- #1
greypilgrim
- 581
- 44
Hi.
If an ideal gas of ##N## particles is allowed to expand isothermically to double its initial volume, the entropy increase is
$$\Delta S=N\cdot k_B \cdot \log\left(\frac{V_f}{V_i}\right)=N\cdot k_B \cdot \log\left(\frac{2V}{V}\right)=N\cdot k_B \cdot \log\left(2\right)\enspace .$$
This can also be derived with Boltzmann's entropy formula ##S=k_B \cdot \log(W^N)##, assuming the particles are independent and all have the same number of microstates ##W## which doubles as ##V## does. So after the expansion, the number of microstates of the whole gas is ##(2W)^N## and
$$S=k_B \cdot \log\left(\left(2W\right)^N\right)=N\cdot k_B \cdot \log\left(2W\right)\enspace ,$$
also leading to the entropy difference ##\Delta S=N\cdot k_B \cdot \log\left(2\right)##.
Now there will be fluctuations where more than ##N/2## particles are in the initial half of the containment (Poincaré's recurrence theorem even states there will be moments where all particles are back there). I read that this implies a (temporary) reduction in entropy. This is somehow what
$$\Delta S=N\cdot k_B \cdot \log\left(\frac{V_f}{V_i}\right)$$
says if ##V_i## and ##V_f## are taken as the volumes the gas actually has at a moment in time – and not the volumes the gas is allowed to have.
This gets even more confusing if we use Boltzmann's formula: The number ##(2W)^N## contains all microstates in ##2V##, including the ones where there's an imbalance of particle numbers in the halves. Hence those microstates are part of the macrostate with entropy
$$S=N\cdot k_B \cdot \log\left(2W\right)\enspace ,$$
so it will not change even when all particles are in one half of the containment.
As mentioned before, I think this discrepancy results in using the volume the gas actually has at a point in time versus the volume it is allowed to have. But is the first even a well-defined property, especially in thermodynamic equilibrium that is needed for entropy to make sense?
If an ideal gas of ##N## particles is allowed to expand isothermically to double its initial volume, the entropy increase is
$$\Delta S=N\cdot k_B \cdot \log\left(\frac{V_f}{V_i}\right)=N\cdot k_B \cdot \log\left(\frac{2V}{V}\right)=N\cdot k_B \cdot \log\left(2\right)\enspace .$$
This can also be derived with Boltzmann's entropy formula ##S=k_B \cdot \log(W^N)##, assuming the particles are independent and all have the same number of microstates ##W## which doubles as ##V## does. So after the expansion, the number of microstates of the whole gas is ##(2W)^N## and
$$S=k_B \cdot \log\left(\left(2W\right)^N\right)=N\cdot k_B \cdot \log\left(2W\right)\enspace ,$$
also leading to the entropy difference ##\Delta S=N\cdot k_B \cdot \log\left(2\right)##.
Now there will be fluctuations where more than ##N/2## particles are in the initial half of the containment (Poincaré's recurrence theorem even states there will be moments where all particles are back there). I read that this implies a (temporary) reduction in entropy. This is somehow what
$$\Delta S=N\cdot k_B \cdot \log\left(\frac{V_f}{V_i}\right)$$
says if ##V_i## and ##V_f## are taken as the volumes the gas actually has at a moment in time – and not the volumes the gas is allowed to have.
This gets even more confusing if we use Boltzmann's formula: The number ##(2W)^N## contains all microstates in ##2V##, including the ones where there's an imbalance of particle numbers in the halves. Hence those microstates are part of the macrostate with entropy
$$S=N\cdot k_B \cdot \log\left(2W\right)\enspace ,$$
so it will not change even when all particles are in one half of the containment.
As mentioned before, I think this discrepancy results in using the volume the gas actually has at a point in time versus the volume it is allowed to have. But is the first even a well-defined property, especially in thermodynamic equilibrium that is needed for entropy to make sense?