0

According to Halliday-Resnick, a free expansion of a gas is an irreversible process. However, the text continues that in a system of particles in a box, it is possible (though very unlikely) for a system of uniformly distributed particles to all cluster in one portion of the box, which is, in essence, the reverse of a free expansion.

Not only does this 'reversal' seem to imply that free expansion is not irreversible, but it also seems to violate the 2nd Law of Thermodynamics, as entropy is decreasing when the particles cluster.

Where am I going wrong here?

spinor
  • 365

3 Answers3

4

Let's consider an example case. Take one sixth of a mole of gas particles; that's $10^{23}$ particles. The probability that all happen to be in one half of a box (assuming equal probabilities for the two halves for each particle) is $$ 2^{-10^{23}}. $$ Suppose each particle takes a millisecond to cross the box, so that the whole gas takes about 1 millisecond to sample a new distribution across the two halves of the box. The number of distributions thus sampled per year is

1 year / 1 millisecond $\simeq 3 \times 10^{10}$

So the expected number of years until a gathering all in one half of the box occurs is $$ T = 2^{10^{23}} / 3\times 10^{10}. $$ To understand this number, let's take the log to base 10: $$ \log_{10} T = 10^{23} \log_{10} 2 - \log_{10} 3 - 10 \simeq 0.3 \times 10^{23} $$ Now compare with the age of the universe, which I will take as 14 billion years: $$ \log_{10} (T / 1.4 \times 10^{10} ) \simeq 0.3 \times 10^{23} - \log_{10} 1.4 - 10 \simeq 0.3 \times 10^{23}. $$ Now to be cautious let's take a third of this, thus getting $10^{22}$. So the estimate is that it will take a time greater than

$10^{10^{22}}$ times the current age of the universe from the Big Bang

for the gas to gather spontaneously in one half of the box, if it is to happen just by random independent motion of each of the particles.

None of our knowledge of physics can be trusted on this large a timescale. What the calculation really means is that the authors were incorrect to assert that the gas "could" spontaneously gather in one side of the box. What the calculation means is that the gas could not gather on one side, merely by independent random motions of the particles, by any reasonable definition of the words "could" and "could not".

Andrew Steane
  • 65,285
1

Not wrong. You're just correctly observing that the arrow of time in thermodynamics is statistical and that any sufficiently up-close view of a system will see occasional local decreases in entropy.

g s
  • 14,249
1

Where am I going wrong here?

You are only considering what is happening locally rather than what is happening overall.

Consider the analogy involving heat transfer by conduction. Per the second law, net heat transfer only occurs from the higher to lower temperature object spontaneously. On average, kinetic energy is transferred by collisions between the higher kinetic energy molecules of the higher temperature body, lowering the average KE of the higher temperature body, to the lower kinetic energy molecules of the lower temperature body, increasing the average KE of the molecules of the lower temperature body.

However, since the KE of individual molecules of both bodies are distributed about an average, at the individual molecular level KE can be transferred from higher KE particles of the lower temperature body to lower KE particles of the higher temperature body, i.e., local energy transfer from low temperature to high temperature in apparent violation of the second law. However there is no violation of the second law because heat is the overall average transfer of KE is from the higher temperature body to the lower temperature body.

Hope this helps.

Bob D
  • 81,786