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Derive Dalton’s law of practical pressures on the basis of kinetic theory of gases.

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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+. . …

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