# What does it take to derive the ideal gas law in themodynamics?

How can the ideal gas law be derived from the following assumptions/observations/postulates, and these only ?

1. I'm able to measure pressure $P$ and volume $V$ for gases.
2. I notices that if two systems of gases come into thermodynamical equilibrium, that the quantities $PV$ for gas 1 and gas 2 coincide.
3. I assume there is an energy function $U$.
4. The first and the second law of thermodynamics hold by axiom, together with all the consequences, which can abstractly derived from the two.
5. For a free expansion of my gas $(\Delta U=\Delta Q=\Delta W=P\Delta V=0)$ I find that the expression $PV$ for my gas is the same before and after the experiment.

If I'd postulate $PV=nRT$, then I'd easily see that $U(T,V)$ is only a function of $T$ and I can derive everything else, pretty much even without the second law. Here I know of the second law, which gives me a bunch of expressions and I wonder if that suffices to identify the temperature.

If it's not possible, are there other experiments I can do in my position which would do the job?

Here is a weaker version of the question:

Can the ideal gas law be derived if I additionally know that $U$ is really only a function of $T$, or even that $U(T,V)=C_V T\$?

And what role does the zeroth law of thermodynamics play here?

The essence of the question is not necessarily gas physics. It's rather about how temperature is defined by the laws of Thermodynamics, if there is merely the assumption that the system can be described by internal energy $U$ and is experimentally accessible only by its parameters like $P,V,M,H,\dots$. And there is no connection to a theory lying above.

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Isn't the second condition (thermodynamical equilibrium), together with 2nd law enough to say that PV only depends on T ? And then, as you said, you can derive everything. – Frédéric Grosshans Mar 6 '12 at 17:07
@FrédéricGrosshans: So you're saying $PV=g(T)$ follows, right? – NikolajK Mar 6 '12 at 17:09
That is what the 2nd condition says, since thermal equilibrium $\Leftrightarrow$ same temperature. – Frédéric Grosshans Mar 7 '12 at 14:29
@FrédéricGrosshans: Yeah, okay. Clearly I know that the two $PV$s are equal and that the temperatures are equal, but how do I conclude that for example $PV$ is linearly related to $T$? – NikolajK Mar 7 '12 at 15:31
I haven't checked, but you said you needed "only a function of $T$", and then, you spoke about $g(T)$. That is not the same thing as linearity. – Frédéric Grosshans Mar 7 '12 at 16:24

According to the second law, thermal equilibrium between two systems means that they both have the same temperature $T$. The fact 2 that $PV$ coincide whenever two gases are at thermal equilibrium (and, I assume, for the same $n$) means that $PV$ is only a function of $T$. In othor words, there is a function $g()$ such that \begin{align}PV&=g(T)& P=\frac{g(T)}{V}\end{align}

The goal is now to show that $g()$ is linear, i.e. that $g(T)=Tg'(T)$. In order to show that, one can use a Maxwell relation, more specifically, the one which is linked to Helmholtz free energy $A=U-TS$ which can be defined because of condition 3. We have then (including condition 4 for the computation of the derivatives) $$\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V = -\frac{\partial^2 A }{\partial T \partial V}$$

From the first equation, we have $$\left(\frac{\partial P}{\partial T}\right)_V = \frac{g'(T)}{V}.$$

Except if $PV=g(T)$ is constant over a range of temperatures (which can be checked experimentally), the condition 5 implies that $U(T,V)=U(T)$. If one make as mall isotherm transformation of our gas, one has \begin{align} 0=dU&=\delta Q - PdV& \delta Q=PdV\end{align}. The second principle tells us then that $$dS=\frac{\delta Q}{T}=\frac{P dV}T=\frac{g(T)}{TV}dV.$$ By definition, in this case $dS=\left(\frac{\partial S}{\partial V}\right)_TdV$, so we get this partial derivative from the last equation.. Equating the two partial derivatives according to the Maxwell equation then gives us $g(T)=T g'(T)$, which implies $g'(T)=nR$, where $nR$ is an "arbitrary" constant. Hence, $$PV=nRT.$$

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The idea to consider a isothermal is quite clever! I was sceptical reading the proof first, because you don't really know what the temperature is, but I guess you don't need it at hand to consider isothermals of the system. Same goes with the $PV$ expression, I don't even think "which can be checked experimentally" is necessary (I don't even know how you'd check it without knowing temperature). I think if you know that $P,V,T$ are the only parameters of the system, then your second key $PV=g(T)$ follows in any case and then $U=U(T)$ (i.e. the $6^{th}$ assumptions) follows from the $5^{th}$. – NikolajK Mar 7 '12 at 21:41
"According to the second law, thermal equilibrium between two systems means that they both have the same temperature T." It's the zeroth law that says that. – Siyuan Ren Mar 8 '12 at 1:01
@Karsus Ren : I checked on wikipedia ( en.wikipedia.org/wiki/Laws_of_thermodynamics ) and you're right. I don't know if I learnt thermodynamics (18 years ago in France) without 0th law, if I forgot it, or if it was implicit (as it was before the 1930s) ! I truly feel ashamed ! To my defence, the 2nd law (phrased according to Clausius statement) implicitly admits the existence of temperature. And almost forbids thermal equilibrium without equality in T. Nick Kidman : was 0th law forgotten in condition 4 on purpose ? – Frédéric Grosshans Mar 9 '12 at 11:25
@FrédéricGrosshans: Yes. – NikolajK Aug 23 '12 at 12:42

Perhaps the following comments might be of interest. A gibbsian thermodynamical system is specified by the equations of state $T=f(p,V)$ and $S=g(p,V)$ for two functions $f$ and $g$ which are constricted to satisfy the Maxwell relations. The relevant thermodynamical quantities can then be expressed in terms of $p$, $V$, $f$, $g$ and their partials. For the present discussion, we require $\left (\dfrac{\partial U}{\partial T}\right )_V$ and $\left (\dfrac{\partial U}{\partial V}\right )_T$. These are can be easily computed to be $\dfrac{f g_1}{f_1}$ and $-p+\dfrac f{f_1}$ respectively (the subscript denotes partial differentiation with respect to the first variable $p$). (This is a simple computation involving the chain rule and inverse function theorem---for a systematic treatment, see the arXiv article 1102.1540). From this we can draw several conclusions.

Firstly, $U$ depends only on $T$ if and only $\dfrac f{f_1}=p$ and this means that $f$ has the form $p\phi(V)$ for some function $\phi$ of one variable.

Secondly, $\left (\dfrac{\partial U}{\partial T}\right )_V$ is a constant if and only if $\dfrac{fg_1}{f_1}=c$, i.e. $g$ has the form $c \ln f+ \psi(V)$ for some function $\psi$ of one variable.

Thirdly, both of the above conditions hold if and only if $f=\phi(V)$ and $g=\ln p + \ln \phi(V)$ for some function $\phi$.

If we combine these with the condition that the isotherms are as in Boyle's law i.e. are the hyperbolae $pV = c$, then we can deduce that the equations of state are those of the ideal gas.

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Assumption 1 is standard.

Assumption 2 is not experimentally true. It holds only in the limit of very low pressure, but in fact holds for no real gas.

Assumption 3 is standard as is Assumption 4.

Assumption 5 is crucial. It also is not experimentally true, but what it says is that the energy is not a function of volume. Or, more exactly, it says that $(\partial U/\partial V)_{T,n}$ is zero. This means that the energy is independent of the volume of the gas and so that the energy is independent of the distance between molecules. If that is true, the molecules cannot interact with each other, i.e. no attractive or repulsive forces. And so such a gas must be ideal.

Going further thermodynamically we end up with

$$\left( \frac{\partial U}{\partial V}\right)_{T,n} = T\left(\frac{\partial p}{\partial T}\right)_{V,n} - p$$

This expression is, of course zero for an ideal gas.

which can be applied to any equation of state to find out what the change the energy of the gas with volume actually is.

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