While reading the second law of thermodynamics, my physics book (Sears and Zemansky's University Physics) said that the total change in entropy can never be negative. Then the authors went on to say that an example of such a process would be all the air in my room compressing spontaneously. It was fine till here, until they went a step further and said that "Strictly speaking, this process is not impossible!" and went on to show how there is a decrease of number of possible microscopic by a factor of $2^N$ where $N$ is the number of moles and went on to prove how there is a non zero probability of all moles of air being found in one half of the room. To illustrate, they found the probability of this happening assuming there are 1000 moles of air in the room, the probability was a number that had more than $10^{26}$ zeroes on the right of the decimal, which of course is not zero. They later said it is a vanishingly small quantity, we may well assume that such a process has never actually taken place in the history of the universe and, thus, for all practical purposes, the second law holds. Now, my question is, since there is a non zero probability, is "free compression", per se, possible, if yes then how can the total change in entropy be negative?
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1$\begingroup$ Does this answer your question? Is a spontaneous decrease in entropy *impossible* or just extremely unlikely? $\endgroup$– Michael SeifertCommented Apr 5, 2022 at 14:47
1 Answer
Now, my question is, since there is a non zero probability, is "free compression", per sé, possible, if yes then how can the total change in entropy be negative?
The book already told you that although it is statistically possible for a gas to compress on its own, it is also highly improbable (probability $\rightarrow0$). But if it did occur there would be a decrease in entropy.
Consider a rigid, thermally insulated vessel containing an ideal gas occupying the entire volume. Half of the gas on one side of the vessel spontaneously compresses itself into the other side leaving a vacuum in one half the vessel. One would never expect to see such an event though, as pointed out in Sears and Zemansky, it is statistically possible (though highly improbable). But let's assume it occurs and determine the total change in entropy.
The change in entropy of an ideal gas (the "system") between any two equilibrium states is given by the following equation (where heat capacities are considered constant) for any process (entropy is a state function and does not depend on the process):
$$\Delta S_{sys}=nC_{v}\ln\frac{T_f}{T_i}+nR\ln\frac{V_f}{V_i}$$
In this example, since the vessel is rigid and insulated it is isolated from the surroundings and there is no work $W$ or heat $Q$ between the gas and its surroundings. From the first law $\Delta U=0$. Since for an ideal gas, any process, $\Delta U$ depends only on temperature, that means $T_{f}=T_i$. Thus the first term to the right of the above equation is zero. In this example the final volume of the gas is less than the initial volume or $V_{f}\lt V_i$, and thus the second term is negative. Consequently for this hypothetical event $$\Delta S_{sys}\lt 0$$
Since the surroundings of the vessel is isolated from the gas in the vessel, the change in entropy of the surroundings is zero, or $\Delta S_{surr}=0$. The total change in entropy is then
$$\Delta S_{tot}=\Delta S_{sys}+\Delta S_{surr}=\Delta S_{sys}+0<0$$
In violation of the second law.
Hope this helps.
