Question

In a cylinder, there are \[60g\] $ Ne$ and $64g$ ${O_2}$. If the pressure mixture of gases in the cylinder is $30$ bar then in this cylinder the partial pressure (in bar) of ${O_2}$ is

A) $30$

B) $20$

C) $15$

D) $12$

Answer 1

Hint: To solve this particular problem let us remember about the mole fraction of gas. The mole fraction of a gas is the ratio of the mole number of the gas and the mole number of all gases present in the gas mixture.

Formula used:

${\text{p = mole fraction} \times {P}}$, $P$ is the total pressure of the gases.

The mole number of gas, \[{\text{n = }}\dfrac{{{\text{amount of the gas in the mixture}}}}{{{\text{atomic mass of the gas}}}}\]

\[{\text{The mole fraction = }}\dfrac{{{\text{mole number of the gas(n)}}}}{{{\text{ total mole number of the gases exist in the mixture}}}}\]

Complete step by step answer:

According to Dalton’s law, the total pressure of a gas mixture is the sum of the partial pressure of each of the gas presents in the mixture.

The mole fraction of a gas is the ratio of the mole number of the gas and the mole number of all gases present in the gas mixture. Such as,

The mole fraction\[{\text{ = }}\dfrac{{{\text{mole number of the gas(n)}}}}{{{\text{ total mole number of the gases exist in the mixture}}}}\]

And the partial pressure of an individual gas is calculated by the following relation,

the partial pressure of one gas ${\text{p = mole fraction} \times {P}}$ (Total pressure of the mixture)

Here given in a gas mixture of $Ne$ and ${O_2}$,

The amount of $Ne = 60g$

And the amount of ${O_2} = 64g$.

The atomic mass of Neon gas is $20$ and the atomic mass of an oxygen molecule is $32$.

Since, The mole number of gas, \[{\text{n = }}\dfrac{{{\text{amount of the gas in the mixture}}}}{{{\text{atomic mass of the gas}}}}\]

The mole number of Ne

$ \Rightarrow {n_{Ne}} = \dfrac{{60}}{{20}}$

$ \Rightarrow {n_{Ne}} = 3$ The mole number of ${O_2}$

$ \Rightarrow {n_{{O_2}}} = \dfrac{{64}}{{32}}$

$ \Rightarrow {n_{{O_2}}} = 2$ 

Since, The mole fraction\[{\text{ = }}\dfrac{{{\text{mole number of the gas(n)}}}}{{{\text{ total mole number of the gases exist in the mixture}}}}\]

The mole fraction of ${O_2}$\[{\text{ = }}\dfrac{{\text{2}}}{{{\text{ 3 + 2}}}}\]

\[=\dfrac{{\text{2}}}{{\text{5}}}\]

Now, the partial pressure of an individual gas is

 ${\text{p = mole fraction} \times {P}}$

Here given, the pressure mixture of gases i.e. $P$= 30 bar.

The partial pressure of ${O_2}$ is,

\[ \Rightarrow {p_{{O_2}}} = \dfrac{2}{5} \times 30\]

\[\therefore {p_{{O_2}}} = 12\]

Therefore The partial pressure of ${O_2}$ is $12$ bar. Hence the right answer is in option (D).

Note: Partial pressure is the notational pressure of a constituent gas of a gas-mixture. An individual gas that exists in a gas-mixture occupies the entire volume of the mixture at the same temperature of the partial pressure of that particular gas that can be represented.