91Ó°ÊÓ

1. Molar mass of second gas, $n_2=0.5\text{mol}$
2. Molar mass first gas, $n_1=2\text{mol}$
3. Molar mass of first gas is $M_1$
4. Molar mass of second gas is $M_2=3M_1$

Consider the formula for the gas law:

$p_i{v}_i=nR{T}_i$

Here, p is pressure, v is volume, T is temperature, R is universal gas constant and n is number of moles.

$p_1=\frac{n_{1}RT}{V}$

Ideal gas law is written as,

$pV=nRT$

For 2 mol of an ideal gas, it changes to,

$p_1{V}_1=n_{1}RT$

Pressure p₁ due to the first gas,

$p_1=\frac{n_{1}RT}{V}$

And for second gas,

$p_2{V}_2=n_{2}RT$

Pressure p₂ due to the second gas,

$p_2=\frac{n_{2}RT}{V}$

Therefore, the total pressure will be,

$$\begin{gathered} p=p_1+p_2=\frac{n_{1}RT}{V}+\frac{n_{2}RT}{V} \\ p=p_1+p_2=\left(n_1+n_2\right)\frac{RT}{V} \end{gathered}$$

The fraction due to the second gas is given as,

$$\begin{gathered} ∴\frac{p_2}{p}=\frac{n_{2}RT/V}{\left(n_1+n_2\right)RT/V} \\ ∴\frac{p_2}{p}=\frac{n_2}{\left(n_1+n_2\right)} \\ ∴\frac{p_2}{p}=\frac{\left(0.5\right)}{\left(2+0.5\right)} \\ ∴\frac{p_2}{p}=0.2 \end{gathered}$$

Hence,thefraction of the total pressure on the container wall which is attributable to the second gas is 0.2.