Question

A mixture containing 1.6 g of O 2 ,1.4 g of N 2 , and 0.4 g of He is occupies a volume of 10 litre at 27◦ C .calculate the total pressure of the mixture and partial pressure of each compound.

Answer 1

Explanation:

PV = nRT = 10 litre T = 27

0C = 300 K

nHe = 0.4/4=0.1

N 2= 1.4/28=0.05

O 2 = 1.6/32=0.05

So the total number of moles = 0.1 + 0.05 + 0.05 ×0.2

P = 0.492 atm

Partial Pressure = Total Pressure× mole fraction

P He = 0.492 ( 0.05/0.2 )

= 0.246 atm is the partial pressure of He gas in then cylinder.

P N 2= 0.492 ( 0.05/0.2)

= 0.123 atm is the partial pressure of nitrogen gas in then cylinder.

PO2 = 0.492 ( 0.05/0.2)

= 0.123 atm is the partial pressure of oxygen gas in then cylinder.

√\______Anushka❤❤