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Unformatted text preview: SOLUTION HOMEWORK 13 1. 59. In the following 3 2 V C R = is the molar specific heat at constant volume, 5 2 p C R = is the molar specific heat at constant pressure, ∆ T is the temperature change, and n is the number of moles. The process 1 → 2 takes place at constant volume. (a) The heat added is ( 29 ( 29 ( 29 3 3 3 1.00mol 8.31J/mol K 600K 300K 3.74 10 J. 2 2 V Q nC T nR T = ∆ = ∆ = = (b) Since the process takes place at constant volume the work W done by the gas is zero, and the first law of thermodynamics tells us that the change in the internal energy is 3 int 3.74 10 J. E Q ∆ = = × (c) The work W done by the gas is zero. The process 2 → 3 is adiabatic. (d) The heat added is zero. (e) The change in the internal energy is ( 29 ( 29 ( 29 3 int 3 3 1.00mol 8.31J/mol K 455K 600K 1.81 10 J. 2 2 V E nC T nR T ∆ = ∆ = ∆ = =  (f) According to the first law of thermodynamics the work done by the gas is 3 int 1.81 10 J. W Q E = ∆ = + × The process 3 → 1 takes place at constant pressure. (g) The heat added is 3 5 5 (1.00 mol)(8.31J/mol K)(300K 455K) 3.22 10 J. 2 2 p Q nC T nR T = ∆ = ∆ = =  (h) The change in the internal energy is 3 int 3 3 (1.00mol)(8.31J/mol K)(300K 455K) 1.93 10 J. 2 2 V E nC T nR T ∆ = ∆ = ∆ = =  (i) According to the first law of thermodynamics the work done by the gas is 3 3 3 int 3.22 10 J 1.93 10 J 1.29 10 J. W Q E = ∆ =  × + × =  × (j) For the entire process the heat added is 3 3 3.74 10 J 0 3.22 10 J 520 J. Q = + = (k) The change in the internal energy is 3 3 3 int 3.74 10 J 1.81 10 J 1.93 10 J 0. E ∆ = z = (l) The work done by the gas is 3 3 0 1.81 10 J 1.29 10 J 520 J. W = + × × = (m) We first find the initial volume. Use the ideal gas law p 1 V 1 = nRT 1 to obtain 2 3 1 1 5 1 (1.00mol)(8.31J / mol K)(300K) 2.46 10 m . (1.013 10 Pa) nRT V p ⋅ = = = × × (n) Since 1 → 2 is a constant volume process V 2 = V 1 = 2.46 × 10 –2 m 3 . The pressure for state 2 is 5 2 2 2 3 2 (1.00 mol)(8.31J / mol K)(600K) 2.02 10 Pa . 2.46 10 m nRT p V ⋅ = = = × × This is approximately equal to 2.00 atm. (o) 3 → 1 is a constant pressure process. The volume for state 3 is 2 3 3 3 5 3 (1.00mol)(8.31J / mol K)(455K) 3.73 10 m . 1.013 10 Pa nRT V p ⋅ = = = × × (p) The pressure for state 3 is the same as the pressure for state 1: p 3 = p 1 = 1.013 × 10 5 Pa (1.00 atm) 2. (a) Let p i , V i , and T i represent the pressure, volume, and temperature of the initial state of the gas. Let p f , V f , and T f represent the pressure, volume, and temperature of the final state. Since the process is adiabatic i i f f pV p V γ γ = , so ( 29 1.4 4.3 L 1.2atm 13.6atm 14 atm....
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 Spring '07
 Giammanico
 Thermodynamics, Entropy

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