Question

calculate the total pressure( in ATM) of a mixture of 0.02 Mol of He(g) and 0.01 mole of H2(g) in 3 L flask at 127 °c​

Answer 1

Answer:

The volume of the gas depends upon the number of moles only. They do not depend on the nature of the gas.

The mixture of gases have ,(0.2+0.3+0.4)= 0.9 mole.

The volume of the gaseous mixture at STP= 0.9 x 22.4= 20.16 litres.

The gaseous mixture is at

Temp= T=25°C=298K

Volume V =10 liters

Pressure=? At STP,

P° =1 atm

V°=20 .16 liters

T°=273

Now PV/T=P°V°/T°

So P= P° V°/T° x (T/V)

= (1 x 20.16 x 295)/(273 x 10)= 2.178 atm.

Partial Pressure of H2= 2/9 x 2.178=0.4356

Partial Press of N2=3/9 x 2.178= 0.6534

Partial Pressure of CH4=4/9 x 2.178=0.8712

The partial pressures of H2, N2 and CH4 are respectively 0.4356,0.6534 and 0.8712 atm