Homework #8 (chapter 24): Entropy and the Second Law of Thermodynamics

Exercise 4 :

a) W = -nRTln(V_f/V_i) = -16.9 kJ

b) ΔS = Q/T = (ΔE_int-W)/T = -W/T = -16.9 kJ / 410 K = 41.2 J/K

c) ΔS = 0 for any reversible process

Exercise 6:

ΔS = Q/T = mL/T = -0.001kg*333kJ*kg^-1/268K = -1.24 J/K

Exercise 22 :

a) W_cycle = W_bc-W_da = -p₁(V₁-V₀) + p₀(V₀-V₁) = -2.28 kJ

b) Q^ab = ΔE_int^ab = 3/2 n R (T_b-T_a) = 3/2 n R [(p₁V₀-p₀V₀)*n^-1*R^-1] = 3/2 (p₁V₀-p₀V₀) = 3/2 p₀V₀

Q^bc = ΔE_int^bc – W^bc = 3/2 n R (T_c-T_b) + p₁(V₁-V₀) = 3/2 p₁V₀ + p₁(V₁-V₀) = 11.4 kJ

Q^abc = Q^ab + Q^bc = 14.8 kJ

c) e = |W^cycle|/|Q^abc| = 2.28 kJ / 14.8 kJ = 0.15

d) e = 1 – T_L/T_H = 1 – T_a/T_c = 1 – p₀ * V₀ * p₁^-1 * V₁^-1 = 0.75

Exercise 29:

a) e = 1 – T_L/T_H = 1 - |Q_L|/|Q_H| è |Q_L| = |Q_H|*T_L/T_H

W = |Q_H| - |Q_L| = |Q_H| (1 - |Q_L|/|Q_H|) = |Q_H| (1 – T_L/T_H) = 568 J (1 – 258K/322K) = 113 J

b) W = |Q_L| (T_H-T_L)/T_L = 1230 J (322K-258K)/258K = 305 J

Problem 8:

V_a=V_b=1.22 m³; V_c=9.13 m³

p_b=10.4 atm = 10.4*1.01*10⁵ Pa=1.05*10⁶ Pa

Calculate p_c=p_a=(V_b/V_c)^γ p_b = 3.64*10⁴ Pa with γ = 1.67

Then calculate T from the equation of state of an ideal gas T = (pV)/(nR):

T_a=2670K, T_b=77100K, T_c=20000K

Calculate changes in internal energy for each segment:

ΔE_int^ab = n c_v ΔT^ab = 2 mol * 3/2 * 8.31 J*K^-1*mol^-1 * (77100-2670)K = 1.86*10⁶ J

ΔE_int^bc = n c_v ΔT^bc = 2 mol * 3/2 * 8.31 J*K^-1*mol^-1 * (20000-77100)K = -1.42*10⁶ J

ΔE_int^ca = - ΔE_int^ab - ΔE_int^bc = -1.86*10⁶ J +1.42*10⁶ J = -4.40*10⁵ J, since ΔE_int^cycle = 0

a) ΔE_int^ab = Q^ab + W^ab = Q^ab = 1.86*10⁶ J, which is the heat added to the gas

b) ΔE_int^ca = Q^ca + W^ac è Q^ca = ΔE_int^ca – W^ca = ΔE_int^ca – p_a(V_a-V_c) = -7.28*10⁵ J, which is the heat leaving the gas

c) W^cycle = W^ab + W^bc + W^ca = W^bc + W^ca = ΔE_int^bc + p_a(V_a-V_c) = -1.13*10⁶ J

d) e = |W^cycle|/|Q^ab| = 1.13*10⁶J / 1.86*10⁶J = 0.61