An ideal gas with =1.4 occupies 4.0 L at 300 K and 130 kPa pressure and is heate

Question

An ideal gas with  =1.4 occupies 4.0 L at 300 K and  130 kPa pressure and is heated at constant volume until its pressure has doubled. It's then compressed adiabatically until its volume is one-fourth its original value, then cooled at constant volume to 300 K , and finally allowed to expand isothermally to its original state.

Part A

Find the net work done on the gas.

Express your answer using two significant figures.

An ideal gas with  =1.4 occupies 4.0 L at 300 K and  130 kPa pressure and is heated at constant volume until its pressure has doubled. It's then compressed adiabatically until its volume is one-fourth its original value, then cooled at constant volume to 300 K , and finally allowed to expand isothermally to its original state.

Part A

Find the net work done on the gas.

Express your answer using two significant figures.

Explanation / Answer

gama = k

P1*V1^k = P2*V2^k

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

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)

P3 is equal to P2. P3 = P1*(V1/V2)^k
T3 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)

Total:
Wnet = W12 + W23 + W31

Data:
P1:=130 kPa; V1:=4.0 Liters; T1:=300 K;
V2:=2.0 Liters; T3:=300 K;
k:=1.4;

P1=130 kPa ...||... T1=300 K ...||... V1=4.0 Liters
P2=343 kPa ||... T2=395.9 K .||... V2=2.0 Liters
P3=343 kPa ||... T3=300 K ...||... V3=1.516 Liters

W12 = -415 J

W23 = -166 J

W31 = +504.51

Wnet = - 76.49 J

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