Compressibility factor, $Z$ of a gas is given as $Z=\frac{p V}{n R T}$
(i) What is the value of $Z$ for an ideal gas?
(ii) For real gas what will be the effect on value of $Z$ above Boyle's temperature?
(i) For ideal gas, compressibility factor, $Z=1$.
(ii) Above Boyle's temperature, real gases show positive deviation.
So, $$Z>1$$
The critical temperature $\left(T_c\right)$ and critical pressure $\left(p_c\right)$ of $\mathrm{CO}_2$ are $30.98^{\circ} \mathrm{C}$ and 73 atm respectively. Can $\mathrm{CO}_2(\mathrm{~g})$ be liquefied at $32^{\circ} \mathrm{C}$ and 80 atm pressure?
The given critical temperature and pressure of $\mathrm{CO}_2$ gas are $30.98^{\circ} \mathrm{C}$ and 73 atm respectively. This suggests that $\mathrm{CO}_2$ gas cannot be liquified above $30.98^{\circ} \mathrm{C}$ and 73 atm hewsoever high temperature and pressure may be applied on $\mathrm{CO}_2$ gas. Hence, $\mathrm{CO}_2$ gas cannot be liquified at $32 .{ }^{\circ} \mathrm{C}$ and 80 atm .
For real gases the relation between $p, V$ and $T$ is given by van der Waals' equation
$$\left(p+\frac{a n^2}{V^2}\right)(V-n b)=n R T$$
where, ' $a$ ' and ' $b$ ' are van der Waals' constants, ' $n b$ ' is approximately equal to the total volume of the molecules of a gas. ' $a$ ' is the measure of magnitude of intermolecular attraction.
(i) Arrange the following gases in the increasing order of ' $b$ '. Give reason.
$$\mathrm{O}_2, \mathrm{CO}_2, \mathrm{H}_2, \mathrm{He}$$
(ii) Arrange the following gases in the decreasing order of magnitude of ' $a$ '. Give reason.
$$\mathrm{CH}_4, \mathrm{O}_2, \mathrm{H}_2$$
(i) Molar volume occupied by the gas molecules $\propto$ size of the molecules and van der Waals' constant ' $b$ ' represents molar volume of the gas molecules. Hence, value of ' $b$ ' increases in the following order
$$\mathrm{H}_2<\mathrm{He}<\mathrm{O}_2<\mathrm{CO}_2$$
(ii) van der Waals' constant ' $a$ ' is the measure of magnitude of intermolecular attraction. The magnitude of intermolecular attractions increases with increase in size of electron cloud in a molecule. Hence, for the given gases magnitude of 'a' decreases in the following order
$$\mathrm{CH}_4>\mathrm{O}_2>\mathrm{H}_2$$
Greater the size of electron cloud, greater is the polarisability of the molecule and greater is the dispersion forces or London forces.
The relation between pressure exerted by an ideal gas $\left(p_{\text {ideal }}\right)$ and observed pressure ( $p_{\text {real }}$ ) is given by the equation,
$$p_{\text {ideal }}=p_{\text {real }}+\frac{a n^2}{V^2}$$
(i) If pressure is taken in $\mathrm{Nm}^{-2}$, number of moles in mol and volume in $\mathrm{m}^3$, calculate the unit of ' $a$ '.
(ii) What will be the unit of ' $a$ ' when pressure is in atmosphere and volume in $\mathrm{dm}^3$ ?
Given that, $p_{\text {ideal }}=p_{\text {real }}+\frac{a n^2}{V^2}$
(i) $a=\frac{p V^2}{n^2}$ If units of $p=\mathrm{Nm}^{-2}$,
units of $V=\mathrm{m}^3$, units of $n=\mathrm{mol}$
then, units of $a=\frac{\mathrm{Nm}^{-2}\left(\mathrm{~m}^3\right)^2}{(\mathrm{~mol})^2}=\mathrm{Nm}^4 \mathrm{~mol}^{-2}$
(ii) If units of $p=\mathrm{atm}$, units of $V=\mathrm{dm}^3$, units of $n=\mathrm{mol}$
then, units of $a=\frac{p V^2}{n^2}=\frac{\mathrm{atm} \cdot\left(\mathrm{dm}^3\right)^2}{(\mathrm{~mol})^2}=\mathrm{atm} \mathrm{~dm}{ }^6 \mathrm{~mol}^{-2}$
Name two phenomena that can be explained on the basis of surface tension.
Two phenomena that can be explained on the basis of surface tension are as following
(i) Rise or fall of the liquid in a capillary-(capillary action).
(ii) Spherical shape of small liquid drops.