# Addition of inert gas at “constant volume”?

My textbook states that

"The addition of inert gases at constant volume does not affect the equilibrium state of the reactants and products contained in that volume."

Although I am not questioning the truth behind this statement, I am very confused by it.

Let us consider a homogeneous gaseous equilibrium system contained within a closed cylinder. The cylinder has a very small hole on its side, through which an inert gas can be pumped in. This is how I visualize the addition of an inert gas to the system. Now, upon introducing a given amount of inert gas into the system, how does the volume remain constant in any way?

While deriving Van der Waals equation for real gases, I learned that the term $V$ in the ideal gas equation stands for the empty space that is available to the gas molecules for movement. Therefore, when we introduce an inert gas into the system, aren't we essentially decreasing this amount of space that is available to the reactant and product molecules for movement?

I am very confused. Please share your insights for it would be tremendously helpful for me. Thanks ever so much in advance :) Regards.

## 2 Answers

The inert gas that you add does not act like a piston, denying part of the volume to the other molecules in the mixture. For a constant volume system comprised of an ideal gas mixture, the partial pressures of the reactants and products do not change when you add the inert gas. It only increases the total pressure, and that change is only because of its own partial pressure. However, if you are not operating in the ideal gas regime, the addition of the inert gas will affect the equilibrium.

• OK, so what you're saying is that according to the assumptions of KTG for ideal gases, the volume of the reaction mixture will remain the volume of the container since the molecules of the inert gas have negligible volumes themselves, yeah? – user33789 Aug 31 '16 at 13:48
• Additionally, if the reaction mixture is at constant pressure instead of constant volume, these inert gas molecules will exert pressure on the piston, moving it up and hence, increasing the volume, yeah? While studying Le Chatelier's principle for changes in volume, I didn't relate the volume change to pressure change. Instead, I imagined that on decreasing the volume, the equilibrium shifted in that direction which produces lesser number of molecules so that the volume would increase and not so pressure would decrease. I am wrong to think of it like this, right? – user33789 Aug 31 '16 at 13:53
• I agree to your first comment. I don't understand your second comment. – Chet Miller Aug 31 '16 at 19:48

You should not :)

"does not affect the equilibrium state" means that the thermodynamical constant is still the same. If you add an inert gas $\ce{G}$ in your system it will not react with other chemical coumponds.

For example at the beginning you have $\ce{A} \rightleftharpoons \ce{B}$ and after you add you inert gas you have $\ce{A + G} \rightleftharpoons \ce{B + G}$. So it not afftects the value of $\ce{K_{eq}}$ because $\ce{K_{eq}}$ depends only of $\ce{T}$ the temperature.

If you use a solid container the volume will not change but the pressure will.

Also we have $$\ce{\mu_i(T,P_i)=\mu_i^°(T)}+\ce{RT\ln\frac{P_i}{P^°}}$$

Then $$\left(\frac{\partial\Delta_r\ce{G}}{\partial n_{inert}}\right)_{\ce{T,V}}=0$$

If you prefer calculus.