Dependence of average speed of molecules of gaseous mixture

We know that the average speed of gases in a single gas chamber is given by $\sqrt{8RT/\pi M}$

where R is universal gas constant,T is temperature,M is molar mass of gas.

But what if we mix two gases in any ratio say 1:1 and then try to find average speed of anyone of the gases.

Will the both gases have have same average speed or different?If same ,how will it be calculated and if different will it be given by same above formula?

• Just to clarify: suppose we call the gases $A$ and $B$, are you asking if the average velocity of the molecules of $A$ is the same in the mixture as it would be in pure $A$ (and likewise for the molecules of $B$)? If so, the answer is yes. Commented Jan 26, 2016 at 12:49
• yes i am asking the same but why doesn't it change as in the mixture there will be more bumping with other molecules? Commented Jan 26, 2016 at 13:04

If the two gasses are mixed then they will reach the same temperature, with the lighter molecules moving faster than heavier ones.

In the initial equation, 8, R, T and pi would all be the same. Since the gas molecules have different masses the speeds must therefore be different.

If a tank hits my car side on, my car goes flying. If my car hits a tank side on, the tank hardly moves. Its high mass causes its velocity to change little. So it's normal that the lighter molecules get banged around faster when gasses are mixed.

The parameter which is important is the average kinetic energy of the molecules.
When the two gases mix they will move to a state where the average kinetic energy of all the molecules is the same.
So go for something like the total kinetic energy before mixing is equal to the total kinetic energy after mixing which will be made easier by having a 1:1 mixture. Since kinetic energy depends on the square of velocity the parameter which is most often used when mentioning the speed of the molecules is the root mean square velocity.

• but what about my question's answer? Commented Jan 26, 2016 at 11:26
• maybe it could be useful to note that the everage kinetic energy is $\left\langle \text{KE}\right\rangle=\frac{n_1}{n_1+n_2}\left\langle\frac{1}{2}m_1 v_1^2\right\rangle+\frac{n_2}{n_1+n_2}\left\langle \frac{1}{2}m_2v_2^2\right\rangle$, where $n_i$ is the number of molecules of each species. Using $\left\langle\text{KE}\right\rangle\propto kT$ it could be possible in principle to solve for each $\langle v_i^2\rangle$. Commented Jan 26, 2016 at 11:31

If the two gases are NOT interacting, then you can treat them independently and use your formula for each of them, weighting the result according to the mean speed, using your formula, will be

$$v=f \sqrt{8RT/\pi M_1}+(1-f)\sqrt{8RT/\pi M_2}$$

where $f$ is the fraction of gas number 1 and $M_1$ and $M_2$ the two molar masses - so you can solve it easily. However, in general, if the two gases are interacting you might a different formula for the mean speed of each of them, or, in other words, your formula for the mean speed is not valid anymore and depends on the interaction itself.