Consider a mixture of gases occupying a volume V. Let m1, m2, m3 be the molecular masses of the gases and n1, n2, n3 be the number of their molecules, P1, P2, P3 … ..the pressure exerted by individual gases and v1, v2, v3 be the r.m.s velocities of the molecules of various gases.
According to kinetic theory,
\(P_1=\frac{1}{3}\frac{m_1n_1}{V}v_1^2,\)
\(P_2=\frac{1}{3}\frac{m_2n_2}{V}v_2^2,\)
\(P_3=\frac{1}{3}\frac{m_3n_3}{V}v_3^2\)
Adding,
P = P1+P2+P3
\(=\frac{1}{3}\frac{m_1n_1}{V}v_1^2+\) \(\frac{1}{3}\frac{m_2n_2}{V}v_2^2+\) \(\frac{1}{3}\frac{m_3n_3}{V}v_3^2\)
As the temperature of all the gases in the mixture is same, so their average K.E. will be equal that is,
\(\frac{1}{2}m_1v_1^2=\frac{1}{2}m_2v_2^2=\frac{1}{2}m_3v_3^2=.......=\frac{1}{2}mv^2\) (say)
or \(m_1v_1^2=M_2v_2^2=m_3v_3^2=.....=mv^2\)
Now,
P = P1 + P2 + P3+. . . .
\(=\frac{1}{3V}\)(n1+n2+n3+. . . . )mv2
= \(\frac{1}{3}\frac{mn}{V}v^2\)
Here n = n1 + n2 + n3+ = total no. of molecules in the mixture.
But \(\frac{1}{3}\frac{mn}{V}v^2=P\)
= total pressure exerted by the mixture.
It proves Dalton’s Law of particular pressure.
∴ P = P1 + P2 + P3+. . …