Question

What will be the total pressure when partition is removed if a box of 1 – liter capacity is divided into two equal compartments by thin partition which is filled with $6g$ of ${H_2}$ and $16g$ $C{H_4}$ respectively and pressure in each compartment is recorded as $P$ atm?
A.$P$
B.$2P$
C.$\dfrac{P}{2}$
D.$\dfrac{P}{4}$

Answer 1

Calculate the number of moles of ${H_2}$ and $C{H_4}$ respectively. Then, find the pressures for ${H_2}$ and $C{H_4}$ in the terms of the pressure in each compartment and then use Dalton's law of partial pressure.

Complete step by step answer:
In the question, it is given that, box of 1 – liter capacity is divided into two equal compartments by thin partition is filled with $6g$ of ${H_2}$ and $16g$ $C{H_4}$ respectively. Therefore, finding the number of moles of ${H_2}$ and $C{H_4}$ -
Let the number of moles of ${H_2}$ be ${n_{{H_2}}}$ and number of moles of $C{H_4}$ be ${n_{C{H_4}}}$.
$\therefore {n_{{H_2}}} = \dfrac{6}{2} = 3$ , and
$\Rightarrow {n_{C{H_4}}} = \dfrac{{16}}{{16}} = 1$
Now, calculating the pressure for ${H_2}$ and $C{H_4}$ in the terms of the pressure in each compartment –
Let the pressure for ${H_2}$ and $C{H_4}$ be ${P_{{H_2}}}$ and ${P_{C{H_4}}}$ respectively.
We know that, from question, pressure in each compartment is recorded as $P$ atm –
$\therefore {P_{{H_2}}} = \dfrac{3}{{3 + 1}} \times P = \dfrac{3}{4}P \\\ \Rightarrow {P_{C{H_4}}} = \dfrac{1}{{3 + 1}} \times P = \dfrac{P}{4} \\\$
Now, use the Dalton’s Law of partial pressure, which states that, “total pressure exerted by the mixture of gases is equal to the sum of partial pressure of each individual gas present”. It is used for the determination of pressure of dry gas and for the calculation for partial pressures.
Let the total pressure exerted on the gas be ${P_{total}}$.
Therefore, the total pressure exerted can be calculated as –
${P_{total}} = {P_{{H_2}}} + {P_{C{H_4}}}$
Putting the partial pressure values of ${H_2}$ and $C{H_4}$ respectively in the above equation, we get –
${P_{total}} = \dfrac{3}{4}P + \dfrac{P}{4} \\\ \Rightarrow {P_{total}} = \dfrac{{3P + P}}{4} \\\ \Rightarrow {P_{total}} = \dfrac{{4P}}{4} \\\ \therefore {P_{total}} = P \\\$
Hence, the total pressure exerted is $P$.
Therefore, the correct option is (A).

Note:
The mole fraction of a specific gas in a mixture of gases is equal to the ratio of the partial pressure of that gas to the total pressure exerted by the gaseous mixture. This mole fraction can also be used to calculate the total number of moles of a constituent gas when the total number of moles in the mixture is known.