Question

An ideal gas with ?=1.4 occupies 3.0L at 300 K and 100kPapressure 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 state A.

Part A

Find the net work done on the gas.

Express your answer using two significant figures.

Part B

Find the minimum volume reached.

Express your answer using two significant figures.

Answer 1

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:=100 kPa; V1:=3.0 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=100 kPa ...||... T1=300 K ...||... V1=3.0 Liters
P2=290.3 kPa ||... T2=395.9 K .||... V2=2.0 Liters
P3=290.3 kPa ||... T3=300 K ...||... V3=1.516 Liters

A check-point of work process by process:
W12 = -351.5 J
W23 = -140.6 J
W31 = +427.0 J

Thus our final result:
Wnet = -65.06 Joules

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

When operating a refrigerator or air conditioner designed to remove heat from a desired region, its coefficient of performance is 6.563.

When operating a heat pump designed to add heat from a desired region, its coefficient of performance is 7.563.