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.
Viscosity of a liquid arises due to strong intermolecular forces existing between the molecules. Stronger the intermolecular forces, greater is the viscosity. Name the intermolecular forces existing in the following liquids and arrange them in the increasing order of their viscosities. Also give reason for the assigned order in one line.
Water, hexane $\left(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3\right)$, $\left(\mathrm{CH}_2 \mathrm{OHCH}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{OH}\right)$ glycerine
In water and glycerine - Hydrogen bonding and dipole-dipole interactions exists as both are polar molecules.
hexane - Dispersion forces/London forces are present due ts non-polar nature. The order of intermolecular forces among the given liquids is, hexane < water < glycerine.
Since, hexane has weakest intermolecular forces and glycerine the strongest (three OH groups) intermolecular forces. Therefore, hexane has minimum viscosity and glycerine has maximum viscosity.
Explain the effect of increasing the temperature of a liquid, on intermolecular forces operating between its particles. What will happen to the viscosity of a liquid if its temperature is increased?
As the temperature of a liquid increases, kinetic energy of the molecules increases which can overcome intermolecular forces. So, the liquid can flow more easily, this results in decrease in viscosity of the liquid.
The variation of pressure with volume of the gas at different temperatures can be graphically represented as shown in figure. On the basis of this graph answer the following questions.
(i) How will the volume of a gas change if its pressure is increased at constant temperature?
(ii) At a constant pressure, how will the volume of a gas change if the temperature is increased from 200 K to 400 K ?
(i) In accordance to Boyle's law, pressure of a gas is inversly proportional to its volume if temperature is kept constant. Thus, the volume of a gas will decrease if the pressure on the gas is increased keeping the temperature constant. e.g., at 200 K when pressure increases from $p_1$ to $p_2$, volume of the gas decreases, $V_2
(ii) In accordance to Charles law, volume of a gas is directly proportional to its temperature if pressure is kept constant.
Thus, on increasing temperature, the volume of a gas will increase if pressure is kept constant.
At constant $p$ when we increase the temperature from 200 K to 400 K , the volume of the gas increases, $V_4>V_3$.
