Ideal Gas Law Multiple Gases Mixed Together (Ex5)
You may have a mixture of gases that are mixed together, and this mixture of gases might have multiple chemical species. Each chemical species in a mixture will have its own pressure.
There can be a single “total pressure” that can be calculated, but that total pressure will be the sum of all the partial pressures, where the partial pressures are the pressures exerted by each chemical species (this is called Dalton’s Law).
Here’s a practice example we made, where multiple gases and vessels are mixed together:
Example problem: A 1 L volume bulb of methane at 10 kPa is connected to a 3 L volume bulb of helium gas at 20 kPa. Find the total pressure afterwards, and also find the mole fractions of hydrogen and methane after the mixing. Use the ideal gas law.
When you have a “change in situation” you can equate two ideal gas laws at once. You can think of the left-side as “before combining” and the right-side as “after combining”.
\[\frac{{{P_1}{V_1}}}{{nRT}} = \frac{{{P_2}{V_2}}}{{nRT}}\]
How did I get this above?
You need to consider one of the species at a time for the equation above.
The volume changes because you’re connecting two volumes at once, which means the pressure will also change because pressure and volume are inversely related.
BUT, considering one species, the number of moles n will NOT change. R will also always be constant and can be cancelled out.
Temperature WON’T change here because it’s just inert gases, there’s no reaction no burning, etc. Cancelling out, the above expression is greatly simplified.
\[{P_1}{V_1} = {P_2}{V_2}\]
And you use this equation twice: once for methane, and once for hydrogen.
These will get you the partial pressures, which can then be summed to provide the total pressure after the mixing.
So for V2, which is volume 2, add 1 L and 3 L to get 4 L.
Volume 1 value is dependent on which chemical species we’re using. Methane is initially at 1 L, while helium is initially at 3 L.
For methane:
\[\begin{array}{l}(10kPa)(1L) = {P_2}(4L)\\{P_2} = 2.5kPa\end{array}\]
For hydrogen:
\[\begin{array}{l}(20kPa)(3L) = {P_2}(4L)\\{P_2} = 15kPa\end{array}\]
And sum together the partial pressures from each chemical species to get the total pressure:
\[{P_{total}} = 2.5kPa + 15kPa = 17.5kPa\]
That’s done. To get the mole fractions, you can just use the partial pressures over the total pressure.
\[\begin{array}{l}{x_{C{H_4}}} = \frac{{2.5kPa}}{{17.5kPa}} = 0.143\\{x_{{H_2}}} = \frac{{15kPa}}{{17.5kPa}} = 0.875\end{array}\]
And to check your mole fractions are correct, you know that all of the mole fractions will sum up to 1.