A gas that follows Boyle's law, Charle's law and Avogadro's law is called an ideal gas. Under what conditions a real gas would behave ideally?
At low pressure and high temperature, a real gas behaves as an ideal gas. Almost all gases are real gas.
Two different gases ' $A$ ' and ' $B$ ' are filled in separate containers of equal capacity under the same conditions of temperature and pressure. On increasing the pressure slightly the gas ' $A$ ' liquefies but gas $B$ does not liquify even on applying high pressure until it is cooled. Explain this phenomenon.
The temperature above which a gas cannot be liquefied howsoever high pressure may be applied on the gas is called critical temperature. Since, gas 'A' liquifies easily, this suggests gas ' $A$ ' is below its critical temperature. On the other hand, gas 'B' does not liquefy easily even on applying high pressure. This suggests that gas ' $B$ ' is above its chitical temperature.
Value of universal gas constant $(R)$ is same for all gases. What is its physical significance?
Unit of $R$ depends upon those units in which $p, V$ and $T$ are measured as, $R=\frac{p V}{n T}$. If pressure is measured in Pascal, per mole volume is measured in $\mathrm{m}^3$ and temperature is measured in Kelvin then units of ' $R$ ' are $\mathrm{Pam}^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ or $\mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$. Since, joule is the unit of work done, so ' $R$ ' is work done by the gas per mole per kelvin.
One of the assumptions of kinetic theory of gases states that "there is no force of attraction between the molecules of a gas." How far is this statement correct? Is it possible to liquefy an ideal gas? Explain.
This statement is correct only for ideal gases. It is not possible to liquefy an ideal gas because there is no intermolecular forces of attractions between the molecules of an ideal gas.
The magnitude of surface tension of liquid depends on the attractive forces between the molecules. Arrange the following in increasing order of surface tension:
Water, alcohol $\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)$ and hexane $\left.\left[\mathrm{CH}_3\left(\mathrm{CH}_2\right)_4 \mathrm{CH}_3\right)\right]$.
In the above given molecules, only hexane $\left(\mathrm{CH}_3\left(\mathrm{CH}_2\right)_4 \mathrm{CH}_3\right)$ is a non-polar molecule in which only Landon dispersion forces exist. These forces are very weak while both water and alcohol are polar molecules in which dipole-dipole interactions as well as H -bonding exists.
However, H -bonding interactions are much stronger in water than $\mathrm{H}_2 \mathrm{O}$, therefore, it possesses stronger intermo lecules than alcohol and hexane. Hence, the increasing order of surface tension is
$$\text { hexane }<\text { alcohol }<\text { water }$$
Greater is the attractive forces between the molecules, greater is the magnitude of surface tension of liquid.