Question

An ideal gas with γ=1.4 occupies 4.5 L at 300 K and 110 kPa pressure and is compressed adiabatically until its volume is 2.0 L. It's then cooled at constant pressure until it reaches 300 K, then allowed to expand isothermally back to stateA.

Find the net work done on the gas.

Find the minimum volume reached.

Answer 1

Total energy is a state function, and internal energy is a state function. "State function" means independent of path.

BUT, heat and work are definably NOT state functions. This cycle trades heat and work in order that it operates.

If you draw the cycle on a P-V diagram, the area enclosed is the net-work done per cycle.

If the cycle goes clockwise, it is a power cycle (heat engine) with positive work done.

If the cycle goes counter-clockwise, it is a refrigeration cycle (or heat pump) with negative work done.
State 1: Pre-adiabatic compression
State 2: Post-adiabatic compression, pre-isochoric cooling/compression
State 3: Post-isochoric cooling/compression, pre-isothermal heat absorption/expansion.

What conditions are known?
State 1: P1, T1, V1
State 2: V2
State 3: T3

First goal: complete P, V, and T for all states.

Use adiabatic condition to get pressure at state 2.
P1*V1^k = P2*V2^k

I call it k, not gamma.

Solve for P2:
P2 = P1*(V1/V2)^k

Use ideal gas law to get T2:
P1*V1 = n*R*T1
P2*V2 = n*R*T2

T2 = (P2*V2)/(n*R)
n*R = P1*V1/T1

Thus:
T2 = T1*(P2*V2)/(P1*V1)

We know P3 because it is equal to P2. P3 = P1*(V1/V2)^k
We also know T3, because it is given.

Solve for V3 with ideal gas law:
V3 = n*R*T3/P3

Thus:
V3 = V1*P1*T3/(P3*T1)

Thus:
V3 = V1*(V2/V1)^k * T3/T1

Summary:
State 1: P1, T1, V1 is given
State 2: V2 given, P2 = P1*(V1/V2)^k, T2 = T1*(V1/V2)^(k-1)
State 3: T3 given, P3 = P1*(V1/V2)^k, V3 = V1*(V2/V1)^k * T3/T1

Now to find work associated with each process:
W12 = (P2*V2 - P1*V1)/(1 - k)
W23 = P2*(V3 - V2)
W31 = P1*V1*ln(V1/V3)

Make substitutions:
W12 = (P1*(V1/V2)^k*V2 - P1*V1)/(1 - k)
W23 = P1*(V1/V2)^k*(V1*(V2/V1)^k * T3/T1 - V2)
W31 = P1*V1*ln(T1/((V2/V1)^k)/T3)

Simplify:
W12 = P1*(V2*(V1/V2)^k - V1)/(1 - k)
W23 = P1*(V1/V2)^k*(V1*(V2/V1)^k * T3/T1 - V2)
W31 = P1*V1*ln(T1/((V2/V1)^k)/T3)

Total:
Wnet = W12 + W23 + W31

Data:
P1:=110 kPa; V1:=4.5 Liters; T1:=300 K;
V2:=2.0 Liters; T3:=300 K;
k:=1.4;

Unit note: kPa*Liters = Joules

Complete state-by-state table:
P1=110 kPa ...||... T1=300 K ...||... V1=4.5 Liters
P2=290.3 kPa ||... T2=395.9 K .||... V2=2.0 Liters
P3=290.3 kPa ||... T3=300 K ...||... V3=1.516 Liters

calculate work for each state and add them to have the answer.

The negative sign indicates that work is done on the gas, thus having it operate as a refrigeration cycle.

minimum volume reached = V3 =1.516 Liters