Question

At room temperature, ammonia gas at 1 atm pressure and HCl gas at pressure P atm are allowed to effuse through identical pin holes from opposite ends of a glass tube of 1 meter length and uniform area of cross section. ${\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{Cl}}$ is first formed at a distance of 60 cm from the end through which HCl gas was sent in. Calculate the value of P (in atm). (Write your answer to the nearest integer)

Answer 1

We know that diffusion is the process of mixing of gases irrespective of the density relationship and without the effect of external energy. Here, we have to use the formula of rate of diffusion of two gases namely a and b, $\dfrac{{{r_a}}}{{{r_b}}} = \dfrac{{{p_a}}}{{{p_b}}}\sqrt {\dfrac{{{M_b}}}{{{M_a}}}}$.

Complete step by step answer:

Let’s first derive the formula of rate of diffusion.

We know that rate of diffusion is directly proportional to pressure and inversely proportional to square of molar masses. So,

$r\alpha p$ …… (1)

$r\alpha \dfrac{1}{{\sqrt M }}$ …… (2)

On combining equation (1) and (2), we get,

$r\alpha \dfrac{p}{{\sqrt M }}$

$r = k\dfrac{p}{{\sqrt M }}$

So, form two gases (a and b) rate of diffusion is,

For gas a, ${r_a}\alpha \dfrac{{{p_a}}}{{\sqrt {{M_a}} }}$ …… (3)

For gas b, ${r_b}\alpha \dfrac{{{p_b}}}{{\sqrt {{M_b}} }}$ …… (4)

Dividing equation (3) by (4), we get

$\dfrac{{{r_a}}}{{{r_b}}} = \dfrac{{{p_a}}}{{{p_b}}}\sqrt {\dfrac{{{M_b}}}{{{M_a}}}}$

Now, come to the question. Two gases namely hydrochloric acid (HCl) and ammonia $\left( {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right)$
are effused at different pressure and constant temperatures. So, the rate of diffusion of ammonia and hydrochloric acid is,

$\dfrac{{{r_{{\rm{HCl}}}}}}{{{r_{{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}} = \dfrac{{{p_{{\rm{HCl}}}}}}{{{p_{{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}}\sqrt {\dfrac{{{M_{{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}}{{{M_{{\rm{HCl}}}}}}}$
…… (5)

Here, pressure of HCl gas is given P and pressure of ammonia gas is given as 1 atm.

The molar mass of ${\rm{N}}{{\rm{H}}_{\rm{3}}} = 14 + 3 = 17\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$
and the molar mass of ${\rm{HCl}} = {\rm{1 + 35}}{\rm{.5}} = {\rm{36}}{\rm{.6}}\;{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$
.

Given that the length of the glass tube is 1 metre or 100 cm. The ammonium chloride is produced at a distance of 60 cm from the end through which HCl was sent in. So, the rate of diffusion of HCl is 60.

Rate of diffusion of ammonia (${\rm{N}}{{\rm{H}}_{\rm{3}}}$
)=$100 - 60 = 40\,{\rm{cm}}$

Now, we have to put all the above values in equation (5).

$\dfrac{{60}}{{40}} = \dfrac{P}{1}\sqrt {\dfrac{{17}}{{36.5}}}$

$\Rightarrow P = \dfrac{{60}}{{40}} \times \sqrt {\dfrac{{36.5}}{{17}}}$

$\Rightarrow P = \dfrac{{6.041}}{{4.12}} \times \dfrac{{60}}{{40}}$

$\Rightarrow P = 2.19\;{\rm{atm}}\,$

Hence, the value of P is 2 atm.

Note:
Always remember that effusion is a particular case of diffusion as a result of which a gas present in a container is allowed to escape through a small hole or aperture. Laws of diffusion are also applicable to effusion.