(a) For the process AB,
`" "dV=0rArrdW=0" "(because"volume is constant")` ltBrgt `" "dQ=dU+dW=dU`
`rArr " "dQ=dU="Change in internal energy.")`
Hence, in this process heat supplied is utilised to increase, internal energy of the system.
Since, `p=((nR)/(V))T`, in isochoric process, `Tpropp.` So tempereature increases with increases
of pressure in process AB which inturn increases internal energy of the system i.e.,
dU`gt` 0. This imply that dQ`gt`0. So heat is supplied to the system in process AB.
(b) For the process CD, volume is constant by pressure decreases.
Hence, temperature also decreases so heat is given to surroundings.
(c ) To calculate work done by the enging in one cycle, we calculate work done in each part separately.
`" "W_(AB)=int_(A)^(B)pdV=0, W_(CD)=int_(V_(C))^(V_(D))pdV=0" "(becausedV=0)`
`" "W_(BC)=int_(V_(B))^(V_(C))pdV=kint_(V_(B))^(V_(C))(dV)/(V^(gamma))=(k)/(1-gamma)[V^(1-gamma)]_(V_(B))^(V_(C))`
`" "=(1)/(1-gamma)[pV]_(V_(B))^(V_(C))=((p_(C)V_(C)-p_(B)V_(B)))/(1-gamma)`
Similarly, `" "W_(DA)=(p_(A)V_(A)-p_(D)V_(D))/(1-gamma)" "[becauseBC" is adiabtic process"]`
`because B and C` lies on adiabatic curve BC. ltBrgt `therefore" "p_(B)V_(B)^(gamma)=p_(C)V_(C)^(gamma)`
`" "p_(C)=p_(B)((V_(B))/(V_(C)))^(gamma)=p_(B)((1)/(2))^(gamma)=2^(-gamma)p_(B)`
Similarly, `" "p_(D)=2^(-gamma)p_(A)`
Total work done by the engine in one cycle ABCDA.
`" "W=W_(AB)+W_(BC)+W_(CD)+W_(DA)=W_(BC)+W_(DA)`
`" "=((p_(C)V_(C)-p_(B)V_(B)))/(1-gamma)+((p_(A)V_(A)-p_(D)V_(D)))/(1-gamma)`
`" "W=(1)/(1-gamma)[2^(-gamma)p_(B)(2V_(B))-p_(B)V_(B)+p_(A)V_(A)-2^(-gamma)p_(B)(2V_(B))]`
`" "(1)/(1-gamma)[p_(B)V_(B)(2^(-gamma+1)-1)-p_(A)V_(A)(2^(-gamma+1)-1)`
`" "=(1)/(1-gamma)(2^(1-gamma)-1)(p_(B)-p_(A))V_(A)`
`" "(3)/(2)[1-((1)/(2))^(2//3)](p_(B)-p_(A))V_(A)`