Please show work! State Variable Table must have values for Pressure, Volume, Te
ID: 1596511 • Letter: P
Question
Please show work! State Variable Table must have values for Pressure, Volume, Temperature and Internal Energy at each point. Thermodynamic Variable Table must have values for Change in Internal Energy, Work, and Change in Heat for each process (1->2, 2->3, 3->4, 4->1) Thank you!
A 0.040 mole sample of an ideal monatomic gas evolves through the process 1 rightarrow 2 rightarrow 3 rightarrow 4 rightarrow 1 represented on the pV diagram shown in the figure. Make State Variable table and a Thermodynamic Variable Table. Draw thick arrows on your diagram indicating the heat transfers that occur during the cycle. show that net change in internal energy per cycle is zero. Also, show that the net heat transferred equals the work done each cycle. Determine the actual efficiency of the heat engine based on this cycle. Compute the actual coercion of performance of a refrigerator based on the cycle. 4 rightarrow 3 rightarrow 2 rightarrow 1 rightarrow 4.. Compute the actual coefficient of performance of a heat pump based on the cycle 4 rightarrow 3 rightarrow 2 rightarrow 1 rightarrow 4 Does the reverse process make a better heat pump or a better fridge? Why?Explanation / Answer
number of moles, n = 0.04 mol
at 1
P1 = 8 atm = 8*1.10*10^5 Pa
V1 = 100 cm^3 = 0.1*10^-3 m^3
using PV = nRT
T1 = 243.080 K
for a monoatomic gas
gamma = 1 + 2/3 = 5/3
and Cv = R/(gamma - 1) = 12.465
so, IE1 = nCvT1 = 121.1996 J
step 1->2
isobaric process
P2 = 8 atm = 1.01*10^5*8 Pa
V2 = 0.4*10^-3 m^3
as P is constant
V/T = constant
V1/T1 = V2/T2
T2 = 972.32 K
IE2 = nCvT2 = nCv4T1 = 4E1 = 484.7984 J
Change in internal energy = IE2 - IE1 = 363.5988 J
Work Done = PdV = P2(V2 - V1) = 242.4 J [ from conservation of energy]
Heat input to the system = change in internal intergy - work done = 121.1988 J [ from conservation of energy]
step 2-> 3
T3 = P3V3/nR = 759.626 K
IE3 =nCvT3= 378.75 J
then change in internal energy = IE3 - IE2 = -106.048 J
Work done = area under curve = [5*100 + 0.5*100*3]*1.01*10^-1 = 65.65 J
Heat input to the system
Heat input to the system = change in internal intergy - work done = -171.698 J [ from conservation of energy]
step 3-> 4
isobaric process
T4 = P4V4/nR = 151.92 K
IE4 = nCvT4 = 75.75 J
change in IE = IE4 - IE3 = -303 J
work done = area under curve = -5*400*1.01*10^-1 = -202 J
eat input to the system = change in internal intergy - work done = -101 J [ from conservation of energy]
step 4->1
work done = 0 (isochoric process)
Ichange in IE = IE1 - IE4 = 45.4496 J
Heat provided = change in IE = 45.4496 J
From the graph, and the calculations abpve the state variable table can be constructed as below:
step P (atm) V(cm^3) T(k) IE(J)
1 8 100 243.080 121.1996
2 8 400 972.32 484.7984
3 5 500 759.626 378.75
4 5 100 151.92 75.75
Thermodynamic Variable table
step dW(J) dH(J) dIE(J)
1->2 242.2 121.1988 363.5988
2->3 65.65 -171.698 -106.048
3->4 -202 -101 -303
4->1 0 45.4496 45.4496
total 105.85 -106.04 0
b) Efficiency = work/heat = 0.9982