Question

A box of $1L$ 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

Hint:Use The Raoult’s law for Partial pressure of gases which states that the partial pressure of a gas is directly proportional to its mole fraction in the mixture of the gases, to separately determine partial pressure of the two gases. Partial pressure of the two gases can be then added using scalar addition to determine the value of total pressure.
Formula Used:
1. ${P^ \circ } = xP$
2. $P = \mathop \sum \limits_{i = 1}^n \,P_i^ \circ $
3. ${x_a} = \dfrac{{moles\,of\,a}}{{total\,moles\,of\,gas\,in\,mixture}}$

Complete step by step answer:
Let us first understand the terms which have been used in the question.
Partial pressure of gases can refer to the individual pressure of gases in a mixture of gases. It is given a mathematical value using the Raoult’s law which state that the partial pressure of a gas is directly proportional to its mole fraction in the mixture of the gases
${P^ \circ } = xP$
where, ${P^ \circ } = Partial\,pressure,x = mole\,fraction\,of\,gas,P = Total\,pressure$
To start, let us find out the individual values for Hydrogen gas and Methane.
Raoult’s law needs mole fraction in the formula and for that we need the number of moles:
$Number\,of\,moles = \,\dfrac{{given\,mass}}{{molecular\,mass}}$
For Hydrogen gas: $given\,mass = 2g,Molecular\,mass = 2g$
Hence substituting this in above equation we get:
$\dfrac{2}{2} = 1mole$
For Methane gas: $given\,mass = 16g,molecular\,mass = 16g$
Substituting these values in the above equation:
$\dfrac{{16}}{{16}} = 1mole$
Next, we need to find out the value for mole fraction
${x_a} = \dfrac{{moles\,of\,a}}{{total\,moles\,of\,gas\,in\,mixture}}$
\[total{\text{ }}moles{\text{ }}of{\text{ }}a{\text{ }}gas{\text{ }}in{\text{ }}a{\text{ }}mixture\,of\,gas = {\text{ }}moles{\text{ }}of{\text{ }}Hydrogen{\text{ }}gas{\text{ }} + {\text{ }}moles{\text{ }}of{\text{ }}methane\]
Which will be equal to: $1 + 1 = 2moles$
Mole fraction for Hydrogen gas: $\dfrac{1}{2} = 0.5$
Mole fraction for Methane gas: $\dfrac{1}{2} = 0.5$

Applying Raoult’s law of partial pressure here:
${P^ \circ } = xP$
where, ${P^ \circ } = Partial\,pressure,x = mole\,dfraction\,of\,gas,P = Total\,pressure$
Value of Pressure is given to be as $P$ atm for each gas when partition is there
Substituting these values in the above equation:
For hydrogen gas: $0.5P$
For methane gas: $0.5P$
Total pressure when partition is removed can be calculated using the formula:
$P = \mathop \sum \limits_{i = 1}^n \,P_i^ \circ $
$Total\,pressure = \,Partial\,pressure\,of\,hydrogen + partial\,pressure\,of\,methane$
Substituting the above values we get
$total\,pressure = 0.5P + 0.5P$
which will be equal to:
$Total\,pressure = P$
Hence option A is correct.

Note:
Raoult’s law could only be used in this question because we had non-volatile gases such as hydrogen and methane. Raoult’s law is not applicable to mixtures containing volatile gases.
It is also only applicable for very dilute solutions, and if the mixture becomes concentrated, the law starts to show deviation.