Consider a mixture of non-interacting ideal gases with n_{1} moles of gas 1, n_{2} of gas 2 and so on

Gases are enclosed in a container with volume V, temperature T and pressure P.

Equation of state of mixture
PV = (n_{1}+ n_{2}) RT
or P = n_{1}RT/V + n_{2}RT/V + - - - -
= P_{1} + P_{2} + - - --
where,
P_{1} = n_{1}RT/V
is pressure the gas 1 would exert at same V and T if no other gases were present in the enclosure. This is known as law of partial pressure of the gases.

The total pressure of mixture of ideal gases is sum of partial pressures of individual gases of which mixture is made of.

Partial pressure in terms of mole fraction
Here
Total number of moles of gas
n=n_{1} + n_{2} + n_{3} ….
Total Pressure
P=n RT/V
Partial Pressure
P_{1} = n_{1} RT/V
Therefore
P_{1} = (n_{1} /n) P
Now n_{1} /n = Mole Fraction of Gas 1(χ_{1})
P_{1} = χ_{1} P
Similarly
P_{2} = χ_{2} P

Kinetic Theory of an ideal gas

Following are the fundamental assumptions of kinetic theory of gases.

Gas is composed of large number of tiny invisible particles know as molecules

These molecules are always in state of motion with varying velocities in all possible directions.

Molecules traverse straight line path between any two collisions

Size of molecule is infinitely small compared to the average distance traverse by the molecules between any two consecutive collisions.

The time of collision is negligible as compared with the time taken to traverse the path.

Molecules exert force on each other except when they collide and all their molecular energy is kinetic.

Intermolecular distance in gas is much larger than that of solids and liquids and the molecules of gas are free to move in entire space free to them.

Question
What will be the minimum pressure required to compress 500 dm^{3} of air at 1 bar to 200 dm^{3} at 30^{o} C. Answer
P_{1} V_{1} = P_{2} V_{2}
1 ×500 = P × 200
P=2.5 bar Question
A vessel of 120 ml capacity contains a certain amount of gas at 35^{o }C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 ml at 35^{o} C. What will be the pressure? Answer
P_{1} V_{1} = P_{2} V_{2}
120 ×1.2= 180×P
P=.8 bar

Question
At 0^{0} C the density of gaseous oxide at ‘2’ bar is same as that of nitrogen at 5 bars. What is the molecular mass of oxide? Answer
We know that
PM= d RT
At constant T. R. n and d
P M = constant
P_{1}M_{1} = P_{2} M_{2}
2 M_{1} = 5 28
M_{1} = 70