A monatomic ideal gas undergoes the shown cyclic process in which path of the process `2 rarr 3` is a semicircle. If `2 mol` of gas is taken, find

Total heat rejected in one cycle

Options

A. `(P_(0)V_(0)(32-pi))/( 4)`
B. `(P_(0)V_(0)(32+pi))/( 4)`
C. `(P_(0)V_(0)(32+pi))/( 8)`
D. `(P_(0)V_(0)(32-pi))/( 8)`

Answer

D

Explanation

Process `1 rarr 2` ( Isochoric process )
`W_(12)=0`
`Q_(12)=DeltaU_(12)=nC_(V)(T_(2)-T_(1))`
`=n(3)/(2) R(T_(2)-T_(1))`
`=(3)/(2)[P_(0)V_(0)-(P_(0)//2)V_(0)]`
`=(3P_(0)V_(0))/(4)` (heat absorbed )
Process `3 rarr 1`
`W_(3 rarr 1)=` Area under process `3 rarr 1`
`=-((P_(0))/(2)+P_(0)).(V_(0))/(2)=-(3P_(0)V_(0))/(4)`
Work done during `3 rarr 1` will be negative as volume is decreasing.
`DeltaU_(3 rarr1)=n(3R)/( 2)(T_(1)-T_(2))(3)/(2)((P_(0)V_(0))/(2)-P_(0)2V_(0))=-(9P_(0)V_(0))/(4)`
`Q_(3 rarr1)=DeltaU_(3 rarr1)+W_(3 rarr1)=-3 P_(0) V_(0)` ( heat rejected )
process ` 2 rarr 3`
The temperature will be maximum at `x=2 P_(0), T_(x)=(3P_(0)V_(0))/(2R)`
`W_(r rarrx)=W_(x rarr3)=(1)/(4)pi(P_(0)V_(0))/(2)+(P_(0)V_(0))/(2)=(P_(0)V_(0))/(2)((pi)/(4)+1)`
`W_(2 rar3)=(P_(0)V_(0))/(4)(pi+4)`
`DeltaU_(2 rarrx)=n(3)/(2)R(T_(x)-T_(2))=(3)/(2)(3P_(0)V_(0)-P_(0)V_(0))=3P_(0)V_(0)`
`Delta U _(x rarr 3)=n(3)/(2)R(T_(x)-T_(2))=(3)/(2)(2P_(0)V_(0)-3P_(0)V_(0))`
`=-(3)/(2)P_(0)V_(0)`
`:.DeltaQ_(2 rarr x)=Delta U_(2rarrx)+DeltaW_(2 rarr x)`
`=3 P_(0)V_(0)+(P_(0)V_(0))/(2)((pi)/(4)+1)`
`=P_(0)V_(0)(3+(1)/(2)+(pi)/( 2))=(P_(0)V_(0))/( 2)(7+(pi)/(4))`
`Delta Q_(x rarr 3)=Delta U_(x rarr 3)+ W_(xrarr 3)`
`=-(3)/(2)P_(0)V_(0)+(P_(0)V_(0))/(2)((pi)/(4)+1)=(-P_(0)V_(0))/(2)(2-(pi)/(4))`
Total heat rejected in one cycle
`=3P_(0)V_(0)+(P_(0)V_(0))/(2)(2-(pi)/(4))=4 P_(0)V_(0)-(pi P_(0)V_(0))/(8)`
`=P_(0)V_(0)(4-(pi)/(8))=(P_(0)V_(0)(32-pi))/(8)`