# Calculation of the efficiency of a Carnot engine for photon gas

I am working on a problem considering the Carnot cycle for a photon gas. The cycle has two isothermic processes at $T_{H}$ and $T_{C}$ and two adiabatic processes.

The entropy of the photon gas is $S(U,V)=bU^{0.75}V^{0.25}$ from which I have derived $$U=aT^{4}V$$ and $$P=\dfrac{1}{3}aT^{4}$$

where $a=(\dfrac{3b}{4})^{4}$.

I want to calculate the work done and heat absorbed in each of the four processes of the Carnot cycle, then calculate the efficiency $\eta$ of the engine (which, since it is a Carnot engine, will be $\dfrac{T_{H}-T_{C}}{T_{H}}$).

I have labelled my cycle with 1 in the bottom right corner, 2 in the bottom left, 3 in the upper left, and 4 in the upper right.

Then, along the adiabatic curves(2--> 3 and 4 --> 1): $\Delta Q=0$. So for the 2--> 3$$W= U_{3}-U_{2}=a(T_{H}^{4}V_{3}-T_{C}^{4}V_{2})$$

And for the 4-->1: $$W= U_{1}-U_{4}=a(T_{C}^{4}V_{1}-T_{H}^{4}V_{4})$$

Along the isothermal paths, since $dU=aT^{4}dV+4aT^{3}VdT=at^{4}dV$, we know from the first law that $$dU=\Delta Q- p dV=aT^{4}dV$$ .

So along 3 --> 4: $W= \dfrac{1}{3}aT_{H}^{4}(V_{4}-V_{3})$ and $\Delta Q=\frac{4}{3}aT^{4}(V_{4}-V_{3})$.

Along 1 --> 2, $W=\dfrac{1}{3}aT_{C}^{4}(V_{2}-V_{1})$.

Then $$\eta=\frac{a\left[T_{H}^{4}\left(V_{3}-V_{4}\right)+T_{c}^{4}\left(V_{1}-V_{2}\right)\right]+\dfrac{1}{3}aT_{H}^{4}(V_{4}-V_{3})+\dfrac{1}{3}a(T_{C}^{4}(V_{2}-V_{1}))}{\dfrac{4}{3}aT_{H}^{4}(V_{4}-V_{3})}$$

Which, using $T^{3}V=const.$ to show $V_{1}=(\frac{T_{H}}{T_{C}})^{3}V_{4}$ and $V_{2}=(\frac{T_{H}}{T_{C}})^{3}V_{3}$, I plug in to find: $$\eta=\frac{T_{C}-T_{H}}{2T_{H}}$$

Which is wrong by a factor of -2!

How do I eliminate this incorrect factor of -2?

Also, I have done some other work and found that that $\dfrac{V_{1}}{V_{2}}=\dfrac{V_{4}}{V_{3}}$

• Though I don't know the Carnot Cycle for Photons , I think it's true. Because it is just the case for ideal gas. – Vim Jan 24 '15 at 2:53
• I'd made an error before (I used the ideal gas equation to express p, instead of the photon gas equation) but have now fixed it. I am still trying to eliminate the dependence on volumes to get the Carnot efficiency. Is there any way I should adjust my question to encourage an answer? – user44816 Jan 26 '15 at 0:48
• I have been using the assumption that, along the isothermal paths, $dU=0$, but given the formula for $U$ given above, is this a valid assumption? And, if it is not valid, then how do I calculate the heat absorbed/expelled, along isothermic curves? – user44816 Jan 26 '15 at 1:38
• Sorry I'm not familiar with the case of photons. But I think the $\Delta U$ during a particular process depends completely on $U$'s equation. And as you have shown, $U=aT^4 V$ , I think along isothermal paths $V$ will also play a part in $\Delta U$. – Vim Jan 26 '15 at 10:12

## 1 Answer

Along the adiabatic paths, $$\Delta Q=0$$ so $$dU=-pdV$$ meaning the work done is given by $$W=-\Delta U$$.

This changes the values I proposed in my question.

In fact:

Along the 2 --> 3 transition $$W=U_{2}-U_{3}=a(T_{C}^{4}V_{2}-T_{H}^{4}V_{3})$$

Along the 4 --> 1 transition $$W=U_{4}-U_{1}=a(T_{H}^{4}V_{4}-T_{C}^{4}V_{1})$$

This correction yields $$\eta=\frac{a\left[T_{H}^{4}\left(V_{4}-V_{3}\right)+T_{c}^{4}\left(V_{2}-V_{1}\right)\right]+\dfrac{1}{3}aT_{H}^{4}(V_{4}-V_{3})+\dfrac{1}{3}a(T_{C}^{4}(V_{2}-V_{1}))}{\dfrac{4}{3}aT_{H}^{4}(V_{4}-V_{3})}$$

And since $T^{3}V=constant$ along adiabatic curves, we can use the substitutions: $$V_{1}=(\frac{T_H}{T_{C}})^{3}V_{4}$$ and $$V_{2}=(\frac{T_H}{T_{C}})^{3}V_{3}$$ to solve for our final answer, which is the expected Carnot efficiency: $$\eta=\frac{T_{H}-T_{C}}{T_{H}}$$