Solving Heat Engine Questions: 156 Moles of Monatomic Ideal Gas

[neibegafig]

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?

 

[Chestermiller]

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

 

[neibegafig]

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?

 

[Chestermiller]

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

 

[HalfLostAndSearching]

Hello, it seems like you have made good progress on this question so far. Let me provide some guidance on how to approach 2, 3, and 4.

2) To calculate the energy transferred into the gas from the hot reservoir, you can use the formula Q = nCvdT, where n is the number of moles, Cv is the specific heat at constant volume (12.47 J/K/mol for a monatomic ideal gas), and dT is the change in temperature. In this case, the temperature increases from 384 K to 1112 K, so dT = 728 K. Plugging in the values, we get Q = (156 mol)(12.47 J/K/mol)(728 K) = 1.45 x 10^6 J.

3) Similarly, to calculate the energy transferred out of the gas into the cold reservoir, we can use the same formula Q = nCvdT. In this case, the temperature decreases from 1112 K to 384 K, so dT = -728 K (negative because the temperature is decreasing). Plugging in the values, we get Q = (156 mol)(12.47 J/K/mol)(-728 K) = -1.45 x 10^6 J. Note that the negative sign indicates that energy is leaving the gas and being transferred to the cold reservoir.

4) To calculate the work done by the gas during the cycle, we can use the formula W = nRTln(Vf/Vi), where n is the number of moles, R is the gas constant (8.314 J/mol/K), T is the temperature at which the gas is expanding or compressing, and Vf and Vi are the final and initial volumes, respectively. In this case, we have two stages where the gas is expanding (steps 2 and 4) and two stages where it is compressing (steps 1 and 3). For step 2, T = 1112 K and Vf = 2.5 m^3, so W = (156 mol)(8.314 J/mol/K)(1112 K)ln(2.5/0.5) = 8.79 x 10^5 J. For step 4, T = 384 K and Vf = 0.5 m^3, so W = (156 mol)(8.314 J/mol/K)(