A 1.5 mol sample of ideal gas, with C_P, m = 2.75 R, initially at a pressure of
ID: 1063248 • Letter: A
Question
A 1.5 mol sample of ideal gas, with C_P, m = 2.75 R, initially at a pressure of 1.5 atm and a temperature of 298.15 K undergoes the following 3-step cycle: I rightarrow II Adiabatic reversible expansion to twice its initial volume II rightarrow III Reversible isothermal compression III rightarrow I Constant volume heating, returning it to its initial conditions Assume heat capacities are constant as a function of temperature. Determine P, V, and T at the end of each step. Determine q, w, delta U, and delta H for each step. Also determine w_total and comment on whether work is being done on the system or on the surroundings.Explanation / Answer
1. For the given cycle
a) determination of P, V and T
at I
P1 = 1.5 atm
V1 = 1.5 x 0.08205 x 298.15/1.5 = 24.46 L
T1 = 298.15 K
at II
P2 = 1.5 x 22.46/48.92 = 0.75 atm
V2 = 2V1 = 48.92 L
T2 = 0.75 x 48.92/0.08205 x 1.5 = 298.11 K
at III
P3 = 0.75 atm
V3 = 48.92 L
T3 = 298.11 K
b) From I --> II
dU = w = -nRln(V/V1) = -1.5 x 8.314 ln(2) = -8.64 J
dH = nCpdT = 1.5 x 2.75 x 8,314(298.11 - 298.15) = -1.372 J
q = 0
From II ---> III
w = -nRTln(V2/V1) = -1.5 x 8.314 x 298.11 ln(2) = -2576.934 J
dU = 0
dH = 0
q = -w = 2576.934 J
From III ---> I
w = 0
dU = dq = nCvdT = 1.5 x 1.75 x 8.314 x (198.15 - 198.11) = 0.873 J
dH = nCpdT = 1.5 x 2.75 x 8.314(298.15 - 298.11) = 1.382 J
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.