We began this section looking for a way to quantify the dispersal of energy. The multiplicity $\text{Ω}$ does this; each microstate provides a different way to spread the energy, and maximal dispersion—that is, equilibrium—occurs for the macrostate that has the largest value of $\text{Ω}$. We need to find a way to compute the multiplicity $\text{Ω}$ for a physical system such as a gas or a liquid.

To begin, consider a container divided into 1000 small bins that can each “store” a molecule; that is, there are 1000 distinct spatial locations for each molecule. Suppose the container holds one molecule. No macroscopic measurements can reveal which of the 1000 bins the molecule is in, so each of the 1000 possible locations is a microstate corresponding to the same one-molecule macrostate. The multiplicity of this macrostate is $\text{Ω}=1000$.

If there are two molecules, each placed randomly into a bin, then the number of possible arrangements—each a microstate—is $1000\times 1000={1000}^2$. For three molecules it is $1000\times 1000\times 1000={1000}^3$, and for N molecules ${1000}^N$. We might insist on physical grounds that there can be no more than one molecule per bin—two molecules can’t occupy the same point in space—but as long as N is much less than the number of available bins (a dilute system), then this constraint has little effect on the calculation and the multiplicity of a macrostate with N molecules is $\text{Ω}={1000}^N$.

Now suppose we allow the container volume V to vary without changing the bin size. For example, doubling the volume would result in 2000 bins, and then the multiplicity would be $\text{Ω}={2000}^N$. Because the number of bins is proportional to V, the multiplicity of a system that consists of N molecules in volume V must be

where c is an unknown proportionality constant. (Don’t confuse this use of c with the system’s specific heat or concentration.) It might depend on the temperature or other state variables, but c does not depend on the system’s volume; the dependence of $\text{Ω}$ on V is now fully specified.

It’s not immediately obvious how Equation 20.44 is useful, so let’s consider a specific example. FIGURE 20.20 shows a container of gas with initial volume $V_{\text{i}}$ that is undergoing a slow isothermal expansion (no change in temperature). Recall that a process that happens so slowly that it’s always essentially in internal equilibrium is called a quasi-static process. You learned in Chapter 19 that an isothermal expansion requires a heat input, so we’ll let the environment supply an infinitesimal amount of heat $dQ$ to cause an infinitesimal volume expansion to $V_{\text{f}}=V_{\text{i}}+dV$.

FIGURE 20.20 The multiplicity of a gas increases as it undergoes a slow isothermal expansion.

For infinitesimal changes, the first law, $\text{Δ}{E}_{\text{th}}=W+Q$, becomes $d{E}_{\text{th}}=dW+dQ$, where $dW$ and $dQ$ are infinitesimal amounts of energy transferred to or from the system in mechanical and thermal interactions. The thermal energy of a gas does not change in an isothermal process, so $d{E}_{\text{th}}=0$. We found in Chapter 19 that the expansion work done on an ideal gas is $dW=-pdV$. Thus the isothermal expansion is characterized by

where, in the last step, we used the ideal-gas law for p. We can rearrange Equation 20.45 to show that the fractional change in volume during the isothermal heating is directly proportional to the heat $dQ\text{:}$

As the gas expands, its multiplicity—the number of microstates—increases from ${\text{Ω}}_{\text{i}}=c{V}_{\text{i}}^N$ to ${\text{Ω}}_{\text{f}}=c{V}_{\text{f}}^N=c{(V_{\text{i}}+dV)}^N$. We don’t know the proportionality constant c, but we don’t need it if we form a ratio in which c cancels:

We dropped the subscript on V at the end because, with the volume changing only infinitesimally, there’s essentially no difference between $V_{\text{i}}$ and $V_{\text{f}}$.

The exponent N is extremely large. One way to deal with large exponents is to use logarithms. Suppose we take the natural logarithm of Equation 20.47 and use the logarithm properties $\text{ln}(a/b)=\text{ln }a-\text{ln }b$ and ${\text{ln }a}^n=n\text{ln }a$ to write

Now $dV/V\ll 1$ because it’s an infinitesimal expansion, and a very useful approximation is $\text{ln}(1+a)\approx a$ when $a\ll 1$. You can use your calculator to check that, for example, $\text{ln}(1.001)=0.0009995\approx 0.001$. (This approximation is the first term in the Taylor-series expansion of $\text{ln}(1+x)$, which you may have studied in calculus.) We can use this approximation and Equation 20.46 to write Equation 20.48 as

Equation 20.49 is an important micro/macro connection because the measurable macroscopic quantities $dQ$ and T allow us to calculate $\text{Δ}(\text{ln}\text{Ω})=\text{ln}{\text{Ω}}_{\text{f}}-\text{ln}{\text{Ω}}_{\text{i}}$, the change in the logarithm of the multiplicity.