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ID: 1898544 • Letter: L
Question
Life saver rating for a detailed answer. I will be very grateful. Thank you so much.
Four moles of an ideal nonatomic gas, with a ratio of heat capacities gamma = 1.67, initially have a pressure p1 = 2.00 atm. a volume V1 = 45.0 L and a temperature T1. Then the gas undergoes the following three-step cyclic process, as shown in the p-V diagram below. Adiabatic expansion from the temperature T1, pressure p1, and volume V1 to a temperature T2, volume V2 = 65.0 L and pressure p2. Increase in pressure at constant volume to a pressure p3 and a temperature T3. Isothermal compression to the original starting pressure, volume, and temperature. What is T1? What are T2 and p2? What are T3 and p3? From the figure, is the net work done by the gas during the cycle positive or negative? What is the work done by the gas during each of the three steps of the cycle? What is the net work done by the gas during the entire cycle? How much heat is absorbed by the gas during each of the three steps of the cycle? What is the net heat absorbed by the gas during the entire cycle? Why should this last result be exactly the same as the last result in part (e)?Explanation / Answer
P1 = 2 atm = 2*(1.01325*10^5) Pa
a) P1*V1 = nRT1 where R = universal gas constant = 8.314 J/K-mol
2*(1.01325*10^5)*(45*10^-3) = 4*8.314*T1
T1 = 274.2 K
b) For adiabatic process P2/P1 = (V1/V2)^
P2/2 = (45/65)^1.67
P2 = 1.082 atm
For adiabatic process, T2/T1 = (P2/P1)^(1-1/)
T2/274.2 = (1.082/2)^(1-1/1.67)
T2 = 214.3 K
c) For constant temperature process 3-1, T3 = T1
So, T3 = 274.2 K
For constant volume process 2-3: P2/T2 = P3/T3
1.082/214.3 = P3/274.2
P3 = 1.384 atm
d) Since the cycle is clockwise, work done is negative.
e) Work done in adiabatic process, W12 = nR(T2-T1)/(1-) = 4*8.314*(214.3-274.2)/(1-1.67) = 2973.2 J
Work done in constant volume process W23 = 0
Work done in isothermal process W31 = P3*V3 ln (V1/V3) = 1.384*(1.01325*10^5)*(65*10^-3) ln(45/65) = -3351.9 J
Work done during entire cycle = 2973.2 + 0 + (-3351.9) = -378.7
f) Heat transfer in adiabatic process 1-2, Q12 = 0
Heat transfer in constant volume process 2-3 is Q23 = W23 + nR(T3-T2)/(-1) = 0 + 4*8.314*(274.2 - 214.3)/(1.67-1) = 2973.2 J
Heat transfer in isothermal process Q31 = W31 = -3351.9 J
Net heat transfer in cycle = Q12 + Q23 + Q31 = -378.7 J
Results of parts e and f are same because for a cycic process, net heat transfer = net work done. (No net change in internal energy)
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