# Do all ideal gases have the same elasticity when compared at the same barometric pressure?

By using the equation of state for an ideal gas (aka ideal gas law)

$$PV = mRT$$

with gas volume bounded by a rigid container, one can derive an expression for the compliance (reciprocal of elasticity) of the gas.

Let $V = V_{container}$ be the fixed volume of the container and the mass $m$ is what changes. In other words we are dealing with an open system. $R$ is the specific gas constant. Solving for $m$

$$m = \frac{PV}{RT}$$

Assume that at the initial state $m_i$ is the initial mass of the gas in $V_{container}$ and that it’s at pressure $P$ at equilibrium with the ambient gas pressure at $P_{ambient}$ of density $\rho_{ambient}$. Then introduce a change in mass $\Delta m = m_{final} – m_{ambient}$

$$m_{final} – m_{ambient} = \frac{P_{final} V_{container} }{RT} - \frac{P_{ambient}V_{container}}{RT}$$

$$m_{final} – m_{ambient} = [P_{final} - P_{ambient}]\frac{ V_{container} }{RT}$$

Express the change in mass in terms of volume introduced at ambient pressure Since

$$m=V\rho$$

$$[V_{final} – V_{ambient} ]\rho_{ambient}= [P_{final} - P_{ambient}]\frac{ V_{container} }{RT}$$ Divide both sides by $\rho_{ambient}$

$$[V_{final} – V_{ambient} ]= [P_{final} - P_{ambient}]\frac{ V_{container} }{\rho_{ambient}RT}$$ But $\rho_{ambient}RT = P_{ambient}$ And so $$\Delta V= \Delta P\frac{V_{container}}{P_{ambient}RT}$$

like $$\frac{\Delta P}{\Delta V} = \frac{P_{ambient}}{V_{container}}= E$$ which is sometimes referred to as the elasticity of the gas since it can behave in the same manner as a spring if additional incremental mass is either added or removed from the container; there is a resemblance, an analogy to a linear spring equation $$\Delta P= E \Delta V$$

where volume change is like displacement and change in pressure like force.

But in deriving this expression it seems that the composition of the gas plays no role in determining this constant of elasticity, only the size of the container and the initial pressure.

But (at least my own nagging) intuition says that composition of the gas should make a difference in the 'springiness', the elasticity of the gas. Say for example a one liter bottle of helium (monotomic) vs a one liter bottle of nitrogen (diatomic). Avogadro says same number of particles, and the derivation above the same elasticity.

But I would expect the two experiments to have different elasticity.

I'm assuming an adiabatic process, so no heat created or lost.

So is my intuition wrong; is the elasticity independent of the specific gas?

• No idea if it goes anywhere, but if you say $PV = nRT \rightarrow \Delta P \Delta V = nRT$ and use $\Delta P/ \Delta V = E$, you end up with $E = nRT/\Delta V^2$ which shows that the composition of the gas does come into play through $R$. In your equations you've written, there is no appeal to the equation of state. – tpg2114 Feb 27 '17 at 23:21
• I think the issue/confusion is that you give: $\frac{\Delta P}{\Delta V} = \frac{P_{ambient}}{V_{container}}= E$ which implicitly assumes that at zero volume, there is zero pressure (that's why the $\Delta$'s went away I'm guessing). But at zero volume, pressure isn't really defined (and neither is temperature) and so I don't think that's a good reference point to pick. Instead, if you took a container at ambient conditions and changed the volume by a $\Delta V$ while measuring the $\Delta P$, you would find it changes based on the gas. – tpg2114 Feb 27 '17 at 23:26
• @tpg2114 it can be confusing, but turns out the $\Delta V$ in my expression comes about from the $n$ factor rather than the $V$ which winds up being the fixed container volume. It's an open system. I just translate the mass back into volume based on the ambient gas density. That's the change in volume. – docscience Feb 27 '17 at 23:54
• @tpg2114 at zero pressure you have zero molecules in the container. Well before that point I suspect the ideal gas law does not apply any more, and for really high pressures you have to introduce the compressibility factor, $Z$ into the model. So yes I'm considering more moderate pressure conditions near earth's atmospheric pressure. – docscience Feb 27 '17 at 23:56
• @tpg2114 to better convey, I've added the details of my derivation. Note that R is eliminated, so no specificity to gas. – docscience Feb 28 '17 at 0:39