A cyclic process abcd is given for a monoatomic gas (`C_V=3/2R` and `C_p=5/2R`) as shown in figure. Find Q, W and `DeltaU` in each of the four processes separately. Also find the efficiency of cycle.

Answer

A, C

Explanation

Process ab
`V=` constant (`:.` Isochoric process)
`W_(ab)=0`
`:.` `Q_(ab)=DeltaU_(ab)=nC_VDeltaT`
`=n(3/2R)(T_b-T_a)`
`=3/2(nRT_b-nRT_a)`
`=3/2(p_bV_b-p_aV_a)`
`=3/2(2p_0V_0-p_0V_0)`
`=1.5p_0V_0`
Process bc
`p=` constant (`:.` Isobaric process)
`Q_(bc)=nC_pDeltaT`
`=n(5/2R)(T_c-T_b)`
`=5/2(nRT_c-nRT_b)`
`=5/2(p_cV_c-p_bV_b)`
`=5/2(4p_0V_0-2p_0V_0)`
`=5p_0V_0`
`DeltaU_(bc)=nC_VDeltaT`
`=n(3/2R)(T_c-T_b)`
`=3/2(nRT_c-nRT_b)`
`=3/2(p_cV_c-p_bV_b)`
`=3/2(4p_0V_0-2p_0V_0)`
`=3p_0V_0`
`W_(bc)=Q_(bc)-DeltaU_(bc)=2p_0V_0`
Process cd
Again an isochoric process.
`:.` `W_(cd)=0`
`Q_(cd)=DeltaU_(cd)=nC_VDeltaT`
`=n(3/2R)(T_d-T_c)`
`=3/2(nRT_d-nRT_c)`
`=3/2(p_dv_d-p_cV_c)`
`=3/2(2p_0V_0-4p_0V_0)`
`=-3p_0V_0`
Process da
This is an isobaric process.
`:.` `Q_(da)=nC_pDeltaT`
`=n(5/2R)(T_a-T_d)`
`=5/2(nRT_a-nRT_d)`
`=5/2(p_aV_a-p_dV_d)`
`=5/2(p_0V_0-2p_0V_0)`
`=-2.5 p_0V_0`
`DeltaU_(da)=nC_VDeltaT`
`=n(3/2R)(T_a-T_d)`
`=3/2(nRT_a-nRT_d)`
`=3/2(p_aV_a-p_dV_d)`
`=3/2(p_0V_0-2p_0V_0)`
`=-1.5 p_0V_0`
`W_(da)=Q_(da)-DeltaU_(da)`
`=-p_0V_0`
Efficiency of cycle
In the complete cycle,
`W_(n et)=W_(ab)+W_(bc)+W_(cd)+W_(da)`
`=0+2p_0V_0+0-p_0V_0`
`=p_0V_0`