Efficiency of an Ideal Gas Cycle: Monoatomic or Polyatomic?

In summary: But as I said before, I stopped there because of the temperature.In summary, the conversation discusses the use of an ideal gas in an engine operating on a specific cycle. It is determined that the gas is monoatomic and the efficiency of the engine is to be calculated. The efficiency is found by calculating the net work done by the gas, which is the area within the graph. The temperature changes in each part of the cycle need to be calculated in order to find the net work and efficiency. Different equations and concepts, such as the ideal gas law and adiabatic condition, are used to solve the problem.
  • #1
fluidistic
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Homework Statement


One mole of an ideal gas is used as the working substance of an engine that operates on the cycle shown in the figure.
B-C and D-A are reversible adiabatic processes.
1) Is the gas monoatomic or polyatomic?
2)What is the efficiency of the engine?

Homework Equations


None given. This is the exercise 48, page 530 of Fundamentals of physics, 5th edition.

The Attempt at a Solution


1)I used the formula [tex]PV^{\gamma}=\text{a constant}[/tex] which gave me that the gas is monoatomic.
2)I'm stuck here. I think that [tex]\varepsilon=\frac{W_{\text{out}}}{Q_{\text{in}}}[/tex]. I think that there's work done by the system in A-B while W is done on the system in C-D.
For A-B, [tex]\Delta E =\frac{3R\Delta T}{2}[/tex] but I didn't succeed in getting rid of the temperature...
I also found the work in A-B to be [tex]RT \ln 2[/tex], where T is the temperature in point A... I still have the same problem with the temperature.

Feel free to help me, I truly appreciate it.
 

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  • #2
In an actual engine, the work done on the gas is done by the engine itself: eg the output work of the A-B and B-C cycles in one cylinder is actually used to perform the work in the C-D and D-A cycles of another cylinder. So the net output work of the engine is what you want in order to measure efficiency.

The net work done by the gas is the area under A-B and B-C LESS the area under C-D and D-A. So, the net work is the area inside the graph.

AM
 
  • #3
This is a monoatomic ideal gas, só

[tex]PV=nRT [/tex]

where n is the number of moles, and the internal energy U is

[tex] U= \frac{3}{2}nRT [/tex].

Express all temperatures with T0 (the temperature at A).

Both A-->B and C-->D are at constant pressure so the work done by the gas is

[tex] W=P \Delta V [/tex].

From the first law of Thermodynamics, you know that

[tex] \Delta U = Q-W [/tex].

You can express all work done and heat absorbed by the gas with nRT0. The efficiency is the work done / heat absorbed during the whole cycle.

ehild
 
  • #4
You have to calculate Q and W for each part of the cycle.

For A-B, the Qin is:

[tex]Q_{in} = C_v\Delta T + P\Delta V = C_p\Delta T[/tex]

where [tex]\Delta T = P\Delta V/R[/tex]. Since [itex]C_p = 5R/2[/itex],

[tex]Q_{in} = \frac{5}{2}P\Delta V[/tex]

For B-C, there is no heat flow. The change in temperature from B-C is determined by the adiabatic condition:

[tex]TV^{\gamma - 1} = K[/tex]

Since it is adiabatic, the work done is the same as the change in internal energy: [itex]W = -\Delta U = -C_v\Delta T[/itex]

AM
 
  • #5
Thanks to both. I realized that the net work was done by the machine and it was the area between the curves. I also realize that if the sense was reversed, then the net work would have been done on the machine, also the area between the curves. However I don't think it helps me to calculate it, unless the graph shows what is the area between the curves.

Andrew Mason said:
You have to calculate Q and W for each part of the cycle.
Ok, that's what I had started but stopped because of the temperature I couldn't get rid of.

Andrew Mason said:
For A-B, the Qin is:

[tex]Q_{in} = C_v\Delta T + P\Delta V = C_p\Delta T[/tex]

where [tex]\Delta T = P\Delta V/R[/tex]. Since [itex]C_p = 5R/2[/itex],
Ok here I understand that [tex]Q_{in}=C_p\Delta T[/tex] since it's a constant pressure process. But why it is also worth [tex]C_v\Delta T + P\Delta V[/tex]? I've no idea here, I'd love to know the explanation.
Also, why does [tex]\Delta T = P\Delta V/R[/tex]? I'm also at a loss here.


Andrew Mason said:
[tex]Q_{in} = \frac{5}{2}P\Delta V[/tex]
Ok, I can follow you if I assume the anterior.

Andrew Mason said:
For B-C, there is no heat flow. The change in temperature from B-C is determined by the adiabatic condition:

[tex]TV^{\gamma - 1} = K[/tex]

Since it is adiabatic, the work done is the same as the change in internal energy: [itex]W = -\Delta U = -C_v\Delta T[/itex]

AM
I can follow you here. I'd have to work out [tex]\Delta T[/tex] I think, but I'll try.
 
  • #6
fluidistic said:
However I don't think it helps me to calculate it, unless the graph shows what is the area between the curves.
The area inside the path is (the area under B-C + area under C-D) minus (the area under D-A + area under A-B). Just calculate the area under each part of the curve and then do the arithmetic.

fluidistic said:
Ok, that's what I had started but stopped because of the temperature I couldn't get rid of.
You don't want to get rid of the temperature.

fluidistic said:
Ok here I understand that [tex]Q_{in}=C_p\Delta T[/tex] since it's a constant pressure process. But why it is also worth [tex]C_v\Delta T + P\Delta V[/tex]? I've no idea here, I'd love to know the explanation.
The first law: dQ = dU + dW. Change in internal energy is: [itex]\Delta U = nC_v\Delta T[/itex] and here n = 1. Since P is constant, [itex]W = P\Delta V[/itex]. So [itex]Q = C_v\Delta T + P\Delta V[/itex]. But by definition, [itex]Q = C_p\Delta T[/itex] for a constant pressure process. So [itex]C_p\Delta T = C_v\Delta T + P\Delta V[/itex].

fluidistic said:
Also, why does [tex]\Delta T = P\Delta V/R[/tex]? I'm also at a loss here.
It is just the ideal gas law: PV = nRT. Since P is constant from B-C and from D-A, applying PV = nRT at the B and at C (or at D and at A) and subtracting gives [itex]P\Delta V = nR\Delta T[/itex].


fluidistic said:
I can follow you here. I'd have to work out [tex]\Delta T[/tex] I think, but I'll try.
You have to work out the temperature changes to solve this.

AM
 
  • #7
Andrew Mason said:
The area inside the path is (the area under B-C + area under C-D) minus (the area under D-A + area under A-B). Just calculate the area under each part of the curve and then do the arithmetic.
Are you sure? There's no area under C-D, and I'm tempted to say that the area between the curves is the area under A-B + area under B-C - area under D-A.
I don't see how it could be "area under B-C + area under C-D) minus (the area under D-A + area under A-B)"

Andrew Mason said:
You don't want to get rid of the temperature.
Ok, thanks.
Andrew Mason said:
The first law: dQ = dU + dW. Change in internal energy is: [itex]\Delta U = nC_v\Delta T[/itex] and here n = 1. Since P is constant, [itex]W = P\Delta V[/itex]. So [itex]Q = C_v\Delta T + P\Delta V[/itex]. But by definition, [itex]Q = C_p\Delta T[/itex] for a constant pressure process. So [itex]C_p\Delta T = C_v\Delta T + P\Delta V[/itex].

It is just the ideal gas law: PV = nRT. Since P is constant from B-C and from D-A, applying PV = nRT at the B and at C (or at D and at A) and subtracting gives [itex]P\Delta V = nR\Delta T[/itex].

You have to work out the temperature changes to solve this.

AM
Ok, it makes perfectly sense to me now. Thanks you a million.
I'd appreciate if you could say a word about my thoughts for the area between the curves. That is, the net work done by the system.
 
  • #8
fluidistic said:
Are you sure? There's no area under C-D, and I'm tempted to say that the area between the curves is the area under A-B + area under B-C - area under D-A.
I don't see how it could be "area under B-C + area under C-D) minus (the area under D-A + area under A-B)"
The area within the path is what you have to quantify. Divide the path into 4 segments and calculate the area under each segment.

I may have confused you. I thought A was at the lower left corner when I wrote post #6. My post #2 had it right. The area within the path is [itex](A_{A-B} + A_{B-C}) - (|A_{C-D}| + |A_{D-A}|)[/itex].

I am not sure why you do not think there is an area under C-D:

[tex]A_{C-D} = \frac{P_0}{32}(8-16)V_0 = -P_0V_0/4[/tex]

AM
 
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  • #9
Andrew Mason said:
The area within the path is what you have to quantify. Divide the path into 4 segments and calculate the area under each segment.

I may have confused you. I thought A was at the lower left corner when I wrote post #6. My post #2 had it right. The area within the path is [itex](A_{A-B} + A_{B-C}) - (|A_{C-D}| + |A_{D-A}|)[/itex].

I am not sure why you do not think there is an area under C-D:

[tex]A_{C-D} = \frac{P_0}{32}(8-16)V_0 = -P_0V_0/4[/tex]

AM
Thank you very much once again, you've helped me a lot. I'm sorry for the area under C-D, I guess I was so tired I was thinking about another problem involving the Diesel cycle...
 

FAQ: Efficiency of an Ideal Gas Cycle: Monoatomic or Polyatomic?

What is a cycle in thermodynamics?

A cycle in thermodynamics refers to a series of thermodynamic processes that occur in a closed system and return the system to its original state. This allows for the transfer of energy and work within the system.

What is the purpose of a cycle in thermodynamics?

The purpose of a cycle in thermodynamics is to allow for the conversion of energy from one form to another. This is essential for the functioning of many devices, such as engines and refrigerators.

What are the four types of cycles in thermodynamics?

The four types of cycles in thermodynamics are Carnot cycle, Rankine cycle, Otto cycle, and Brayton cycle. Each cycle has its own unique characteristics and is used for specific applications.

What is the difference between an open and closed cycle in thermodynamics?

An open cycle in thermodynamics allows for the continuous flow of matter into and out of the system, while a closed cycle does not. Closed cycles are typically more efficient as they do not require the constant intake of new matter.

What is the efficiency of a cycle in thermodynamics?

The efficiency of a cycle in thermodynamics is a measure of how much of the input energy is converted into useful work. It is typically expressed as a percentage and varies depending on the type of cycle and the conditions under which it operates.

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