Thermodynamic Systems with Two Independent Variables: The Flaw in the 2nd Law

In summary, the thermodynamic system of two independent variables, like a gas inside a recipient, at equilibrium, will change their final state if the same amount of heat is given to the system in two different ways, with one being an irreversible process and the other a reversible process. The difference between the final states is due to the difference in heat transfer for an irreversible process being strictly less than TdS.
  • #1
oramao
1
0
Consider a thermodynamic system of two independent variables, like a gas inside a recipient,
at equilibrium.

We are now going to carry out two different transformations at constant volume:

Transformation 1) A reversible process where an amount of heat Q is "given" to the system
Transformation 2) An irreversible process where the same amount of heat Q is "given" to the system.

In both transformations the change of Internal energy (E) is the same, once volume is constant for both (Q=change of Internal energy)

Therefore if we choose as independent variables, Internal Energy and volume, we obtain:

Transformation 1) Initial state E1 and V and final state E2 and V
Transformation 2) Initial state E1 and V and final state E2 and V

As there are only two independent variables, the final state of the system is the same for transfomations 1 and 2 and therefore the final entropy is also the same in both cases. This is not what the 2nd law tells us
 
Physics news on Phys.org
  • #2
How can volume be both a constant and a variable?
 
  • #3
oramao said:
As there are only two independent variables, the final state of the system is the same for transfomations 1 and 2 and therefore the final entropy is also the same in both cases. This is not what the 2nd law tells us
How are you imagining that Q can be added to the system reversibly with no change in volume? What could be changing such that Q should reversibly enter this system? The reason Q moves around in reversible equilibria is that something is changing that forces Q to move to keep the entropy the same, so if nothing changes, there's no reason for Q to go anywhere-- unless it is not a reversible equilibrium (like a T difference).
 
  • #4
A gas in a fixed volume container, with added heat to it, will change the final energy of the system. Remember, entropy relations for fixed volume have to account for the increase in pressure inside the container. Thus, the changing pressure keeps the entropy constant, thus not violating the 2nd law!
 
  • #5
oramao said:
This is not what the 2nd law tells us

I think you have specified inconsistent assumptions. Kind of like proving that all triangles are isosceles: you start by picking any vertex and dropping the perpendicular to the midpoint of the other side ;-)

The "catch" in your assumptions, in addition to the comments provided by the others above, is that the heat transfer for an irreversible process is strictly less than TdS. Only for a reversible process does δQ=TdS. So the end states are not the same.

BBB
 

1. What is the 2nd law of thermodynamics?

The 2nd law of thermodynamics states that the total entropy of a closed system will always increase over time. This means that energy will naturally flow from high energy states to low energy states, resulting in a decrease in usable energy and an increase in disorder.

2. How does the 2nd law apply to systems with two independent variables?

In systems with two independent variables, such as temperature and pressure, the 2nd law of thermodynamics still holds true. The change in entropy for such systems can be calculated using the Gibbs free energy equation, which takes into account the effects of both variables.

3. What is the "flaw" in the 2nd law of thermodynamics?

The flaw in the 2nd law of thermodynamics refers to the fact that it does not account for the possibility of a decrease in entropy for certain systems. This is known as a "negative entropy" or "negative disorder" and can occur in systems with a high degree of organization or complexity.

4. Can a system with two independent variables violate the 2nd law of thermodynamics?

No, the 2nd law of thermodynamics is a fundamental principle that cannot be violated. However, for systems with two independent variables, it is possible for the change in entropy to be negative, which may seem to contradict the 2nd law. This is where the "flaw" in the 2nd law comes into play.

5. How can the flaw in the 2nd law be explained?

The flaw in the 2nd law can be explained by considering the limitations of the law. While it holds true for the majority of systems, there are certain cases where the decrease in entropy is allowed due to the unique properties of those systems. This does not mean that the 2nd law is incorrect, but rather that it is not applicable in all situations.

Similar threads

  • Classical Physics
Replies
3
Views
1K
Replies
5
Views
518
Replies
7
Views
783
Replies
9
Views
1K
  • Classical Physics
Replies
2
Views
763
Replies
18
Views
938
  • Thermodynamics
Replies
8
Views
786
  • Thermodynamics
Replies
2
Views
721
  • Thermodynamics
2
Replies
46
Views
2K
Replies
5
Views
1K
Back
Top