Solving Heat Engine Questions: 156 Moles of Monatomic Ideal Gas

AI Thread Summary
The discussion revolves around solving a heat engine problem involving 156 moles of a monatomic ideal gas undergoing a Stirling cycle. Key calculations include determining the initial pressure, energy transferred from the hot reservoir, energy transferred to the cold reservoir, work done during the cycle, and the efficiencies of the Stirling cycle and Carnot cycle. The first law of thermodynamics is emphasized, particularly in understanding the relationship between heat (Q) and work (W) during constant temperature processes. It is clarified that for an ideal gas at constant temperature, the change in internal energy is zero, leading to the conclusion that Q equals W in those steps. The conversation highlights the importance of applying thermodynamic principles correctly to solve the problem.
neibegafig
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Hi, I am hoping someone may be able to help me as i am quite stuck in this question.

Suppose that 156 moles of a monatomic ideal gas is initially contained in a piston with a volume of 0.5 m3 at a temperature of 384 K. The piston is connected to a hot reservoir with a temperature of 1112 K and a cold reservoir with a temperature of 384 K. The gas undergoes a quasi-static Stirling cycle with the following steps:

1.) The temperature of the gas is increased to 1112 K while maintaining a constant volume.
2.) The volume of the gas is increased to 2.5 m3 while maintaining a constant temperature.
3.) The temperature of the gas is decreased to 384 K while maintaining a constant volume.
4.) The volume of the gas is decreased to 0.5 m3 while maintaining a constant temperature.

It may help you to recall that CV = 12.47 J/K/mole for a monatomic ideal gas, Avagadros number (6.022E23) times the number of moles of the gas.

1) What is the pressure of the gas under its initial conditions?

2) How much energy is transferred into the gas from the hot reservoir?

3) How much energy is transferred out of the gas into the cold reservoir?

4) How much work is done by the gas during this cycle?

5) What is the efficiency of this Stirling cycle?

6) What is the maximum (Carnot) efficiency of a heat engine running between these two reservoirs?



for 1, 5, and 6 i Got just fine, 1 was Pi = nNkTi/Vi = 995604.48 Pa, 5 was stirling energy efficiency of ((Th-Tc)/Th) * NkLn5/(NkLn5 + Cv(Th-Tc)/Th) = 0.4065, and 6 was 1 - (Tc/Th) = 0.6546

I have some idea on getting 2,3, and 4 but I am stuck. I know that Q = nk(dT) + W where n is the number of moles, k is 1.381E-23, and dT is change in Temperature, I am stuck though as to what W is exactly and what the formula for it would be in the two situations (2, and 3). Would someone be able to help?
 
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To answer questions 2 - 4, you need to apply the 1st law. You also need to assume that the expansion in step 2 and the compression in step 4 occurs reversibly. In steps 2 and 4, if you maintain constant temperature, what does that tell you regarding the changes in internal energy? From this information about the changes in internal energy for these process steps, what does that tell you about the relationship between the work done on the surroundings W and the heat added to the system Q in these process steps?

Chet
 
Chestermiller said:
To answer questions 2 - 4, you need to apply the 1st law. You also need to assume that the expansion in step 2 and the compression in step 4 occurs reversibly. In steps 2 and 4, if you maintain constant temperature, what does that tell you regarding the changes in internal energy? From this information about the changes in internal energy for these process steps, what does that tell you about the relationship between the work done on the surroundings W and the heat added to the system Q in these process steps?

Chet

If i am getting this correctly then the change in internal energy is based on the pressure times the changes in volume (integral of pdv). If i got this right, the relationship between W and Q is that W is that integral of pdv (the change in volume from 0.5 to 2.5 from step 2) for energy going into gas? Or am i missing one more component?
 
neibegafig said:
If i am getting this correctly then the change in internal energy is based on the pressure times the changes in volume (integral of pdv). If i got this right, the relationship between W and Q is that W is that integral of pdv (the change in volume from 0.5 to 2.5 from step 2) for energy going into gas? Or am i missing one more component?

Yes. You're missing an important component. For an ideal gas, the internal energy is a function only of temperature, and, if the temperature is held constant, the change in internal energy is zero. That means the Q = W.

Chet
 
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