$$\eqalign{ & \frac{{{{\left( {PV} \right)}_{Observed}}}}{{{{\left( {PV} \right)}_{Ideal}}}} < 1 \cr & \Rightarrow {V_{obs}} < {V_{ideal'}}\,{V_{obs}} < 22.4\,\,{\text{litre}}{\text{.}} \cr} $$

At higher temperature and low pressure real gas acts as an ideal gas and obey $$pV = nRT$$   relation.

One mole of a substance contains the number of molecules which is independent of pressure.

When we add some of $$In$$  or $$Ga$$  ( group 13 element ) into $$Ge$$ ( group number 14 element ), then electron vacancy is created which is called holes. In this type of semiconductor the charge carrier are holes (positive), so it is called $$p$$ - type semiconductor.

According to van der Waals' equation for $$1\,mole$$   of a real gas, $$\left( {P + \frac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT$$
If $$a$$  and $$b$$  are small, $$\frac{a}{{{V^2}}}$$  and $$b$$  can be neglected as compared to $$P$$  and $$V$$  and the equation reduces to $$PV = RT.$$   Hence, a real gas will resemble an ideal gas when constants $$a$$  and $$b$$  are small.

The particles of a gas are infinitely small and stay apart as far as possible.

$$\eqalign{ & M = \frac{{mRT}}{{PV}} \cr & \,\,\,\,\,\,\, = \frac{{0.0625 \times 0.083 \times 819}}{{0.1 \times \left( {\frac{{34.05}}{{1000}}} \right)}} \cr & \,\,\,\,\,\,\, = 1247.74\,\,g\,\,mo{l^{ - 1}} \cr} $$

TIPS/Formulae:
Find the volume by either
$$V = \frac{{RT}}{P}\left( {PV = RT} \right)\,{\text{or}}\,{P_1}{V_1} = {P_2}{V_2}$$        and match it with the values given in graph to find correct answer. Volume of $$1$$ $$mole$$  of an ideal gas at $$273K$$  and $$1$$ $$atm$$  is $$22.4L$$  and that at $$373K$$  and $$1$$ $$atm$$  pressure is calculated as;
$$V = \frac{{RT}}{P} = \frac{{0.082 \times 373}}{1} = 30.58L \simeq 30.6\,L$$

$${U_{rms}}:{U_{av}} = \sqrt {\frac{{3RT}}{M}} :\sqrt {\frac{{8RT}}{{\pi M}}} \,or\,\sqrt 3 :\sqrt {\frac{8}{\pi }} = 1.086:1$$

$$\eqalign{ & {U_{av}} = \sqrt {\frac{{8RT}}{{\pi M}}} ;\,\,\,\left\{ {_{{T_2} = 927 + 273 = 1200\,K}^{{T_1} = 27 + 273 = 300K}} \right\} \cr & \frac{{{U_{a{v_1}}}}}{{{U_{a{v_2}}}}} = \sqrt {\frac{{{T_1}}}{{{T_2}}}} \,or\frac{{0.3}}{{{U_{a{v_2}}}}} = \sqrt {\frac{{300}}{{1200}}} \,or\frac{{0.3}}{{{U_{a{v_2}}}}} = \sqrt {\frac{1}{4}} \cr & or\,{U_{a{v_2}}} = 0.6\,\,m/\sec . \cr} $$