ch19-p063 - 63. In the following CV = 3 R is the molar...

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3 int 3.74 10 J. EQ Δ= = × (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 () ( ) ( ) 3 int 33 1.00mol 8.31J/mol K 455K 600K 1.81 10 J. 22 V En C Tn R T Δ = = − × (f) According to the first law of thermodynamics the work done by the gas is 3 int 1.81 10 J. WQ E =− + × The process 3 1 takes place at constant pressure. (g) The heat added is 3 55 (1.00 mol)(8.31J/mol K)(300K 455K) 3.22 10 J. p Qn CT n RT = Δ = = −× (h) The change in the internal energy is 3 int (1.00mol)(8.31J/mol K)(300K 455K) 1.93 10 J. V C T n R T = × 63. In the following 3 2 V CR = is the molar specific heat at constant volume, 5 2 p = is the molar specific heat at constant pressure, Δ 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 ( ) ( ) 3 1.00mol 8.31J/mol K 600K 300K 3.74 10 J. V C T n = Δ = (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
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33 0 1.81 10 J 1.29 10 J 520 J. W =+ × × = (m) We first find the initial volume. Use the ideal gas law
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ch19-p063 - 63. In the following CV = 3 R is the molar...

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