The density `(rho)` versus pressure (p) graph of one mole of an ideal monoatomic gas undergoing a cyclic process is shown in figure. Molecular mass of gas is M.

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

Answer

A, B, C

Explanation

For Process 1-2 `rho prop p`
`:.` `1/Vprop p`
`:.` Process is isothermal.
`DeltaU_1=0`
`Q_1=W_1=nRTIn((p_i)/(p_f))`
`=(p_0M)/(rho_0)In(1/2)` (as `n=1` and `RT=(PM)/(rho)`)
`=-(p_0M)/(rho_0)In(2)`
For Process 2-3
`Q_2=Q_p=nC_pDeltaT`
`=(1)(5/2R)(T_3-T_2)`
`=5/2((p_3M)/(rho_3)-(p_2M)/(rho_2))`
`=5/2((2p_0M)/(rho_0)-(2p_0M)/(2rho_0))`
`=2.5(p_0M)/(rho_0)`
`DeltaU_2=nC_VDeltaT`
Substituting the values like above we get, `Delta_2=(1.5p_0M)/(rho_0)`
`W_3=Q_2-DeltaU_2=(p_0M)/(rho_0)`
For Process 3-1
Density is constant. Hence,
volume is constant.
`:.` `W_3=0`
`:.` `Q_3=DeltaU_3=nC_VDeltaT`
`=(1)(3/2R)(T_1-T_3)`
`=3/2((p_0M)/(rho_0)-(2p_0M)/(rho_0))`
`=-(1.5p_0M)/(rho_0)`
`sumQ_(-ve)=|Q_1+Q_3|`
`=(p_0M)/(rho_0)(3/2+In2)(+p_0M//rho_0)+((-p_0M)/(rho_0)In2)`
(c) `eta=(W_(n et))/(sumQ_(+ve))=((+p_0M//rho_0)+((-p_0M)/(rho_0)In2))/((2.5p_0M//rho_0))`
`=2/5(1-In2)`