Question

A pot, whose volume is 2.461 L contains 0.3 mol N₂, 0.5 mol He and 6.2 mol O₂ at 27°C. Calculate the partial pressure of gases in mixture?

Answer 1

First Method

From ideal gas equation,

\(p = \frac{nRT}{V}\)

T 27 + 273 = 300 K 

R = 0.0821 L atm K^-1mol^-1 

V = 2.461 L

Partial pressure of Nitrogen: 

\(P'_{N_2} = \frac{0.3\times 0.0821\times 300}{2.461} = 3.002\ atm\)

Partial pressure of Helium 

\(P'_{He} = \frac{0.5\times 0.0821\times 300}{2.461}\)

= 5.004 atm

Partial pressure of Oxygen

\(P'_{O_2} = \frac{6.2\times 0.0821\times 300}{2.461}\)

= 62.050 atm 

Second Method:

Total number of moles in mixture 

0.3 + 0.5 + 6.2= 7 moles

Total pressure \('PT' = \frac{n\times R\times T }{V}\)

\( = \frac{7\times 0.0821\times 300}{2.461}\)

= 70.06 atm 

Partial pressure = Total pressure × mole fraction Now, partial pressure of Nitrogen 

P'N₂ = PT × Mole fraction of N₂ gas

\( = 70.06\times \frac{moles \ of\ N_2\ gas}{Tootal\ moles}\)

= 3.002 atm 

Partial pressure of Helium 

\(P_{He} = \frac{70.06\times 0.5}{7}\)

= 5.004 atm 

Partial pressure of Oxygen 

\(P'_{O_2} = \frac{70.06\times 6.2}{7}\)

= 62.050 atm