Box containing particles and be left alone

[touqra]

Suppose I have a box containing particles and be left alone, with a total amount of energy, E. Hence, at equilibrium, these particles will obey the Maxwell-Boltzmann distribution. The entropy of the system is,

$$S=NkT lnV$$

where T is the temperature of the system. V = volume of box.

Suppose now, I divide this box equally into two, hence, now, each new box's volume is V/2, and the number of particles in each box is just N/2. But the temperature of these two new systems will stay the same, since they were in equilibrium initially.
Now, the new entropy of both systems would be:

$$S_{new}=\frac{N}{2}kTln V/2 + \frac{N}{2}kTln V/2$$
$$S_{new} = NkT ln V - NkTln 2$$

But, $$S_{new} < S$$

Where did I go wrong?

 

[dextercioby]

You're adressing the famous "Gibbs paradox" in statistical mechanics (there's actually no paradox), but you're using an incorrect formula for entropy...

Daniel.

 

[sir-pinski]

Your calculation for the new entropy is correct, but the assumption that the temperature will stay the same is incorrect. When the box is divided into two, the temperature of each new system will actually decrease. This is because the total energy, E, is now divided between two systems instead of one, meaning each system will have half the original energy. According to the equation E=3/2NkT, a decrease in energy will result in a decrease in temperature.

Therefore, the correct calculation for the new entropy would be:

S_{new} = \frac{N}{2}kTln\frac{V}{2} + \frac{N}{2}kTln\frac{V}{2} = NkTln\frac{V}{2} - NkTln2

This shows that the new entropy is actually less than the original entropy, which is in line with the second law of thermodynamics. The second law states that the entropy of an isolated system will either remain constant or increase over time, never decrease.

In summary, the mistake in your calculation was assuming that the temperature would stay the same when the box was divided into two. In reality, the decrease in energy would result in a decrease in temperature, leading to a decrease in entropy.