1
$\begingroup$

I'm pretty new to Thermodynamics, but I've been playing around with some different computations to reinforce my understanding. I came across the result $$\Omega \propto \left( PV^\gamma \right)^{fN/2},$$ which I'm not familiar with from my reading. Here, $P$ is pressure, $V$ is volume, $\gamma = C_P / C_V$ is the heat capacity ratio, $f$ is the number of degrees of freedom, $N$ is number of particles, and $\Omega$ represents the number of microstates.

So, I'm curious if (1) the relationship is correct and (2) if it appears to have any utility and (3) if it is possible to turn this into an equality instead of just a proportionality?


Derivation: Starting from the Second Law of Thermodynamics, we have $$dU = dQ - dW,$$ which can be rearranged to $$dQ = dU + dW.$$ Substituting in the pressure $P$ and volume $V$ of an ideal gas we have $$dQ = \frac{f}{2} d(PV) + P dV, $$ where $f$ represents number of degrees of freedom. By the product rule, $$dQ = \frac{f}{2} V dP + \frac{f}{2} P dV + P dV, $$ $$dQ = \frac{f}{2} V dP + \left( \frac{f}{2} + 1 \right) P dV.$$ Dividing both sides by temperature $T$ gives $$\frac{dQ}{T} = \frac{f}{2} \frac{V}{T} dP + \left( \frac{f}{2} + 1 \right) \frac{P}{T} dV.$$ The right-hand side can be rewritten with the ideal gas law $PV = NkT$, where $N$ represents the number of particles and $k$ is the Boltzmann constant. $$\frac{dQ}{T} = \frac{f}{2} Nk \frac{dP}{P} + \left( \frac{f}{2} + 1 \right) Nk \frac{dV}{V}.$$ Integrate both sides from some initial state to some final state in a particular process where pressure and volume are continually well-defined. $$\int \frac{dQ}{T} = \frac{f}{2} Nk \int \frac{dP}{P} + \left( \frac{f}{2} + 1 \right) Nk \int \frac{dV}{V}$$ The left-hand side is simply the change in entropy $\Delta S = \int dQ / T$ and the right-hand side can be expressed in terms of the initial and final pressure and volume values. $$\Delta S = \frac{f}{2} Nk \ln \left( \frac{P_f}{P_i} \right) + \left( \frac{f}{2} + 1 \right) Nk \ln \left( \frac{V_f}{V_i} \right)$$ However, $S = k \ln \Omega$, where $\Omega$ represents the number of gas microstates, so $\Delta S = k \ln (\Omega_f / \Omega i)$, and thus $$k \ln \left( \frac{\Omega_f}{\Omega_i} \right) = \frac{f}{2} Nk \ln \left( \frac{P_f}{P_i} \right) + \left( \frac{f}{2} + 1 \right) Nk \ln \left( \frac{V_f}{V_i} \right),$$ $$\frac{\Omega_f}{\Omega_i} = \left( \frac{P_f V_f^\gamma}{P_i V_i^\gamma} \right)^{fN/2}.$$ In this final step, I have substituted the heat capacity ratio $\gamma = C_P / C_V = 1 + 2/f$ for simplicity. This suggests a proportionality between the state variables $P$ and $V$ and the number of microstates of an ideal gas. $$\boxed{\Omega \propto \left( PV^\gamma \right)^{fN/2}}$$

$\endgroup$
  • $\begingroup$ Are you asking whether you did the math correctly? $\endgroup$ – Chet Miller Dec 23 '18 at 13:51
  • $\begingroup$ In part, yes, but more importantly I'm curious if this formula has any utility and if there is a way to turn the proportionality into an equality. $\endgroup$ – Trevor Kafka Dec 24 '18 at 4:27
  • $\begingroup$ The math looks correct, but you could have done it much more simply by starting with $$dS=C_vd\ln{T}+Rd\ln{V}$$in conjunction with $$d\ln{P}+d\ln{V}=d\ln{T}$$ $\endgroup$ – Chet Miller Dec 24 '18 at 12:48
1
$\begingroup$

This answer addresses only the first question. The answer is yes, the given expression for $\Omega$ is correct. The purpose of this answer is to provide additional confidence in the correctness of that expression, by checking it in two different ways.

One way to check it is to deduce it using a statistical-mechanics argument, without relying on thermodynamics at all. This will be only a heuristic argument, and it will be specific to a monatomic ideal gas. Notation:

  • $D$ is the number of spatial dimensions.

  • $N$ is the number of atoms.

  • $\Omega(E,V)$ denote the number of microstates as a function of the total energy $E$ and volume $V$.

I'll use units in which Boltzmann's constant is equal to $1$. To get a finite value for $\Omega$, we need to take quantum physics into account, but we can do this in a heuristic way: just suppose that the number of possible locations of an individual atom is proportional to $V$, as though space were partitioned into discrete cells. Then the number of possible location-combinations of $N$ atoms is proportional to $V^N$. The quantity $\Omega(E,V)$ must be dimensionless. Since $V$ has dimensions of length$^D$, and since the only relevant constants are Planck's constant $\hbar$ and the single-atom mass $m$, the only possible dimensionless combination that is proportional to $V^N$ is $$ \Omega(E,V)\propto \left(\frac{mE}{\hbar^2}\right)^{ND/2}V^N\propto E^{ND/2}V^N. \tag{1} $$ Therefore, the entropy is $$ S(E,V)\equiv\log \Omega(E,V)=\frac{ND}{2}\log E+N\log V+\text{constant}. \tag{2} $$ The temperature $T$ and pressure $p$ are defined by $$ \frac{1}{T}\equiv\frac{\partial S}{\partial E} = \frac{ND}{2E} \hskip2cm \frac{p}{T}\equiv\frac{\partial S}{\partial V} = \frac{N}{V}. \tag{3} $$ Use these relationships to deduce $E\propto pV$, and use this in (1) to get $$ \Omega(p,V)\propto (pV^\gamma)^{ND/2} \hskip2cm \gamma = \frac{D+2}{D}. \tag{4} $$ This agrees with the result shown by the OP, in the special case of a monatomic ideal gas. Specializing to three-dimensional space ($D=3$) gives the familiar result $\gamma=5/3$.

Here's another check. Generalize (2) to $$ S(E,V)=\frac{Nf}{2}\Big(\log E+(\gamma-1)\log V\Big)+\text{constant} \tag{5} $$ with arbitrary $f$ (instead of $D$) and arbitrary $\gamma$. The motive for considering this generalization is simply that it is consistent with the OP's expression for $\Omega$. To see this, use the definitions of $T$ and $p$ to get $$ \frac{1}{T}\equiv\frac{\partial S}{\partial E} = \frac{Nf}{2E} \hskip2cm \frac{p}{T}\equiv\frac{\partial S}{\partial V} = \frac{Nf}{2V}(\gamma-1). \tag{6} $$ Use these relationships to deduce $E\propto pV$, and use this in (5) along with $\Omega=e^S$ to get $$ \Omega(p,V)\propto (pV^\gamma)^{Nf/2}. \tag{7} $$ This shows that (5) is consistent with the OP's equation, rewritten here as equation (7). The check consists of using (5) to show that $\gamma$ really is the heat-capacity ratio. The heat capacity at constant volume is $$ C_V = T\left(\frac{\partial S}{\partial T}\right)_V = -E\left(\frac{\partial S}{\partial E}\right)_V = \frac{Nf}{2}. \tag{8} $$ where the first equation in (6) was used to change the variable to $E\propto T$. The heat capacity at constant pressure is $$ C_p = T\left(\frac{\partial S}{\partial T}\right)_p. \tag{9} $$ To evaluate this, first we need to rewrite the entropy (5) in terms of $T$ and $p$ instead of $E$ and $V$. Equations (6) imply $E\propto T$ and $V\propto T/p$, so the definition (9) gives $$ C_p = \frac{Nf}{2}\gamma. \tag{10} $$ The results (8) and (10) show that $\gamma=C_p/C_V$, as desired.

Determining the proportionality factor in (7) would probably have little utility, because we rarely need to add the number of microstates of one system to the number of microstates of another system. When we combine two systems, we multiply their individual $\Omega$s (that is, add their entropies) to get the total $\Omega$, at least in the simplest case where the mutual interactions have negligible effect on the result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.