Question

Q.41*
At room temperature, A, gas (vapour density = 40) at 1 atm pressure and B gas (vapour density = 10) at p
atm pressure are allowed to diffuse through identical pinholes from opposite ends of a glass tube of 1m
length and of uniform cross-section. The two gases first meet at a distance of 60 cm from the A end. The
value of p is​

Answer 1

Answer:

Pa/pb=da/db...

1/p=60/40

p=2/3

Answer 2

$1.3334 atm$is the value of p. Hence the pressure of the B gas is $1.3334 atm .$

Explanation:

The Pressure of A gas $=1 atm$

The pressure of B gas $=p$

The length of the glass tube $=1m=100cm$

Distance of the two gas from A end$=60 cm$

Distance of the gases from another end$=100-60=40cm$

According to Graham's law

$\[\frac{{{r}_{A}}}{{{r}_{B}}}=\frac{{{p}_{A}}}{{{p}_{B}}}\times \sqrt{\frac{{{M}_{A}}}{{{M}_{B}}}}\]$

The Pressure of gases is different and the temperature is constant.

Where $r=\frac{d}{t}$

the rate of diffusion is directly proportional to the distance traveled by  the gas.

The Molar mass of gas A $\[=2\times 40=80\]$

The Molar mass of gas B $\[=2\times 10=20\]$

Hence,

$\[\frac{{{d}_{A}}}{{{d}_{B}}}=\frac{{{p}_{A}}}{{{p}_{B}}}\times \sqrt{\frac{{{M}_{A}}}{{{M}_{B}}}}\]$

$\[\frac{60}{40}=\frac{1}{p}\times \sqrt{\frac{80}{20}}\]$

$\[ & p=\frac{4}{6}\times \sqrt{4} \\ \\ & p=\frac{2}{3}\times 2 \\\\ & p=1.3334 \\ \end{align}\]$atm

Hence p value is $1.3334 atm$.