The pressure versus volume graph of one mole of an ideal monatomic gas undergoing a cyclic process is shown in figure. The molecular mass of the gas is M.

(a) Find the work done in each process.
(b) Find heat rejected by gas in one complete cycle.
(c) Find the effiency of the cycle.

Answer

A, B, C

Explanation

(a) Given, `n=1` `:.` `m=M`
Process AB `rho prop p`, i.e. It is an isothermal process (T=constant), because `rho=(pM)/(RT)`.
`:.` `W_(AB)=RT_AIn((p_(A))/(p_(B)))=RT_AIn(1/2)`
`=-(p_0M)/(rho_0)In(2)`
`DeltaU_(AB)=0`
and `Q_(AB)=W_(AB)=-(p_0M)/(rho_0)In(2)`
Process BC is an isobaric process (p=constant)
`W_(BC)=p_B(V_(C)-V_(B))=2p_(0)((M)/(rho_(C))-(M)/(rho_B))=2p_0(M/rho_0-M/(2rho_0))=(p_0M)/(rho_0)`
`DeltaU_(BC)=C_VDeltaT`
`=(3/2R)[(2p_0M)/(rho_0R)-(2p_0M)/(2rho_0R)]=(3p_0M)/(2rho_0)`
`Q_(BC)=W_(BC)+DeltaU_(BC)=(5p_0M)/(2rho_0)`
Process CA As `rho=` constant
`:.` `V=` constant
So, it is an isochoric process.
`W_(CA)=0`
`DeltaU_(CA)=C_(V)DeltaT`
`=(3/2R)(T_A-T_C)`
`=(3/2R)[((p_0M)/(rho_0R)-(2p_0M)/(rho_0R)]`
`=-(3p_0M)/(2rho_0)`
`Q_(CA)=DeltaU_(CA)=-(3p_0M)/(2rho_0)`
(b) Heat rejected by gas `=|Q_(AB)|+|Q_(CA)|`
`=(p_0M)/(rho_0)[3/2+In(2)]`
(c) Efficiency of the cycle (in fraction)
`eta`=" Total work done"/"Heat supplied"=W_(T otal)/(Q_(+ve))`
`=((p_0M)/(rho_0)[1-In(2)])/(5/2((p_0M)/(rho_0)))`
`=2/5[1-In(2)]`