This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
This note was uploaded on 04/07/2008 for the course PHYSICS AN 1100, 2101 taught by Professor Giammanico during the Spring '07 term at LSU.
 Spring '07
 Giammanico

Click to edit the document details