Schottky defect in crystals is observed when equal number of cations and anions are missing from the lattice. So, the crystal remains neutral, e.g. $$NaCl.$$

Here, volume is constant and mass of hydrogen gas is also fixed, therefore the number of moles remains same. Now as the temperature increases, then pressure also increases. This will lead to more collisions among the gaseous molecules and hence, the energy of molecules increases.

Let $$m$$  be the mass of $$S{O_3}$$  and $${O_2}$$  enclosed in the vessel.
Number of moles of $$S{O_3} = \frac{m}{{80}}$$
Number of moles of $${O_2} = \frac{m}{{32}}$$
Partial pressure of $$S{O_3},{P_A} = \frac{m}{{80}} \times \frac{{R \times T}}{V}$$
PartiaI pressure of $${O_2},{P_B} = \frac{m}{{32}} \times \frac{{R \times T}}{V}$$
Now, $$\frac{{{P_A}}}{{{P_B}}} = \frac{m}{{80}} \times \frac{{32}}{m} = \frac{2}{5}$$
Hence, ratio of partial pressure of $$m\,g$$  of $$S{O_3}$$  and $${O_2}$$  is 2 : 5.

$$\eqalign{ & {\text{Moles of}}\,A,\left( {{n_A}} \right) = \frac{{{P_A}{V_A}}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{8 \times 12}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{96}}{{RT}} \cr & {\text{Moles of}}\,B,\left( {{n_B}} \right) = \frac{{{P_B}{V_B}}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{8 \times 5}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{40}}{{RT}} \cr} $$
$${\text{Total pressure}} \times {\text{total volume}}$$       $$ = \left( {{n_A} + {n_B}} \right) \times RT$$
$$\eqalign{ & p \times \left( {12 + 8} \right) = \frac{1}{{RT}}\left( {96 + 40} \right)RT \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,P = 6.8 \cr} $$
$${\text{Partial pressure of }}A$$     $$ = P \times {\text{mole fraction of }}A$$
$$\eqalign{ & = 6.8\left( {\frac{{\frac{{96}}{{RT}}}}{{\frac{{96 + 40}}{{RT}}}}} \right) \cr & = 4.8\,atm \cr} $$
$${\text{Partial pressure of}}\,B$$     $$ = 6.8 - 4.8 = 2\,atm.$$

$$\eqalign{ & u:v:\alpha = \sqrt {\frac{{3RT}}{M}} :\sqrt {\frac{{8RT}}{{\pi M}}} :\sqrt {\frac{{2RT}}{M}} \cr & = \sqrt 3 :\sqrt {\frac{8}{\pi }} :\sqrt 2 \cr} $$

According to Charle’s law “the volume of a fixed mass of a gas increases or decreases by $$\frac{1}{{273.15}}$$  of its volume at $${0^ \circ }C$$  for each degree rise or fall of temperature, if pressure is kept constant”.

$$PV = nRT.$$    Keeping other conditions same, if $$n$$ becomes half, the pressure will become half.

A liquid below its melting point, is present in solid state and above its melting point, is present in vapour (gaseous) state, so a liquid can exist between melting point and boiling point.

$$\eqalign{ & {\text{For simple cube,}} \cr & {\text{Radius}}\,\left( r \right) = \frac{a}{2}\,\,\,\left[ {a = {\text{edge length}}} \right] \cr & {\text{Volume of the atom}} = \frac{4}{3}\pi {\left( {\frac{a}{2}} \right)^3} \cr & \therefore \,\,{\text{Packing fraction}} = \frac{{\frac{4}{3}\pi {{\left( {\frac{a}{2}} \right)}^3}}}{{{a^3}}} = \frac{\pi }{6} \cr} $$

The root mean square velocity of gas molecule at $$STP$$  is given by
$${U_{rms}} = \sqrt {\frac{{3RT}}{M}} $$
∴ As the molar mass of gas increases, then $${U_{rms}}$$  will decrease, so the order of $${U_{rms}}$$  of these gases is
$$HBr < {O_2} < {N_2} < {H_2}$$