A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3. The propane un

A piston–cylinder assembly contains propane, initially at
27°C, 1 bar, and a volume of 0.2 m3. The propane undergoes
a process to a final pressure of 4 bar, during which the
pressure–volume relationship is pV^1.1 = constant. For the
propane, evaluate the work and heat transfer, each in kJ.
Kinetic and potential energy effects can be ignored.

The final volume V2 is found from
(1 bar)(0.2 m^3)^(1.1) = (4 bar)(V2)^(1.1)
(V2)^(1.1) = (1/4) (0.2 m^3)^(1.1)
V2 = 0.056716 m^3
W = integral of p dV
but p/p1 = (V1/V)^(1.1)
W = integral of (p1 (V1)^(1.1)) dV/(V^(1.1))
After integration you have
(p1 (V1)^(1.1))(1/1.1)(-1/V^(0.1)) to be evaluated at
V = 0.056716 m^3 and V = 0.2 m^3
W = 1*0.2^1.1*(1/1.1)(1/0.2-1/0.056716) bar*m^3
= -1.955 bar*m^3 = -195.5 kJ
I.e., 195.5 kJ of work are done ON the gas.

The final temperature is
T2 = T1 p2 V2 / (p1 V1)
= (300.15 K) (4)(0.056716/0.2) = 340.47 K
The change in internal energy would be Cv delta-T
if Cv were constant; at 25C it's 0.066 kJ/(mol K).
The number of moles of propane here are
(100000 Pa)(0.2 m^3)/
((8.314 J/molK)(300.15K))
= 8.0146 (moles), so the change in internal energy is about (8.0146 mol)(0.066 kJ/molK)(40.32K)
= about 21.3 kJ

Therefore, the heat transferred out was about
195.5 kJ - 21.3 kJ = 174 kJ.