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

Hint: 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.