Question

A gas mixture contains nitrogen and helium in 7:4 ratio by weight. The pressure of the mixture is 760mm. The partial pressure of Nitrogen is

Answer 1

Partial pressure of a gas is defined as the pressure that would be exerted by one of the gases in a mixture if it occupied the same volume on its own in given area.

Dalton's Law of partial pressures: 

Dalton's law states that; the total pressure of a mixture of gases, is the result of the sum total of the individual gases that make up the mixture.

Therefore, in the question above, partial pressure of the helium plus that of nitrogen = 760 mm-Hg

That is;
   
         PHe₂  + PN₂   =  760 mm-Hg

The partial pressure of gases is related to the pressure of the mixture and there mole fractions.

Calculate the mole fractions:

let use 7g of nitrogen and 4 g of helium that are in the ratio 7:4.

Find moles of 7 grams of Nitrogen:

moles = mass/molar mass   =  7g/14gmol⁻
                                             = 0.5moles

Find moles of    He
  mass/ molar mass   = 4g/4gmol⁻
                                  = 1 mole

Mole ratio = 0.5:1 
                 = 1:2

Relationship between mole ratio and partial pressure according to Dalton's law:

Mol₁/mol₂ = P₁/P₂ =P₁/total pressure=P₂/total pressure

Therefore,    PN₂/ total pressure = mol₁/mol₂


    PN₂/760mm-Hg = 1/2

Therefore PN₂  = 380mm-Hg

The partial pressure of nitrogen thus is 380mmHg

Answer 2

Answer:answer will be 152mm or 0.2atm

Explanation:firstly, for partial pressure of N2 the formula is

Pa=Pt * mole fraction

moles of N2 = 7/28=1/4

moles of he = 4/4= 1

mole fraction of N2 = 1/4/1/4+1 = 1/5

Partial pre. of N2 = 760 * 1/5 = 152 mm

1 atm = 760 mm of hg

so 152 mm hg will be 0.2 atm

hope it helps