Understanding a proof of Carnot's theorem.

  • Thread starter Thread starter Fallen Seraph
  • Start date Start date
  • Tags Tags
    Proof Theorem
AI Thread Summary
The discussion revolves around understanding the proof of Carnot's theorem, which asserts that no engine can be more efficient than a Carnot engine operating between the same heat reservoirs. A hypothetical engine is proposed to have greater efficiency, leading to a contradiction when it is shown that this would result in a violation of the second law of thermodynamics. The confusion arises regarding the efficiency definitions and the role of reversible processes, with clarification that any engine with efficiency less than that of a Carnot engine is not reversible. The conversation emphasizes the distinction between efficiency and the coefficient of performance, noting that proper conversion between these metrics is essential for accurate comparisons. Ultimately, the proof stands firm, reinforcing the validity of Carnot's theorem.
Fallen Seraph
Messages
32
Reaction score
0
[SOLVED] Understanding a proof of Carnot's theorem.

I'm having trouble understand this proof of Carnot's theorem, and I would appreciate it if someone could point out where my reasoning goes wrong.

The proof reads thusly:


Suppose there exists a a hypothetical engine with a greater efficiency than a Carnot engine.

Consider this engine working from the same hot and cold reservoirs as a Carnot engine.

Adjust the cycle of the Carnot engine such that its work output == that of the hypothetical engine == W

Since the Carnot engine is reversable, we can turn it into a refrigerator that takes in work W from the hypothetical engine and energy Q2 from the cold reservoir and then outputs energy Q1 into the hot reservoir.

Let the energy taken from the hot reservoir by the hypothetical engine == P1.

We have that the efficiency of the hypothetical engine is greater than that of the Carnot one, so

W/P1>W/Q1

=>

Q1>P1.

This means that our construction is taking heat from the cold reservoir, and depositing it in the hot one. Which violates the 2nd law, and thus proves the theorem.


My problem with it is that is seems to imply that either the second law is wrong, or the Carnot engine is not the most efficient because:

Why not just get an engine that is less efficient than the Carnot one and construct the same device as was made in the proof except with the Carnot engine in the place that the hypothetical one occupied in the proof, and the less efficient engine in the place of the Carnot?

This, as far as I can see, would also move heat from the cold reservoir to the hot one.

What am I missing?

Thanks in advance.
 
Science news on Phys.org
Fallen Seraph said:
What am I missing?

Any process with efficiency less than Carnot efficiency is not reversible, for example.
 
Ah! So it is!

Thank you.
 


This Carnot engine works as refrigerator. So, shouldn't it eficiency be Q2/W and not W/Q1?
 
that's not efficiency,that's co-efficient of performance..
we are working with efficiency here so if we take coefficient of performance we must have to convert it into efficiency through the relation between them finally..:)
 
I was watching a Khan Academy video on entropy called: Reconciling thermodynamic and state definitions of entropy. So in the video it says: Let's say I have a container. And in that container, I have gas particles and they're bouncing around like gas particles tend to do, creating some pressure on the container of a certain volume. And let's say I have n particles. Now, each of these particles could be in x different states. Now, if each of them can be in x different states, how many total...
Thread 'Why work is PdV and not (P+dP)dV in an isothermal process?'
Let's say we have a cylinder of volume V1 with a frictionless movable piston and some gas trapped inside with pressure P1 and temperature T1. On top of the piston lay some small pebbles that add weight and essentially create the pressure P1. Also the system is inside a reservoir of water that keeps its temperature constant at T1. The system is in equilibrium at V1, P1, T1. Now let's say i put another very small pebble on top of the piston (0,00001kg) and after some seconds the system...
Back
Top