91Ó°ÊÓ

If there are no intermolecular interactions that take place between the molecules of the gas and if the total volume of the constituent molecules is negligible as compared to the volume of gas, then the gas is called an ideal gas. The equation relating various parameters of the ideal gas is called the ideal gas equation. It is given as-

$\mathbf{PV}\mathbf{=}\mathbf{nRT}$

Here, P is the pressure on the gas, T is the temperature of the gas, V is the volume of the gas, n is the number of moles and R is the gas constant

1. The temperature of the air is$T=20°C=273\text{‿é}+20\text{‿é}=293\text{‿é}$
2. The pressure at air is$P=1.0\text{ a³Ù³¾}=1.01×{10}^5\text{ }\text{Pa}$
3. The molecules of nitrogen are$75\%$
4. The molecules of oxygen are$25\%$

(a)

The ideal gas equation is-

$PV=nRT$

$\frac{n}{V}=\frac{P}{RT}$

Multiply $N_A$on both sides,

$\frac{n{N}_A}{V}=N_A\frac{P}{RT}$

Where$N_A$is Avogadro’s number and R is gas constant hence,$N_A=6.02×{10}^{23}{\text{mol}}^{\text{-1}}$and$R=8.31\text{ J}/\text{mol.K}$

The number of molecules per unit volume is

$\frac{N}{V}=N_A\frac{P}{RT}$

For the given values the equation becomes-

$\frac{N}{V}=(6.02×{10}^{23}\text{ }{\text{mol}}^{\text{-1}})×\frac{1.01×{10}^5\text{ }\text{Pa}}{8.31\text{ }\frac{\text{J}}{\text{mol}}\text{.K}×293\text{‿é}}$

$\frac{N}{V}=2.5×{10}^{25}{\text{molecules/cm}}^{\text{3}}$

(b)

Three fourth of the number of molecules per unit volume are nitrogen molecules and one fourth of the value are oxygen molecules.

The masses of unit mole of the gas are given as-

$$\begin{gathered} M_N=28.0\text{ g/³¾´Ç±ô} \\ {M}_O=32.0\text{ g/³¾´Ç±ô} \end{gathered}$$

The number of moles contained in a sample of mass $M_{sam}$, consisting of N molecules, is related to the molar mass M of the molecules and to Avogadro’s number $N_A$as

$\begin{array}{l}n=\frac{N}{N_A}\\ n=\frac{M_{sam}}{M}\\ M_{sam}=\frac{NM}{N_A}\\ M_{sam}=\frac{3}{4}\frac{N}{V}\frac{M_N}{N_A}+\frac{1}{4}\frac{N}{V}\frac{M_O}{N_A}\end{array}$

For the given values, the above equation can be written as-

$\begin{array}{c}{M}_{sam}=\left(\begin{array}{l}\frac{3}{4}(2.5×{10}^{25}{\text{molecules/cm}}^{\text{3}})\frac{(28.0\text{ g}/\text{mol})}{(6.022×{10}^{23}\text{″¾olecules})}\\ +\frac{1}{4}(2.5×{10}^{25}{\text{molecules/cm}}^{\text{3}})\frac{32\text{ g}/\text{mol}}{(6.022×{10}^{23}\text{″¾olecules})}\end{array}\right)\\ =871.80+332.11\\ =1203.9\text{ g}\\ =\text{1.2 k²µ}\end{array}$

The number of molecules per cubic meter of air is $2.5×{10}^{25}{\text{molecules/cm}}^{\text{3}}$and the mass of 1 cubic meter of air is$1.2\text{ k²µ}$ .