# Is the 2nd law wrong?

1. Jan 11, 2012

### oramao

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

2. Jan 11, 2012

### Skrambles

How can volume be both a constant and a variable?

3. Jan 11, 2012

### Ken G

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. Jan 11, 2012

### cagce88

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. Jan 11, 2012

### bbbeard

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

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