Two moles of an ideal monatomic gas, initially at pressure `p_1` and volume `V_1`, undergo an adiabatic compression until its volume is `V_2`. Then the gas is given heat Q at constant volume `V_2`.
(i) Sketch the complete process on a p-V diagram.
(b) Find the total work done by the gas, the total change in its internal energy and the final temperature of the gas. [Give your answer in terms of `p_1,V_1,V_2, Q and R`]

Answer

A, B, C

Explanation

`n=no. of moles=2,`
(A) The complete process is shown on P-V diagram in the
figure.

(B)(i) Total work done
`W=W_(AB)+W_(BC)=(P_1V_1-P_2V_2)/(gamma-1)+0`
`[:' W_(BC)=PDeltaV=Pxx0=0]`
According to Poisson's law, `P_2=P_1((V_1)/(V_2))^(gamma)`
`:. W=1/(gamma-1)[P_1V_1-P_1((V_1)/(V_2))^(gamma)V_2]`
`=1/(gamma-1)[P_1V_1-P_1V_2(V_1)/(V_2)((V_1)/(V_2))^(gamma-1)]`
For monoatomic gas,
`gamma=1+2/3=5/3,`
`w=3/2[P_1V_1-P_1V_1((V_1)/(V_2))^(2//3)]=3/2P_1V_1[1-((V_1)/(V_2))^(2//3)]`
(ii) `DeltaU=DeltaU_(AB)+DeltaU_(BC)=Q-W`
`=Q-3/2P_1V_1[1-((V_1)/(V_2))^(2//3)]`
`[according to first law of thermodynamics]`
`[(BtoC,Q=DeltaU_(BC)+0),(AtoB, Q=DeltaU_(AB)+W)]`
(iii) For process `BC:DeltaU_(BC)=nC_vDeltaT=Q`
`[:' W_(BC)=0]`
For monoatomic gas `C_v=R/(gamma-1)=3/2R,`
`:. DeltaU_(BC)=Q=2xx(3R)/2DeltaT`
Hence `DeltaT=Q/(3R)`
According to Poission's Law:
For the process AB, `T_AV_B^(gamma-1)=T_BV_B^(gamma-1)`
or `T_B=T_A((V_1)/(V_2))^(gamma-1)=(P_1V_1)/(nR)((V_1)/(V_2))^(gamma-1)`
`:. T_B=(P_1)/(2R).V_1^(gamma).V_2^(1-gamma)=(P_1V_1^(5//3)V_2^(-2//3))/(2R)`
Hence, `T_C=T_B+DeltaT=(P_1V_1^(5//3)V_2^(-2//3))/(2R)+Q/(3R)`