Question

A box of $1$ $L$ capacity is divided into two equal compartments by a thin partition which are filled with $2g$ ${H_2}$ and $16g$ $C{H_4}$ respectively. The pressure in each compartment is recorded as $P$ $atm$ The total pressure when partition is removed will be:
(A) $P$
(B) $2P$
(C) $\dfrac{P}{2}$
(D) $\dfrac{P}{4}$

Answer 1

According to Dalton’s law of partial pressure the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each of the individual gases.

Formula used:
${P_{total}} = {P_1} + {P_2} + ......{P_n}$
No. of moles $= \dfrac{\text{given mass}}{{molar mass}}$
Mole fraction ${X_i} = \dfrac{{{n_i}}}{n} = \dfrac{{{P_i}}}{P}$
${P_i} = {X_i} \times P$
Where ,
${X_i} =$ Mole fraction of individual gas component
${n_i} =$ No of moles of individual gas component
$n =$ Total no of moles of the gas mixture
${P_i} =$ Partial pressure of the individual gas component
$P =$ Total pressure of the gas mixture.

Complete step by step solution:
According to Dalton’s law of partial pressure, in a mixture of non-reacting gases the total pressure exerted is equal to the sum of each of the individual gases.
Here it is given the total volume of the containers as $1$ $L$ also the container is divided in to two partition where each partition is filled with $2g$ ${H_2}$ and $16g$ $C{H_4}$ respectively.
First, we have to calculate the moles of hydrogen as well as methane
Molar mass of hydrogen= $1g/mole$
Molar mass of methane= $16g/mole$
Moles of hydrogen= given mass of hydrogen/molar mass of hydrogen = $\dfrac{{2g}}{{2g/mole}} = 1mole$
Moles of methane= given mass of methane/molar mass of methane = $\dfrac{{16g}}{{16g/mole}} = 1mole$
Now we have to calculate the mole fraction of hydrogen as well as methane,

Mole fraction of hydrogen $= \dfrac{{{n_{{H_2}}}}}{{{n_{{H_2}}} + {n_{C{H_4}}}}}$
${X_{{H_2}}} = \dfrac{1}{{1 + 1}} = \dfrac{1}{2}$
Mole fraction of methane $= \dfrac{{{n_{C{H_4}}}}}{{{n_{C{H_4}}} + {n_{{H_2}}}}}$
${X_{C{H_4}}} = \dfrac{1}{{1 + 1}} = \dfrac{1}{2}$
Now we have to calculate the partial pressure of hydrogen and methane,
Partial pressure of hydrogen $= {X_{{H_2}}} \times {P_T}$
${P_{{H_2}}} = \dfrac{1}{2} \times P = \dfrac{P}{2}$
Partial pressure of methane $= {X_{C{H_4}}} \times {P_T}$
${P_{C{H_4}}} = \dfrac{1}{2} \times P = \dfrac{P}{2}$
Now we have to calculate the total pressure when the partition is removed,
${P_T} = {P_{{H_2}}} + {P_{C{H_4}}}$
${P_T} = \dfrac{P}{2} + \dfrac{P}{2} = P$
Therefore, the total pressure of the system when the partition is removed will be $P$ $atm$ .
$\therefore$ Option A is the correct answer.

Note:
In ideal condition the ratio of partial pressure equals the number of moles. Also, the mole fraction ${X_i}$ of an individual gas component in an ideal gas mixture can be expressed in terms of the components partial pressure or moles of the component.