After doing a long winded derivation about collisions against walls, one ends up with the equation

$$ p= \frac{Nmv^2}{3V}$$

As an expression for pressure of an ideal gas

Now say you have a mixture of two gases then according to my textbook, the average kinetic energy of each gas should be equal to each. Ok, so keeping this premise in mind how would the previous equation we derived under the premise of a single gas in container change?

My attempts:

I know of daltons law and I know that pressures will add additively but I'm not sure of the kinetic energy do we just add both the kinetic energies?

Then it'll be $$p_1 + p_2 = \frac{2Nmv^2}{3V}$$


1 Answer 1


First, you find the pressure by each gas molecules. Following Dalton, as you mention, then sum up them. They have different mass, velocity, and the number of molecules as well.

$$ p=p_1+p_2= \frac{N_1m_1v_1^2}{3V}+\frac{N_2m_2v_2^2}{3V}$$

  • $\begingroup$ If the mixture of gasses is at thermal equilibrium, then the average mv^2 is the same for both species of gas, and the N/V is the density of each gas. Yes, the pressures add. $\endgroup$
    – R.W. Bird
    Commented Apr 19, 2020 at 19:29

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