Two moles of a monatomic ideal gas undergo a cyclic process ABCDA as shown in figure. BCD is a semicircle. Find the efficiency of the cycle.

Answer

B

Explanation

Process AB is isochoric (V=constant).
Hence,
`DeltaW_(AB)=0`
`DeltaW_(BCD)=p_0V_0+pi/2(p_0)(V_0/2)`
`=(pi/4+1)p_0V_0`
`DeltaW_(DA)=-1/2(p_0/2+p_0)(2V_0-V_0)`
` =-3/4p_0V_0`
`DeltaU_(AB)=nC_VDeltaT=(2)(3/2R)(T_B-T_A)` `(n=2,C_V=3/2R)`
`=3R((p_0V_0)/(2R)-(p_0V_0)/(4R))`
`=3/4p_0V_0=DeltaQ_(AB)` `(T=(pV)/(nR))`
`DeltaU_(BCD)=nC_VDeltaT=(2)(3/2R)(T_D-T_B)`
`=(3R)((2p_0V_0)/(2R)-(p_0V_0)/(2R))=3/2p_0V_0`
Hence, `DeltaQ_(BCD)=DeltaU_(BCD)+DeltaW_(BCD)`
`(pi/4+5/2)p_0V_0`
`DeltaU_(DA)=nC_VDeltaT`
`=(2)(3/2R)(T_A-T_D)`
`=(3R)((p_0V_0)/(4R)-(2p_0V_0)/(2R))`
`=-9/4p_0V_0`
`:.` `DeltaQ_(DA)=DeltaU_(DA)+DeltaW_(DA)`
`=-9/4p_0V_0-3/4p_0V_0`
`=-3p_0V_0`
Net work done is,
`W_(n et)=(pi/4+1-3/4)p_0V_0`
`=1.04p_0V_0`
and heat absorbed is
`Q_(ab)=DeltaQ_(+ve)`
`=(3/4+pi/4+5/2)p_0V_0=4.03p_0V_0`
Hence, efficiency of the cycle is
`eta=(W_(n et))/(Q_(ab))xx100`
`=(1.04p_0V_0)/(4.03p_0V_0)xx100`
`=25.8%`