A $3$ grams of ${H_2}$ and $24$ grams ${O_2}$ are present in a gaseous mixture at a constant temperature and pressure. The partial pressure ${H_2}$ is
(A) $\dfrac{1}{3}$ of total pressure
(B) $\dfrac{2}{3}$ of total pressure
(C) $\dfrac{3}{2}$ of total pressure
(D) $\dfrac{1}{2}$ of total pressure
Answer
589.5k+ views
Hint: Partial pressure is a type of pressure exerted by the gas if it alone occupies the whole volume at the same temperature. We know that partial pressure exerted by the individual gas is equal to the total pressure multiplied by the mole fraction. So we will calculate the number of moles of each gas and consequently the partial pressure occupied by hydrogen gas.
Complete Step by step answer:
We know that the partial pressure exerted by the gas is equal to the total pressure multiplied by the mole fraction.
Here the temperature and pressure are kept constant. We will at first calculate the number of moles of hydrogen gas which is the mass of hydrogen gas divide by the molecular mass
$n = \dfrac{m}{M}$
${n_{{H_2}}} = \dfrac{3}{2}$
$\Rightarrow {n_{{H_2}}} = 1.5$
Therefore the number of moles of hydrogen gas is $1.5$
Similarly, we calculate the number of moles of oxygen.
$ \Rightarrow {n_{{O_2}}} = \dfrac{{24}}{{32}}$
$ \Rightarrow {n_{{O_2}}} = 0.75$
Therefore the number of moles of oxygen is $0.75$.
Now the mole fraction of hydrogen gas is given by
${X_{{H_2}}} = \dfrac{{{n_{{H_2}}}}}{{{n_{{H_2}}} + {n_{{O_2}}}}}$
So ${X_{{H_2}}} = \dfrac{{1.5}}{{1.5 + 0.75}}$
$ \Rightarrow {X_{{H_2}}} = \dfrac{{1.5}}{{2.25}}$
Therefore ${X_{{H_2}}} = \dfrac{2}{3}$
Hence the mole fraction of hydrogen gas is $\dfrac{2}{3}$.
Now we know that the partial pressure of a gas is given by total pressure times mole fraction of that particular gas.
$\therefore {P_{{H_2}}} = {X_{{H_2}}} \times {P_T}$
$ \Rightarrow {P_{{H_2}}} = \dfrac{2}{3} \times {P_T}$
$\therefore {P_{{H_2}}} = \dfrac{2}{3} \times {P_T}$ which can be written as
${P_{{H_2}}} = \dfrac{2}{3}$ of total pressure
Hence the correct answer is option B.
Note: The constant pressure and temperature conditions mentioned here correspond to NTP conditions. Also, the partial pressure exerted by a particular gas is independent of the partial pressure exerted by other gases. The total pressure of a mixture solely depends on the number of moles of the gas and not on the kinds of molecules.
Complete Step by step answer:
We know that the partial pressure exerted by the gas is equal to the total pressure multiplied by the mole fraction.
Here the temperature and pressure are kept constant. We will at first calculate the number of moles of hydrogen gas which is the mass of hydrogen gas divide by the molecular mass
$n = \dfrac{m}{M}$
${n_{{H_2}}} = \dfrac{3}{2}$
$\Rightarrow {n_{{H_2}}} = 1.5$
Therefore the number of moles of hydrogen gas is $1.5$
Similarly, we calculate the number of moles of oxygen.
$ \Rightarrow {n_{{O_2}}} = \dfrac{{24}}{{32}}$
$ \Rightarrow {n_{{O_2}}} = 0.75$
Therefore the number of moles of oxygen is $0.75$.
Now the mole fraction of hydrogen gas is given by
${X_{{H_2}}} = \dfrac{{{n_{{H_2}}}}}{{{n_{{H_2}}} + {n_{{O_2}}}}}$
So ${X_{{H_2}}} = \dfrac{{1.5}}{{1.5 + 0.75}}$
$ \Rightarrow {X_{{H_2}}} = \dfrac{{1.5}}{{2.25}}$
Therefore ${X_{{H_2}}} = \dfrac{2}{3}$
Hence the mole fraction of hydrogen gas is $\dfrac{2}{3}$.
Now we know that the partial pressure of a gas is given by total pressure times mole fraction of that particular gas.
$\therefore {P_{{H_2}}} = {X_{{H_2}}} \times {P_T}$
$ \Rightarrow {P_{{H_2}}} = \dfrac{2}{3} \times {P_T}$
$\therefore {P_{{H_2}}} = \dfrac{2}{3} \times {P_T}$ which can be written as
${P_{{H_2}}} = \dfrac{2}{3}$ of total pressure
Hence the correct answer is option B.
Note: The constant pressure and temperature conditions mentioned here correspond to NTP conditions. Also, the partial pressure exerted by a particular gas is independent of the partial pressure exerted by other gases. The total pressure of a mixture solely depends on the number of moles of the gas and not on the kinds of molecules.
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