Question

What is the partial pressure of ${O_2}$ in air at one atm given that $21\% $ of air molecules are ${O_2}$ ?

Answer 1

Hint: We have to know that, in a combination of gases, every constituent gas has an incomplete pressing factor which is the notional pressing factor of that constituent gas in the event that it alone involved the whole volume of the first blend at a similar temperature. The absolute pressing factor of an ideal gas blend is the amount of the incomplete pressing factors of the gases in the combination.

Complete answer:
First, of all, you realize that the partial pressing factor of a gas that is important for a vaporous blend can be determined utilizing the accompanying condition believe Dalton's Law of partial pressures here,
${P_i} = {X_i} \times {P_{total}}$
Where,
The partial pressure of gas $i$ is ${P_i}$ ,the mole fraction of gas $i$ in the mixture is ${X_i}$ , The total pressure of the mixture ${P_{total}}$ .
Presently, you realize that you have an example of air at an all-out pressing factor of one atm also, that $21\% $ of the relative multitude of atoms of gas that make up this example are particles of oxygen gas.
To have the option to compute the halfway pressing factor of oxygen gas, you need to sort out the mole part of oxygen gas in the example. The mole part of oxygen gas is determined by partitioning the quantity of moles of oxygen gas by the complete number of moles of gas present in the example.
$X{O_2} = \dfrac{{Moles{\text{ of }}{{\text{O}}_2}}}{{Total{\text{ moles of gas}}}}$
Then, one mole of gas is equal to the Avogadro’s constant.
$1{\text{ mole gas = 6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{23}}{\text{ molecules of gas}}$
To calculate the number of moles of ${O_2}$ in $21\% $ ,
$N{\text{ molecules gas}} \times \dfrac{{21{\text{ molecules }}{{\text{O}}_2}}}{{100{\text{ molecules gas}}}} \times \dfrac{{1{\text{ mol }}{{\text{O}}_2}}}{{6.022 \times {{10}^{23}}molecules{\text{ }}{{\text{O}}_2}}} = \dfrac{{21N}}{{6.022 \times {{10}^{23}}}}{\text{ }}mol{\text{ }}{{\text{O}}_2}$
Then, the total number of moles of gas,
$N{\text{ molecules gas}} \times \dfrac{{1{\text{ mol gas}}}}{{6.022 \times {{10}^{23}}{\text{ molecules gas}}}} = \dfrac{N}{{6.022 \times {{10}^{23}}}}mol{\text{ gas}}$
Applying both the values in the $X{O_2}$ equation,
$X{O_2} = \dfrac{{\dfrac{{0.21N}}{{6.022 \times {{10}^{23}}}}}}{{\dfrac{N}{{6.022 \times {{10}^{23}}}}}} = 0.21$
Therefore, the partial pressure of ${O_2}$ in one atm is $0.21$ atm.

Note:
We need to know that the fractional pressing factor of a gas is a proportion of the thermodynamic movement of the gas' atoms. Gases break down, diffuse, and respond as indicated by their fractional pressing factors, and not as per their fixations in gas blends or fluids.