Question

A mixture of He, Ne, and N2 gases are a pressure of 1.348. If the pressures of He and Ne are 0.124 atm, what is the partial pressure of N2 in the mixture ?

Answer 1

The Best Answer: 1 - (.47+.23) = 0.30

If Ne has a mole fraction of 0.47 (or 47/100) and Ar is 0.23, then H2(or He) has a mole fraction of 0.30

This means the gas mixture is 30/100 H2(or He).

7.85 x 0.30 = 2.355 atm

Answer 2

Answer :  The partial pressure of $N_2$ in the mixture is, 1.224 atm

Solution :

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.

$P_T=p_{He}+p_{Ne}+p_{N_2}$

where,

$P_T$ = total partial pressure of $He,Ne\text{ and }N_2$ = 1.348 atm

$P_{He}$ = partial pressure of helium

$P_{Ne}$ = partial pressure of neon

$P_{N_2}$ = partial pressure of nitrogen

As we are given that,

$P_{He}+P_{Ne}=0.124atm$

Now put all the given values in above expression, we get the partial pressure of the nitrogen gas in the mixture.

$1.348atm=0.124atm+p_{N_2}$

$p_{N_2}=1.224atm$

Therefore, the partial pressure of $N_2$ in the mixture is, 66 Kpa