Question

O2 and He are taken in equal weights in a vessel.
The pressure exerted by Helium in the mixture is
1) 1/8 th of total pressure
2) 1/9th of total pressure
3) 2/9 th of total pressure
4) 8/9th of total pressure​

Answer 1

By Dalton's law of partial pressure:-

No. Of moles of He = m/ molar mass

= m/4

No. Of moles of O2 = m/32

Partial pressure exerted by He=

(m/4)/(m/4 + m/32)= 8/9

Hence option 4 is the right option

Answer 2

The pressure exerted by Helium in the mixture is 8/9th of total pressure​

Given : $O_{2}$ and He are taken in equal weights in a vessel.

To Find : The pressure exerted by Helium in the mixture is

1) 1/8th of total pressure

2) 1/9th of total pressure

3) 2/9th of total pressure

4) 8/9th of total pressure​

Solution : The pressure exerted by Helium in the mixture is 8/9th of total pressure​

We know that the partial pressure exerted by the helium gas is

P(He) = Mole fraction of Helium ($X_{He}$) × Total pressure ($P_{total$)

and Mole fraction of Helium is   number of moles of helium /Total number of moles

$X_{He}$ = $\frac{n_{He}}{n_{He}+n_{O_2}}$

So P(He) =   $\frac{n_{He}}{n_{He}+n_{O_2}}$ ×  $P_{total}$

Let m be the total mass of mixture

and number of moles = Total  Mass / Molecular Mass

$n_{He} = \frac{m}{4}$

$n_{O_{2}} = \frac{m}{32}$

P(He) = $\frac{\frac{m}{4} }{\frac{m}{4} +\frac{m}{32} }$  × $P_{total}$

P(He) = $\frac{\frac{m}{4} }{\frac{9m}{32} }$ × $P_{total}$

P(He) = $\frac{8}{9}$ × $P_{total}$

So the pressure exerted by Helium in the mixture is 8/9th of total pressure​.

So option 4) is correct.

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