Question

Two moles of an ideal monoatomic 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 1

Correct Answer -
`W = (3)/(2)P_(1)V_(1) [1-((V_(1))/(V_(2)))^(2//3)]`;
`DeltaU = Q - W, T_(F) = (DeltaU)/(3R) +T_(i) =(DeltaU)/(3R) +(P_(1)V_(1))/(2R)`
`= [(P_(1)V_(1)^(5//3)V_(2)^(-2//3))/(2R)+(Q)/(3R)]`
(a) The `P-V` diagram for the complete process will be as follows:
Process `A rarr B` is adiabatic compression and
Process `B rarr C` is isochoric

(b) Total work done by the gas Process `A- B`
`W_("adiabatic") = (P_(i)V_(i) -P_(f)V_(f))/(gamma-1)`
`= (P_(1)V_(1)-P_(2)V_(2))/((5//3-1)) [gamma = 5//3` for monoatomic gas]
`= (P_(1)V_(1)-P_(1)((V_(1))/(V_(2)))^(gamma))/(2//3) =(3)/(2)P_(1)V_(1)[1-((V_(1))/(V_(2)))^(gamma-1)]`
`[[P_(1)V_(1)^(gamma)=P_(2)V_(2)^(gamma),],[:.P_(2)=P_(1)((V_(1))/(V_(2)))^(gamma),]]`
`W = (3)/(2)P_(1)V_(1) [1-((V_(1))/(V_(2)))^(2//3)]`
Process `B -C W_(BC) = 0 (V =` constant)
`:. W_("Total") = W_(AB) +W_(BC) +(3)/(2) P_(1)V_(1) [1-((V_(1))/(V_(2)))^(2//3)]`
(ii) Total change in internal energy
Process `A -B Q_(AB) = 0` (Process is adiabatic)
`:. DeltaU_(AB) =- W_(AB) = (3)/(2) P_(1)V_(1) [((V_(1))/(V_(2)))^(2//3) -1]`
Process `B-C w_(BC) = 0`
`:. DeltaU_(BC) = Q_(BC) = Q` (Given)
`:. DeltaU_("Total") = DeltaU_(AB) +DeltaU_(BC) =` ltbr `(3)/(2) P_(1)V_(1) [((V_(1))/(V_(2)))^(2//3) -1] +Q`
(iii) Final temperature of gas
`DeltaU_("total") =nC_(v) DeltaT =2 ((R)/(gamma-1)) (T_(C)-T_(A))`
`Q +(3)/(2)P_(1)V_(1) [((V_(1))/(V_(2)))^(2//3) -1]`
`= (2R)/((5//3-1)) (T_(C) -(P_(A)V_(A))/(2R))`
or `Q +(3)/(2)P_(1)V_(1) [((V_(1))/(V_(2)))^(2//3)-1] =3R (T_(C) -(P_(1)V_(1))/(2R))`
`:.T_(C) = (Q)/(3R) +(P_(1)V_(1))/(2R) ((V_(1))/(V_(2)))^(2//3) = T_("final")`