Understanding mathematically the free expansion process of an ideal gas

Question

I'm trying to understand mathematically that for the free expansion of an ideal gas the internal energy $E$ just depends on temperature $T$ and not volume $V$.

In the free expansion process the change in internal energy $\Delta Q = \Delta W = 0$, therefore by the first law $\Delta E=0$. Energy is a function of state so $E=E(V,T)=E(p,T)=E(p,V)$, using $V$ and $T$ as the state variables we have $E=E(V,T)$.

I have the following written mathematically in my notes but do not understand how it is derived.

If

$E=E(V,T)$

then

$dE=\left(\frac{\delta E}{\delta V}\right)_T dV+\left(\frac{\delta E}{\delta T}\right)_V dT$

since we know experimentally that $dT=0$ and $dE=0$ from the first law we can write:

$\left(\frac{\delta E}{\delta V}\right)_T=0$, as required.

I have two main problems with this, firstly I do not understand where

$dE=\left(\frac{\delta E}{\delta V}\right)_T dV+\left(\frac{\delta E}{\delta T}\right)_V dT$

comes from. My guess is that it is the differential of $E(V,T)$ but I do not fully understand the notation of what $E(V,T)$ represents that is, I understand that $E$ is a function of temperature and volume but I do not know how this can be written mathematically other than $E(V,T)$ so I can't differentiate it myself to check this. Again if I were to guess I would get an expression for $V$ and $T$ using the ideal gas equation and try to differentiate that?

Secondly even with the results I do not understand why the end result is

$\left(\frac{\delta E}{\delta V}\right)_T=0$ and not $\left(\frac{\delta E}{\delta V}\right)_T dV=0$.

Has $dV$ simply been omitted because multiplying it by 0 in turn makes it 0?

This question has arisen from me looking back over my lecture notes so it is possible I have written something down incorrectly which is now confusing me. If you could help to clarify any of the points above that would be great. Please let me know if more information is needed.

Edit:

I should also add I am not clear on the significance of the subscripted values $T$ and $V$ on either of the fractions represent.

Update:

I found the suffixes $T$ and $V$ mean that the temperature and volume is constant in the process.

Answer 1

$E(V,T)$ is an unspecified function. You don't need to know the form to get the derivation correct. The equation:

$$dE=\left(\frac{\delta E}{\delta V}\right)_T dV+\left(\frac{\delta E}{\delta T}\right)_V dT$$

is called the differential form of the total derivative and can be written for any function regardless of the form; it's simply a definition. As you noted in your edit, the subscripts on the derivatives indicate that they are to be taken with that variable held constant. So the first term on the RHS is the change in energy with change in volume for a constant temperature multiplied by the incremental change in volume.

If we know $dT = 0$ and we've established from the first law that $dE = 0$ then we are left with:

$$dE=\left(\frac{\delta E}{\delta V}\right)_T dV=0$$

Since the gas is expanding, we know that the incremental change in volume $dV \neq 0$. Since the product is zero (because $dE = 0$) then the only conclusion is that $\left(\frac{\delta E}{\delta V}\right)_T = 0$

Answer 2

The internal energy of an ideal gas only depends on temperature. This result is not restricted to free expansions, but is completely general. One arrives to this conclusion by using the Helmholtz equation

$$\left( \frac{\partial E}{\partial V} \right)_T = T^2 \frac{\partial}{\partial T} \left( \frac{\partial p}{\partial T} \right)$$

If you introduce the ideal gas law $pV = NRT$ you can check that its internal energy does not depend on the volume. This method is more general because does not assume processes at constant temperature.

Answer 3

A mathematician's take on this might be interest. We use the definition in terms of differential forms $dE = T dS -p dV$. In order to do this we have to rewrite the equations of the ideal gas (which we take to be $T=pV$, $S = \frac 1{\gamma-1} \ln (pV^\gamma)$ in the foorm where $S$ and $V$ are the independent variables. In the case of the ideal gas this is a simple computation and we get $p=e^{(\gamma-1)S}V^{-\gamma}$, $T=e^{(\gamma-1)S} V^{1-\gamma}$. We can then use the standard method for solving exact ode's to get $E=\dfrac 1{\gamma-1}e^{(\gamma-1) S}V^{1-\gamma}$ which simplifies to $\dfrac 1{\gamma-1}T$. In the case of the ideal gas, we can compute the three other energy-type functions in the same and we append the results since, curiously, we haven't found them anywhere in the secondary literature.

From $dH=TdS + Vdp$, we get $H=\dfrac \gamma{1-\gamma}T$.

The other two computations are a bit messier and we get: from $dF=-SdT-pdV$, $F=T\ln V-\dfrac 1{\gamma-1}(T \ln T-T)$ and from, $dG=-SdT + Vdp$, $G=\ln p-\dfrac{\gamma}{\gamma-1}(T\ln T-T)$.

This method works for the ideal gas because of the simplicity of the equations---in the arXiv article 1102.1540 a systematic method is developed to handle more elaborate models.

Answer 4

My advice would be that you should probably just take the first law of thermodynamics as an experimental observation at this point in your career, I think feeling satisfied with it takes a very solid grasp of statistical mechanics. But all it says is that energy is conserved in a system unless the system does work (or had work done on it, i.e. does negative work) OR there is a change in the microscopics of the system. So you say that a change in energy is always due to bit of work or a bit of microscopic energy:

$$ dE = dW + dQ.$$

That $dW$ is $-pdV$ just comes from Newtownian ideas of physics. The work done is always a push against some force, differentially its the force projected onto the direction of motion. If you instead consider a pressure, a Force per area, then you get a volume as your differential just by unit analysis. But more intuitively a growth in volume against a pressure is the same type of thing as a push against a force. Since we normally use thermodynamics to think about gasses and stuff, we like to write it/think about it this way.

$dQ=SdT$ is more odd. First you need to ask why should microscopic changes in energy be what we think of as heat? Maybe a nice way to think of it is that hot things can change other objects dramatically without changing their macroscopic properties much. A hot pan can turn water to hot gas, but the pan looks the same the whole time...That motivates why Heat might be thought of as microscopic energy. Then you know the heat something gives off is related to temperature, so you posit they are related by a variable called $S$. Now you've written the first law of thermodynamics.

The other important thing you're missing is that $E(V,T)$ does not need to exist for $dE = a dT + b dV$. This is a mathematically subtle point, that's why we call this a partial differential, or equally say that $E$ is not necessarily integrable. One thing that means is that it might be two or more valued at some point in $(T, V)$ space that depend on where you start and where you finish.

Your intuition is correct though, this is just a more general statement than $E=E(V,T).$ This allows a larger class of relationships between the variables, not just functions.

The point that $$ p = \left(\frac{\partial E}{\partial V}\right)_T$$ and $$S=\left(\frac{\partial E}{\partial T}\right)_V$$

are not derived... they are read directly from that equation. That's precisely why we write it in that notation. If energy enters the system when there is no change in temperature then you set dT=0 and divide through by $dV$, etc... All this equation says is what we said about the two different ways for the system to gain energy.