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What is the final pressure of helium and neon gas mixture in this case?

Dhamnekar Winod

Active member
Nov 17, 2018
Gas laws Helium Neon.png

The answer given to the above question is Final pressure$=\frac{(10.0mol\times 5.00bar) + (5.00 mol \times 20.0 bar)}{(10.0 mol +5.00 mol)}= 10.0 bar$ Is this answer correct? if yes, How and why?

My question is while computing this answer, the volumes of each gas is not considered.

Note:- While computing the volumes of each gas, temperature is not considered as it is a constant.

The computed volume of helium gas $=\frac{(10.0mol \times 8.314472 J/mol\cdot K)}{5.00 bar}=0.16628944 Litres$

The computed volume of neon gas $= \frac{(5.00 mols \times 8.314472 J/mol\cdot K)}{20.0 bar}= 0.02078618 Litres$

How to use this additional information to compute the final pressure of the gas mixture?

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
We can treat noble gasses as ideal gasses, which means we can use the formula $p V = n R T$, which is independent of the actual type of gas.
Let the first container have $n_1$ moles, pressure $p_1$, and volume $V_1$.
Let the second container have $n_2$, $p_2$, and $V_2$.
Let the temperature be $T$, and let the final pressure be $p$.

Then we have:
\begin{cases}p_1V_1 = n_1 R T \\ p_2 V_2 = n_2 R T \\ p(V_1+V_2) = (n_1+n_2) R T \end{cases}
Since we do not know the volumes, we will eliminate them from the equations:
$$\begin{cases}V_1 = \frac{n_1 R T}{p_1} \\ V_2 = \frac{n_2 R T}{p_2} \\ p(V_1+V_2) = (n_1+n_2) R T \end{cases}
\implies p = \frac{(n_1+n_2) R T}{V_1+V_2} = \frac{(n_1+n_2) R T}{\frac{n_1 R T}{p_1}+\frac{n_2 R T}{p_2}}
=\frac{n_1+n_2}{\frac{n_1}{p_1}+\frac{n_2}{p_2}}= \frac{10.0+5.00}{\frac{10.0}{5.00}+\frac{5.00}{20.0}}=6.67\,\text{bar}$$

So it looks as if the given answer is incorrect.
The final pressure should be $p=6.67\,\text{bar}$.