A heat engine takes 0.350 mol of an ideal diatomic gas around the cycle shown in
ID: 1511280 • Letter: A
Question
A heat engine takes 0.350 mol of an ideal diatomic gas around the cycle shown in the pV diagram of (Figure 1) . Process 12 is at constant volume, process 23 is adiabatic, and process 31 is at a constant pressure of 1.00 atm. The value of for this gas is 1.40. The magnitude of the change in internal energy for each process is |U12|=2180J, |U23|=785J, and |U31|=1396J. a-Find the pressure at points 1, 2, and 3. b-Find the volume at points 1, 2, and 3. Express your answers in liters to three significant figures, separated by commas.
Explanation / Answer
= Cp/Cv = 1.4
Cp = 1.4 Cv
Also we know Cp-Cv = R = 8.32
Cv = 8.32/0.4 =20.8 J/(mol.K)
Cp = 29.12 J/(mol.K)
= 0.36 mol
======================1.
1.
From the figure Pressure at point 1 P1 = 1 atm= 2.
P2/T2 = P1/T1 since volume is constant
P2 = P1 (T2/T1) = P1*(600/300) = 2 P1 = 2 atm
P3 =P1 = 1 atm
P1=P3 = 1 atm and P2 = 2 atm
2.
v1 = RT/P = 0.36*8.32*300/ 101325 = 8.87e-3 m^3
v2 = v1 =8.87e-3 m^3
v3 = v1 (T3/T1) = 8.9e-3*(492/300) = 14.6e-3 m^3
v1 =v2 =8.9e-3 m^3 and v3 =14.6e-3 m^3
Answer for 3
Q = du + dw
dw = 0 for constant volume process and
du = Cv T = 0.36*20.8* (600-300) = 2246 J
Q = du = 2246 J and dw =0
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Answer for 5.
Q = du + dw
Q = CpT = 0.36*29.12 *(492-300) = 2010 J
dw = R*T = 0.36*8.32*(492-300) = 575 J
or
dw = P.dv = 101325*(14.6e-3-8.9e-3)= 577J
du = Q-dw = 2010 – 577 = 1433 J
Q = 2010 J, du = 1433 J and dw =577 J
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Answer for 4
From 4,
du = 2246
From 5,
du =1433
The difference is the du is the answer for 4
Q= 0, du = -dw = 813