# Box containing particles and be left alone

1. Feb 8, 2006

### 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?

Last edited: Feb 8, 2006
2. Feb 9, 2006

### 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.