Question

Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape?
A. \[\dfrac{1}{8}\]
B. \[\dfrac{1}{4}\]
C. \[\dfrac{3}{8}\]
D. \[\dfrac{1}{2}\]

Answer 1

Hint: According to Graham’s Law, the rate of effusion of gas is inversely proportional to the square root of molar mass of gas.
\[Rate_{diffusion}\propto \dfrac{1}{{\sqrt {density} }}\]
Since volumes of various gases contain a similar number of particles, the quantity of moles per liter at a given T and P is steady. Thus, the thickness of a gas is legitimately relative to its molar mass (MM).
\[Rat{e_{diffusion}}\propto \dfrac{1}{{\sqrt {MM} }}\]

Complete step by step answer:
Let the number of moles of each gas = x
Fraction of hydrogen escaped = \[\dfrac{1}{2}x\]
\[\dfrac{{{n_O}{t_H}}}{{{n_H}{t_O}}} = \dfrac{{\sqrt {{M_H}} }}{{\sqrt {{M_o}} }}\]
\[\dfrac{{{n_O}}}{{\dfrac{1}{2}}} = \dfrac{{\sqrt 2 }}{{\sqrt {32} }}\]
Therefore, \[{t_O}\]= \[{t_H}\] and \[{n_H}\] = \[\dfrac{1}{2}\]
$N_o$ = \[\dfrac{1}{2} \times \dfrac{1}{4} =\dfrac{1}{8}\]
The part of the oxygen that escapes in the time needed for one-portion of the hydrogen to escape is\[\dfrac{1}{8}\]
Here \[{n_O}\] represents the number of moles of oxygen effused in time to and \[{M_O}\] ​ is the molar mass of oxygen.
Similarly, \[{n_H}\] ​ represents the number of moles of hydrogen effuses in time \[{t_H}\]and \[{M_H}\] ​ is the molar mass of hydrogen. According to our question, the correct answer is \[\dfrac{1}{8}\] as Assume 1 mole of hydrogen and 1 mole of oxygen gases are placed in a container with a pin-hole through which both can escape.

Explanation of Graham’s Law:
Graham's Law which is famously known as Graham's Law of Effusion, was detailed by Thomas Graham in the year 1848. Thomas Graham explored different avenues regarding the radiation cycle and found a significant component: gas particles that are lighter will travel quicker than the heavier gas molecules. The pace of emission of a vaporous substance is contrarily corresponding to the square base of its molar mass. This relationship is alluded to as Graham's law. The proportion of the emanation paces of two gases is the square base of the converse proportion of their molar masses:
\[\dfrac{{rate\, of\, effusion}}{{rate \,of\, effusion}} = \dfrac{{\sqrt {{M_B}} }}{{\sqrt {{M_A}} }}\]
Where \[{M_A}\] is the molar mass of gas 1
\[{M_B}\] is the molar mass of gas 2

Note:
In this question we have given, a number of moles of hydrogen and \[{n_{{H_2}}} \] that of oxygen are \[ {n_{{O_2}}} \] not equal. Gas particles are in a condition of steady movement. Thus, they intermix with one another to shape a homogeneous combination. Dissemination is a cycle by temperance of which at least two gases intermix with one another, autonomous of inclination toward structure a homogeneous blend.