Q.370. First law of thermodynamics - work done problem.

Question 370.
A cylinder/piston setup contains air at 100kPa and 20° C and has a volume of  0.3 m^3. The air is compressed to 800 kPa in a reversible process in which PV^1.2 is held constant, after which it is expanded back to 100 kPa in a reversible adiabatic process.
Calculate the final temperature and the net work.

Answer 170.
Given equation PV^1.2
=> P1V1^1.2 = P2V2^1.2 ... ( 1 )

According to Ideal Gas Law equation,
PV = nRT
=> P1V1/T1 = P2V2/T2
=> P1V1T2 = P2V2T1
=> (P1V1T2)^1.2 = (P2V2T1)^1.2 ... ( 2 )

From ( 1 ) and ( 2 ),
P1^0.2 * T2^1.2 = P2^0.2 * T1^1.2
=> T2 = T1 * (P2/P1)^0.2/(1.2)
=> Final temperature,
T2 = (20 + 273) * (8)^(1/6) = 414.4 K = 141.4° C

Work done, W
= RT1 / (n - 1) * [1 - (p2/p1)^(n-1)/n]
= (8.314) * (293) * [1 - (8)^1/6)
= - 1009 J

Work done in adiabatic expansion, W'
= RT2 / (γ - 1) * [1 - (p2/p1)^(γ-1)/γ] ... γ = 1.41 for air
= (8.314) * (414.4) * [1 - (1/8)^(0.41/1.41)]
= 1563 J

Net work done
= - 1009 + 1563 J
= 554 J.

Link to YA!