What Is the Efficiency of a Monoatomic Ideal Gas Cycle?

[diegzumillo]

Hello again, all! I have another basic thermodynamics question :P This one came from a 2008 admission test. Seems simple enough, but as usual, my answer doesn't match =P

Homework Statement

A monoatomic ideal gas performs, reversibly, the cycle shown in the diagram attached in this post. The values of P_0 and V_0 are, respectively, 1\times 10^5 Pa and 100cm^3. The area of the interior of the cycle is 15J. What is the efficiency of this cycle?

Homework Equations

e=\frac{W}{Q_h}
e is the efficiency, W is work done by the system and Q_h is the energy absorbed by the system (heat).
This is probably the only equation relevant here..

The Attempt at a Solution

The problem seemed straigh forward: The area enclosed by the cycle is the work done. The area under the curve ab is the Q_h, right? If so, calculating this is trivial and results in an efficiency of, approximately, 0,43.
However, this is not right. The correct answer should be 0,136.

 

[diegzumillo]

No ideas? :P

 

[Redbelly98]

The Attempt at a Solution

The problem seemed straigh forward: The area enclosed by the cycle is the work done. The area under the curve ab is the Q_h, right?

Ah, no. That area would be the work done from a to b.

Q is ∫T dS, and W is ∫P dV.

The easiest way to get Q, for any single path of this cycle, is to use

ΔU = Q - W​

since ΔU and W are fairly straightforward to calculate.

 

[abhikesbhat]

Redbelly you are smart!

 

[diegzumillo]

Thanks Redbelly :) I got carried away by the relation W=Q_h-Q_c and assumed something wrong :P

Can I classify this system as cyclic? If so, isn't the internal energy variation supposed to be zero?

 

[Redbelly98]

Redbelly you are smart!

Gosh, thank you.

Thanks Redbelly :) I got carried away by the relation W=Q_h-Q_c and assumed something wrong :P

Can I classify this system as cyclic? If so, isn't the internal energy variation supposed to be zero?

Yes, the entire process is cyclic, and ΔU is zero for the entire cycle.

But ... to calculate Q_h, you'll need to consider each individual subpath, and whether heat is flowing into or out of the system for that path. Heat in contributes to Q_h, while heat out does not.

Since ΔU is not necessarily zero for each subpath, it needs to be considered.

 

[diegzumillo]

Hello again! I let go of this problems for a few days. But today I took another look at it, and I think I have a solution :)

We'll use this equation for heat
Q=C\Delta T
where this C depends on the path.
We can start by calculating the heat for the path BC and CA, we'll call them Q_{bc} and Q_{ca}, respectively. We cannot calculate directly Q_{ab} because we only know c_p=3/2 R, for constant pressure, and c_v=5/2 R, for constant volume. So we have
Q_{bc}=\frac{c_v}{R}(P_c V_c - P_b V_b)=-45J
Q_{ca}=\frac{c_p}{R}(P_a V_a - P_c V_c)=-50J
(I'm calling P_i and V_i for better understanding)

We can see that both of those values represent heat leaving the system. Now we'll use the internal energy relations to find Q_{ab}. The internal energy relations for each path are:
\Delta U_{bc}=Q_{bc}-W_{bc} (Yes, W_{bc} is zero)
\Delta U_{ca}=Q_{ca}-W_{ca}
Since we know that
\Delta U=\Delta U_{ab}+\Delta U_{bc}+\Delta U_{ca}=0
and
\Delta U_{ab}=A_{ab}-W_{ab}
We have
Q_{ab}=W_{ab}-(Q_{bc}-W_{bc}+Q_{ca}-W_{ca})
Q_{ab}=W-(Q_{bc}+Q_{ca})=110J
Wich is our heat transferred into the system! (Q_h)

In possession of these values, we can now calculate the efficiency of this cycle:
e=\frac{W}{Q_h}=13,6%

=D

(Btw, for some strange reason, I can't visualize correctly the formulas. So there may be some errors. I'll correct them as soon as I'm able to read what I wrote :P)

 

[Redbelly98]

Latex equations have been having problems for several days now.

Much of what you wrote can be done without Latex. You can get the Greek letter Δ here:
https://www.physicsforums.com/blog.php?b=347

Also:

[noparse]_a[/noparse] for subscript _a
[noparse]²[/noparse] for superscript ²

 

[diegzumillo]

Yep.. just saw the anouncement. What a bummer!

Since there is no estimate for the return, I'll post the solution again! (But I'll keep that one there... it took me a while to do! :P)

 

[diegzumillo]

The solution again. Simplified notation version :D

We'll use this equation for heat
Q=CΔT
where this C depends on the path.
We can start by calculating the heat for the path BC and CA, we'll call them Q_bc and Q_ca, respectively. We cannot calculate directly Q_ab because we only know c_p=3/2 R, for constant pressure, and c_v=5/2 R constant volume. So we have
Q_bc=c_v/R (P_cV_c-P_bV_b)=-45J

Q_ca=c_p/R (P_aV_a-P_cV_c)=-50J

(I'm calling P_i and V_i for better understanding before using the values given by P₀ and V₀)

We can see that both of those values represent heat leaving the system. Now we'll use the internal energy relations to find Q_ab. The internal energy relations for each path are:
ΔU_bc=Q_bc-W_bc (Yes, W_bc is zero)
ΔU_ca=Q_ca-W_ca

Since we know that
ΔU=ΔU_ab+ΔU_bc+ΔU_ca=0
and
ΔU_ab=Q_ab-W_ab
We have
Q_ab=W_ab-(Q_bc-W_bc+Q_ca-W_ca)
Q_ab=W - (Q_bc+Q_ca) = 110J
Wich is our heat transferred into the system! (Q_h)

In possession of these values, we can now calculate the efficiency of this cycle:
e=W/Q_h=13,6%

=D

Thanks again Redbelly! And btw, that's a really helpful list of symbols! :) Especially for this dark times without latex.. lol

 

[Redbelly98]

... c_p=3/2 R, for constant pressure, and c_v=5/2 R constant volume.

It looks like the values of c_v and c_p have been swapped here, but you did use the correct values in the calculation. Good job, you nailed this one!

Thanks again Redbelly! And btw, that's a really helpful list of symbols! :) Especially for this dark times without latex.. lol

You're welcome. "Dark times", LOL