Four moles of an ideal gas are taken through the cycle ABCA as shown below. For
ID: 2257586 • Letter: F
Question
Four moles of an ideal gas are taken through the cycle ABCA as shown below. For the questions below, assume that P units are in 100,000 Pa and V is in cubic meters. Use the temperature of the system at point A for all of these questions.
B. Find the total work done and identify whether it was done by the gas or on the gas
during the complete cycle. Use the temperature of the system at point B and explain
why this temperature is used. (10 points)
C. What is the value of Q for the entire problem?
D. Find the engine efficiency of the engine and the total work done by the engine.
E. Find the change in internal energy and the net thermal energy transferred or the net
work done and show the relationship between these quantities.
Four moles of an ideal gas are taken through the cycle ABCA as shown below. For the questions below, assume that P units are in 100,000 Pa and V is in cubic meters. Use the temperature of the system at point A for all of these questions. Find the total work done and identify whether it was done by the gas or on the gas during the complete cycle. Use the temperature of the system at point B and explain why this temperature is used. What is the value of Q for the entire problem? Find the engine efficiency of the engine and the total work done by the engine. Find the change in internal energy and the net thermal energy transferred or the net work done and show the relationship between these quantities.Explanation / Answer
The processes are as follows
For A-B
PaVa = PbVb = 0.01
So, it is an isothermal process
B-C is an isobaric process
C-A is an isochoric process
B. Work Done
W = Wab + Wbc + Wca
Wca = 0 since CA is isochoric in nature
Wbc = P(Vc - Vb) = 2 * 10^5(0.001 - 0.005) = -800 J
Wab = nRTbln(Vb/Va) = PbVbln(Vb/Va) as it is an ideal gas
Wab = 2 * 10^5 * 0.005 ln (1/5) = -1609.44 J
Total Work Done, W = -2409.44 J
So work is done on the gas. Also, we use Tb, since it is an isothermal process and Ta = Tb. Also Tb describes AB and BC together.
C. Q = Qab + Qbc + Qca
For AB, Qab = Wab since it is an isothermal process and Uab = 0.
So, Qab = -1609.44 J
Qbc = nCp(Tc - Tb)
Qca = nCv(Ta - Tc) = -nCv(Tc - Tb) ..... [Ta = Tb]
Qbc + Qca = nR(Tc - Tb) = PcVc - PbVb = 200 - 1000 = -800 J ....[Cp - Cv = R]
Q = 1609.44 - 800 = 2409.44 J
D. The engine is not delivering any work to the outside environment, hence it's efficiency = 0.
So total work done by the engine = 0
E. Net Change in Internal Energy = U = 0
Since, this is a cyclic process, net internal energy change = 0.
Internal Energy = Q + W
where, Q = Heat Added to System
W = Work Done on System
U = -2409.44 + 2409.44 = 0
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