)=-P_(1)V_(1) 1n(V_(3))/(V_(1))`
`=-4 xx 1-^(5)xx1xx((4)/(1))`
`=-4xx10^(5)xx1xx1n ((4)/(1))`
`DeltaU=0`
`DeltaQ_(1rarr2rarr3rarr1)=DeltaU+DeltaW=4 xx 10^(5)(sqrt2)-1)`
`+ 12 xx10^(5)(sqrt(2)-1)-4xx10^(5)xx1.386`
`~= 1.08 xx 10^(5)J`
`Delta Q=DeltaW~= 1.08 xx 10^(5)J`