In the process shown in figure, the internal energy of and ideal gas decreases by `(3p_0V_0)/(2)` in going from point C to A. Heat transfer along the process CA is

Options

A. `(+P_(0)V_(0)//2)`
B. `(-5P_(0)V_(0)//2)`
C. `(-3P_(0)V_(0)//2)`
D. 0

Answer

B

Explanation

`T_(B)=T_(C)`
`(P_(B)V_(B))/(nR)=(P_(c)V_(c))/(nR)`
`P_(B)V_(B)=P_(C)V_(C)`
`2P_(0)xxV_(0)=P_(0)xxV_(C)`
`V_(C)=2V_(0)`
`T_(C)=P_(0)xx2v_(0)`
nR
`T_(A)=(P_(0)V_(0))/(nR)`
`DeltaU=-3/2P_(0)V_(0)`
`Deltaomega=-P_(0)V_(0)`
`DeltaQ=-5/2P_(0)V_(0)`