Question

One mol of an ideal gas described by Ti = 300 K, Vi = 2.46 L, and Pi = 10 atm undergoes isobraic expansion (constant pressure) until V = 5 L. Next, it undergoes a reversble isothermal (constant temperature) expansion until P = 1.00 atm followed by isobaric compression to T = 300 K. Finally, it is restored to its original state by reversible, isothermal compression.

a) Depict this cyclic process on a P-V diagram.

b) Calculate w, q, deltaU, and deltaH for each step and for the total cyclic process. (Cv,m=3/2R)

Answer 1

a)

P1 = 10 atm, V1 = 2.46 L and T1 = 300 K

After Isobaric Expansion,

P2 = 10 atm, V2= 5 L

During Isothermal Expansion, using PV = nRT and given P3 = 1 atm:

P2V2 = P3V3

10 * 5 = 1 * V3

V3 = 50 L

After Isothermal Expansion,

P3 = 1 atm, V3 = 50 L

After Isobaric Compression,

P4 = 1 atm, V4 = ?

During Isothermal Compression, using PV = nRT and given P4 = 1 atm; V1 = 2.46 L; P1 = 10 atm:

P4V4 = P1V1

1 * V4 = 10 * 2.46

V4 = 24.6 L

P-V Diagram:

b)

Isobaric Expansion:

W₁₂ = P1 (V2 - V1) = 10 (5 - 2.46) = 25.4 atm-L

Since 1 atm-L = 101.33 J,

W₁₂ = 25.4 * 101.33 = 2573.78 J

DeltaU₁₂ = C_v,m (T2 - T1) = (3/2)nR[(P2V2 - P1V1) / nR] = 1.5 (P2V2 - P1V1) = 1.5 (10*5 - 10*2.46) = 3860.67 J

DeltaH₁₂ = Q₁₂ = W₁₂ + DeltaU₁₂ = 2573.78 + 3860.67 = 6434.45 J

Isothermal Expansion:

W₂₃ = PdV = RT2 ln(V3/V2)

Since T2 = (V1/V2) * T1 (using PV = nRT), T2 = (2.46/5) * 300 = 147.6 K

W₂₃ = 8.314 * 147.6 * ln(50/5) = 2822.44 J

U₂₃ = 0, since the process is isothermal

Q₂₃ = W₂₃ + U₂₃ = 2822.44 + 0 = 2822.44 J

Similaly, W, U and Q can be calculated for Isobaric compression and Isothermal compression.

In cyclic process, for complete cycle, U = 0, Q = - W = - (W₁₂ + W₂₃ + W₃₄ + W₄₁)