What will be the molar volume of nitrogen and argon at 273.15 K and 1 atm?
When temperature and pressure of a gas is 273.15 K (or $0{ }^{\circ} \mathrm{C}$ ) and 1 atm (or 1 bar or $10^5$ pascal), such conditions are called standard temperature and pressure conditions (STP). Under these conditions, the volume occupied by 1 mole of each and every gas is 22.4 L . Hence, the moler volume of $\mathrm{N}_2$ and Ar at 273.15 K and 1 atm is 22.4 L .
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.