A sample of 1 mol of perfect gas atoms (or, in reality, Ne), for which C_V, m =

Question

A sample of 1 mol of perfect gas atoms (or, in reality, Ne), for which C_V, m = 3/2 R, is taken through the cycle shown below. The transition from state 3 to sum 1 happens isothermally and reversibly. Show and explain your work. What are the temperatures at points 1, 2, and 3? Calculate q, w, Delta U and Delta H for each step (1 to 2, 2 to 3, and 3 to 1) as well as for the overall cycle. If a numerical answer cannot be calculated, then indicate whether the quantity would be positive (+), negative (-), zero, or unknowable (?).

Explanation / Answer

cycle of an ideal gas
Step1: Isobaric process (P1,V1,T1) to (P1,V2,T2)
Step2: Isochoric process (P1,V2,T2) to (P2,V2,T3)
Step3: Isothermal process (P2,V2,T3) to (P1,V1,T1): T3=T1

(a) Temperatures, T(1), T(2), T(3)
T3 = T1

(b)
Step1: Isobaric process (P1,V1,T1) to (P1,V2,T2)
W1 = -P*(V2-V1) = R*(T1-T2)
dU1 = Cv*(T2-T1) = 3*R/2*(T2-T1)
Q1 = dU-W = 3*R/2*(T2-T1)-R*(T1-T2)= 5/2*R*(T2-T1)
dH1 = dU+d(P*V) = dU+P*dV =3*R/2*(T2-T1)+P1*(V2-V1) = 3*R/2*(T2-T1)+R*(T2-T1)=5/2*R*(T2-T1)

Step2: Isochoric process (P1,V2,T2) to (P2,V2,T3)
W2 = 0
dU2 = Cv*(T3-T2)=3*R/2*(T1-T2)
Q2 = dU = 3*R/2*(T1-T2)
dH2 = dU+d(P*V)=3*R/2*(T1-T2)+V2*(P2-P1)

Step3: Isothermal process (P2,V2,T3) to (P1,V1,T1): T3 = T2
W3 = -R*T1*Ln(V1/V2)
dU3 = 0
Q3 = -W = R*T1*Ln(V1/V2)
dH3 = dU + d(P*V) = R*(T1-T2)

Overall cycle:
Wcycle = W1+W2+W3 = R*(T1-T2)-R*T1*Ln(V1/V2)
dU = 0
Q = Q1+Q2+Q3 = 5/2*R*(T2-T1)+ 3*R/2*(T1-T2)+ R*T1*Ln(V1/V2) = R*(T2-T1)+ R*T1*Ln(V1/V2)
dH = V2*(P2-P1)

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