$$'a’$$ is directly related to forces of attraction. Hence greater the value of $$'a’,$$ more easily the gas is liquified.

$$\eqalign{ & {\text{Pressure exerted by the gas,}} \cr & {\text{P = }}\frac{1}{3}\,\,\frac{{mn{u^2}}}{V}\,\,\,\,\,....\left( 1 \right) \cr & {\text{Here, }}u = \,\operatorname{root} \,{\text{mean}}\,{\text{square}}\,{\text{velocity}} \cr & m = {\text{mass}}\,{\text{of}}\,{\text{a}}\,{\text{molecule,}}\, \cr & n = {\text{No}}{\text{.}}\,{\text{of}}\,{\text{molecules}}\,{\text{of}}\,{\text{the}}\,{\text{gas}} \cr & {\text{Hence}}\left( {\text{A}} \right){\text{ & }}\left( {\text{B}} \right)\,{\text{are}}\,{\text{clearly}}\,{\text{wrong}}{\text{.}} \cr & {\text{Again}}\,\,\,{u^2} = \frac{{3RT}}{M}\,\,\,\,\,\,\,\,\,\left[ {{\text{explained}}\,{\text{from}}\left( {\text{1}} \right)} \right] \cr & {\text{Here}},\,\,M = \,{\text{Molecular}}\,wt.\,{\text{of}}\,{\text{the}}\,{\text{gas;}} \cr & {\text{Hence}}\,\left( {\text{C}} \right)\,{\text{is}}\,{\text{wrong}} \cr & {\text{Further,}}\,{\text{Average}}\,{\text{K}}{\text{.E}}. = \frac{3}{2}\,KT; \cr & {\text{Hence}}\,\left( {\text{D}} \right)\,{\text{is}}\,{\text{true}}{\text{.}} \cr} $$

In fluorite structure each $$C{a^{2 + }}\,ion$$   is surrounded by eight $${F^ - }\,ions.$$  Thus, the coordination number of $$C{a^{2 + }}$$  is eight.

$$\eqalign{ & {U_1} \propto \sqrt {\frac{1}{{{m_1}}}} \,\,{\text{and}}\,\,{U_2} \propto \sqrt {\frac{1}{{{m_2}}}} \cr & \therefore \,\,\frac{{U_1^2}}{{U_2^2}} = \frac{{{m_2}}}{{{m_1}}}\,\,\,\,\therefore \,\,{m_1}U_1^2 = {m_2}U_2^2 \cr} $$

$$p$$ - type of semiconductor are obtained by doping silicon or germanium with elements of group 13 like $$B,$$ $$Al,$$ $$Ga$$  or $$In$$  so silicon is doped with boron.

At high pressure and low temperature, gaseous atoms or molecules get closer to each other and van der Waal forces operates.
So molecules or atoms start attracting each other. Hence a gas deviate the most from its ideal behaviour. While in ideal behaviour we consider that gaseous molecules do not attract each other, i.e., there is no intermolecular forces of attraction.

For positive deviation : $$PV = nRT + nPb$$
$$ \Rightarrow \frac{{PV}}{{nRT}} = 1 + \frac{{Pb}}{{RT}}$$
Thus, the factor nPb is responsible for increasing the $$PV$$  value, above ideal value. $$b$$ is actually the effective volume of molecule. So, it is the finite size of molecules that leads to the origin of $$b$$ and hence positive deviation at high pressure.

Volume of one mole of a gas at $$STP$$  is called its molar volume.

If temperature, volume and pressure of fixed amount ( say $$n$$ mole ) of a gas vary from $${T_1},{V_1}$$  and $${p_1}$$  to $${T_2},{V_2}$$  and $${p_2}$$ respectively. Then, ideal gas equation for two states can be written as
$$\eqalign{ & {p_1}{V_1} = nR{T_1} \cr & {\text{or}}\,\,\frac{{{p_1}{V_1}}}{{{T_1}}} = nR\,\,\,...{\text{(i)}} \cr & {p_2}{V_2} = nR{T_2} \cr & {\text{or}}\,\,\frac{{{p_2}{V_2}}}{{{T_2}}} = nR\,\,\,...{\text{(ii)}} \cr & {\text{On combining Eqs}}{\text{. (i) and (ii), we get}} \cr & \frac{{{p_1}{V_1}}}{{{T_1}}} = \frac{{{p_2}{V_2}}}{{{T_2}}} \cr & {\text{or}}\,\,\,\frac{{{p_1}{V_1}}}{{{p_2}{V_2}}} = \frac{{{T_1}}}{{{T_2}}} \cr} $$
So, it called combined gas equation.
Hence, (C) answer is correct.

$$\eqalign{ & {\text{By ideal gas equation,}} \cr & {P_1}V = {n_1}RT \cr & {n_1} \propto {P_1}\,{\text{and}}\,{n_2} \propto {P_2} \cr & \frac{{{n_1}}}{{{n_2}}} = \frac{{{P_1}}}{{{P_2}}} \Rightarrow \frac{{{n_1}}}{{{n_2}}} = \frac{{170}}{{570}} = 0.30 \cr} $$