# Partial Pressure Mole

For a mixture of gases in a container of specific volume, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone. These individual pressures exerted by gases in the gaseous mixture that contributes are the partial pressures and the total pressure is expressed as follows:

total= P1+P+ ...Pn

where P1, P2, P3,…is referred to as partial pressure.

total = P1  +  P2  +  P3 + …= n1RT/V  +  n2RT/V  +  n3RT/V + ….
= (n1+n2+n3+…)(RT/V)
=  n total (RT/V)

Thus, for a mixture of ideal gases, it is the total number of moles of particles that is important.  The pressure exerted by an ideal gas is not affected by the identity or composition of the gas particles-so volume of the individual gas particle is not important and the forces among the particles must not be important.

The ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture called the mole fraction can be represented in terms of pressure.

Mole fraction (χ1) = n1/ntotal =  {P1(V/RT)}/{P1(V/RT) + P2(V/RT)+ P3(V/RT) + …}

=P1/P TOTAL

Let us discuss some examples on mole fraction in terms of partial pressure.

## Examples of Mole Fraction in Terms of Partial Pressure

• The partial pressure of argon gas, making up 40% of a mixture, is 43.34 kPa. What is the total pressure of the mixture in kPa?

The generalization of this law is: P total= P1+P2 + ...Pn

We need to calculate the mole fraction of Argon (Ar) from the information given.

P Ar = mole fraction Ar  ×  P total

P Ar = 43.34 kPa (as given)

Mole fraction Ar = χ Ar = 40 moles of Ar/ 100 moles of gas mixture

χ Ar = 0.4

So, P total = P Ar/ mole fraction Ar
= 43.34 kPa/ 0.4
= 108.35 kPa

• A gas mixture is 12% Ne, 23% He, and 65% Rn. If the total pressure is 116 kPa, what is the partial pressure of each gas?

This uses the same concept as the previous question.

The mole fractions of each gas:

χ Ne = 12/100= 0.12

χ He = 23/100 = 0.23

χ Rn = 65/100 = 0.65

Partial Pressures of the gases(moles):

P Ne = χ Ne  × P T
= 0.12  × 116
= 13.92 kPa

P He = χ He  × P T
= 0.23  × 116
= 26.68 kPa

P Rn = χ Rn  × P T
= 0.65 × 116
= 75.4 kPa

• A mixture of nitrogen and carbon dioxide is at a pressure of 1.00 atm and a temperature of 278K. If 30% of the mixture is nitrogen, what is the partial pressure of carbon dioxide?

P T = 1 atm

χN2 = 30/100
= 0.3

P N2 = 0.3 × 1
= 0.3 atm

So,

P CO2 = P T – P N2
= 1 atm – 0.3 atm
= 0.7 atm

So the partial pressure of CO2 is 0.7 atm.