# Why does $\oint\frac{\delta Q}{T} = 0$ when evaluated over a Carnot cycle?

In my thermodynamics schoolbook, it is asked to apply the Clausius inequality to a Carnot cycle. According to the book the answer is to evaluate $$\oint\frac{\delta Q}{T}$$ over the cycle which should yield $$\oint\frac{\delta Q}{T} = \frac{Q_H}{T_H}+\frac{Q_C}{T_C}=0$$

1. $$Q_H$$ is the heat transferred from the hot reservoir (positive)
2. $$T_H$$ is the temperature of the hot reservoir
3. $$Q_C$$ is the heat transferred to the cold reservoir (negative)
4. $$T_C$$ is the temperature of the cold reservoir

While it is easy enough to show this is true using $$\delta Q =TdS$$, my schoolbook doesn't introduce entropy until later. Therefore, my question is how to prove this equality without using entropy ?

• Lay out a closed cycle involving the two adiabats and the two isotherms. You won't be able to design one for which this equation is not satisfied. Try it and see. Commented Jun 14, 2020 at 21:54
• Hey there, is there a general way to prove that without using $pV=RT$ ? I know that from $pV=RT$ one can show that for a Carnot cycle $\frac{Q_C}{Q_H}=\frac{T_C}{T_H}$, which then shows that $\frac{Q_H}{T_H}+\frac{Q_C}{T_C}=0$. Otherwise, I don't know how to show $\frac{Q_H}{T_H}+\frac{Q_C}{T_C}=0$ always holds true for those types of cycles. Commented Jun 14, 2020 at 22:43
• What's wrong with using pV=RT? Commented Jun 14, 2020 at 23:43
• I thought this was valid for any fluid, not just ideal gases. Commented Jun 15, 2020 at 0:10
• It, of course, should be true for any fluid. But solving the equations for, say, a van der Waals gas might be a little challenging. Maybe you are up to such a challenge. Commented Jun 15, 2020 at 0:25

I only know a long ride for this, sorry :/

You have to imagine two processes iabf and if that have the same work $$\Delta W_{iabf}=\Delta W_{if}$$: (Dashed lines are adiabatic curves and solide lines are isothermes)

The internal energy variation is the same for both processes and $$\Delta Q_{ab}=\Delta Q_{if}$$.

Now imagine a system S' that provides energy $$\Delta Q$$ to the cycle C(system S). The only way to enter energy is through an isotherm, with temperature $$T_0=T_H$$. This works as the hot reservoir. The ab isotherm has $$T, which is your $$T_C$$. The system S' transfers heat($$\Delta Q'$$) to S: $$\Delta Q' = T_H \frac{\Delta Q}{T_C}$$

The cd path has the opposite direction. Try drawing one Carnot Cycle above the other and you will see they cancel each other.

Now imagine infine mini carnot cycles inside C, the sum of all cycles cancel each other:

$$\Sigma \Delta Q' = \oint_C T_C \frac{dQ}{T} = 0$$ *T varies in each mini cycle.

Moving $$T_C$$ $$\oint_C \frac{dQ}{T} = 0$$

I believe your book assume reversible processes from the begining. Otherwise, according to the second law:

$$\oint_C \frac{dQ}{T} \leq 0$$

• Hey there, thanks for answering my question. I'm a little bit confused about $\Delta Q'$ though, is it the heat exchanged in the c-d isothermal ? If so, why does $\Delta Q'$ equal $\frac{T_H}{T_C}\Delta Q$ ? Does that come from $pV=RT$ ? Or can we show that $\Delta Q' = \frac{T_H}{T_C}\Delta Q$ without using $pV=RT$ ? Commented Jun 14, 2020 at 21:17
• cd is an isotherm. The relation in $\Delta Q'$ comes from a reversible cycle between two reservatories, one with temperature $T_H$ and other with $T_C$. Or you can think about the thermal effitiency: $\eta=1- \frac{T_C}{T_H}=1-\frac{Q_C}{Q_H}$. Thus, $Q_H=T_C \frac{Q_C}{T_H}$ which is written with a diffferent notation: $\Delta Q'=\frac{T_H}{T_C} \Delta Q$ . Commented Jun 15, 2020 at 18:07

We want to prove that $$\oint_\gamma\frac{\delta Q}{T} = \frac{Q_H}{T_H}+\frac{Q_C}{T_C}=0\tag 1$$

without referencing entropy.

It would be really easy to just say

$$\oint_\gamma\frac{\delta Q}{T}=\oint_\gamma dS=0$$ because a loop integral of a state variable is path invariant, but we don't want to do that.

It seems that we can prove the equality $$(1)$$ if we just prove that:

$$\frac{Q_H}{T_H}+\frac{Q_C}{T_C}=0.$$

Now if $$Q_C$$ is negative lets write it as

$$\frac{Q_H}{T_H}+\frac{-Q_C}{T_C}=0$$

than we have the following expression, which we want to prove:

$$\frac{Q_H}{T_H}=\frac{Q_C}{T_C}\tag 2.$$

Now we can prove the Carnot's theorem and from that proof we conclude that:

$$\frac{Q_C}{Q_H}=f(T_C,T_H)$$

and one funtion $$f$$ which Lord Kelvin choose as appropriate here is:

$$\frac{Q_C}{Q_H}=\frac{T_C}{T_H}$$

and now we just do an algebraic manipulation and end up with:

$$\frac{Q_C}{T_C}=\frac{Q_H}{T_H}$$

and that is exactly the equation $$(2)$$ which we wanted to prove.

Now we input it into $$(1)$$ and get:

$$\oint_\gamma\frac{\delta Q}{T} = \frac{Q_H}{T_H}+\frac{-Q_C}{T_C}=\frac{Q_C}{T_C}+\frac{-Q_C}{T_C}=0.$$ Q.E.D.