91Ó°ÊÓ

If the internal energy of the gas is a function of temperature only, then the gas is called as an ideal gas. The ideal gas obeys the ideal gas law, which relates the state parameters of the gas as-

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

1. The number of moles of gas$n=1\text{″¾ol}$
2. The standardpressure$P=1.00\text{ a³Ù³¾}=1.01×{10}^5\text{ P²¹}$
3. The standard temperature$T=273\text{‿é}$
4. The ideal gas constant$R=8.31\text{J}/\text{mol}â‹\dots \text{K}$

$\begin{array}{l}PV=nRT\\ V=\frac{nRT}{P}\end{array}$

For the given values of state parameters, the equation becomes-

$V=\frac{1\text{″¾ol}×8.31\text{J}/\text{mol}â‹\dots \text{K}×273\text{‿é}}{1.01×{10}^5\text{ P²¹}}$

$\begin{array}{c}V=2.24×{10}^{−2}{\text{m}}^3\\ =22.4\text{ L}\end{array}$

The volume occupied by 1 mole of the gas is$22.4\text{ L}$

b.

The number of molecules in 1 mole of the gas is.$6.02×{10}^{23}$

The number of molecules in$22.4\text{ L}$ is$6.02×{10}^{23}$, so the number of molecules in $1{\text{ c³¾}}^{\text{3}}$of gas is-

$\begin{array}{c}N=\frac{6.02×{10}^{23}}{22.4\text{ L}}×\frac{1{\text{″¾}}^3}{{10}^6{\text{cm}}^3}\\ =\frac{6.02×{10}^{23}}{2.24×{10}^{−2}{\text{m}}^3}×\frac{1{\text{″¾}}^3}{{10}^6{\text{cm}}^3}\\ =2.69×{10}^{19}\end{array}$

Hence, number of molecules in$1{\text{ c³¾}}^{\text{3}}$ of gas$N=2.69×{10}^{19}$.

The volume occupied by the one mole of the gas at standard temperature and pressure is $22.4\text{ L}$and the number of molecules in one cubic centimeter of the gas is.$2.69×{10}^{19}$