# How can we combine all the three gas laws into a single ideal gas equation?

In all texts that I have read, it has been stated that the combined gas law for ideal gases was derived from the individual gas laws proposed by Boyle, Charles and Avogadro.

My confusion is this is that, in each individual law, some variables of the system's state are to be kept constant. However, none of these conditions are considered in the combined gas equation. Also, how is it possible for us to suggest that for two states with variables $P$, $V$ and $T$:

$$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$$

• Suppose you have two states which have nothing in common (that is, nothing is kept constant): $P_1,T_1,V_1$, and $P_2,T_2,V_2$. Well, you may just move from the first state to $P_1,T_2$ (that would be a constant pressure process which you know how to deal with), then from there to $p_2,T_2$ (that's a constant temperature process). – Ivan Neretin Sep 2 '16 at 9:40
• physics.stackexchange.com/questions/99347/… – JM97 Sep 2 '16 at 11:31
• I think the historical answer is that it was messy. We like to think of science as pushing back the darkness in a defined way but that just isn't what happens. It is usually more like "blind men describing an elephant." Then someone with sight comes along and explains that one has the tail, one has a leg, one has the trunk and one is rubbing the side. Then the pieces fit. – MaxW Sep 2 '16 at 15:32

It is indeed the case that the rearrangement of the different gas laws is not very satisfactory as it employs the different variables without any constraints. This issue has been addressed by S. Levine in J. Chem. Educ., 1985, 62 (5), p 399, but in the case you do not have access to this reference, I'll shortly repeat his arguments here.

As $P$, $V$ and $T$ are thermodynamic sate functions, we may write the total differential of $T$ as

$$\mathrm{d}T=\left (\frac{\partial T}{\partial V} \right )_P \mathrm{d}V + \left (\frac{\partial T}{\partial P} \right )_V\mathrm{d}P \equiv X\mathrm{d}V+ Y\mathrm{d}P$$

since this differential must be exact, the following relation must hold

$$\left (\frac{\partial X}{\partial P} \right )_V=\left (\frac{\partial Y}{\partial V} \right )_P$$

We can find simple functions satisfying this condition $$X=cP\text{ and }Y=cV$$ where $c$ is a constant that we may choose to be $c=1/R$ for later convenience. Substitution of these functions in the total differential gives $$\mathrm{d}T=\frac{1}{R}\left (P\mathrm{d}V + V\mathrm{d}P\right )$$ and integrating on both sides from 0K to T yields $$T-T_0=\frac{1}{R}\left [(PV)_T-(PV)_{T_0} \right ]$$ and since $T_0=0$ and $PV=0$ at $\pu{0K}$ we arrive at the ideal gas law.