(a) For process AB,
Volume is constant, hence work done dW = 0
Now, by first law of thermodynamics,
`" "dQ =dU+dW=dU+0=dU`
`" "=nC_(v)dT=nC_(v)(T_(B)-T_(A))`
`" "=(3)/(2)R(T_(B)-T_(A))" "(becausen=1)`
`" "=(3)/(2)(RT_(B)-RT_(A))=(3)/(2)(p_(B)V_(B)-p_(A)V_(A))`
Heat exchanged = `(3)/(2)(p_(B)V_(B)-p_(A)V_(A))`
(b) For process BC, `" "p` = constant
`" "dQ=dU+dW=(3)/(2)R(T_(C)-T_(B))+p_(B)(V_(C)-V_(B))`
`" "=(3)/(2)(p_(C)V_(C)-p_(B)V_(B))+p_(B)(V_(C)-V_(B))=(5)/(2)p_(B)(V_(C)-V_(B))`
Heat exchanged = `=(5)/(2)p_(B)(V_(C)-V_(A))" "(becausep_(B)=p_(C)andp_(B)=V_(A))`
( c ) For process CD, Because CD is adiabatic, dQ = Heat exchanged = 0
(d) DA involes compression of gas from `V_(D)` to `V_(A)` at constant pressure `p_(A)`.
`therefore` Heat exchanged can be calculated by similar way as `BC_(1)`,
Hence, `" "dQ=(5)/(2)p_(A)(V_(A)-V_(D)).`