PV=RT is ideal gas equation and gases behave ideally only at high temperature and low pressure.
Therefore option(B).

$\left( {p + \frac{a}{{{V _2}}}} \right)\left( {v - b} \right) = cT$

In Langmuir's model of adsorption of a gas on solids surfaces. The adsorption at a single site on the surface may invoice multiple molecules at the ame time. The mass of gas striking at a given area of surface is independent of the pressure of the gas.

At low pressure and high temperature real gas obey PV=RT  i.e. they behave as ideal gas because at high temperature we can assume that there is no force of attraction or repulsion works among the molecules and the volume occupied by the molecules is negligible in comparison to the volume occupied by the gas

Given $P=50 \ atm , \ \ V=350 ml =0.350 \ \ \ litre, \ \ n=1 \ \ mole $ and $ T=300 K$

At high temperature and low pressure the real gas behaves as an ideal gas.

Van der Walls equation takes into account inter atomic forces between gas particles which is not considered in the ideal gas model. Since simplicity of  liquification of a gas depends upon forces between its particles, Van der Walls equation is followed by easily liquifieable gases. 

No. of moles, $n=\dfrac{m}{M}=\dfrac{2}{4}=0.5 mol$

$\displaystyle b = 4N \times \frac{4}{3} \pi \frac{d^3}{8}$ (standard definition of boyles constant)

$\displaystyle = \frac{4 \times 6.02 \times 10^{23} \times 3.14 \times 2.94^3 \times 10^{-30}}{3 \times 8}$

$\displaystyle = 32 \times 10^{-6}$

The Vander Wall's equation of state is 
$\left(P+ \displaystyle\ \frac{a}{V^{2}}\right)$ $(V-b)$ = $RT$
$P = \displaystyle\ \frac{RT} {V-b} -\displaystyle\ \frac{a}{V^{2}}$
At the critical point, 
$P = P _{C}, V= V _{C}$ and $T= T _{C}$ 
$\therefore P _{C} = \displaystyle\ \frac{RT _{C}}{V _{C}-b} - \frac{a}{V _{C}^{2}}$ .......(i)
At the critical point on the isothermal,
$\displaystyle\ \frac{dP _{C}}{dV _{C}} = 0$
$\therefore 0 = \displaystyle\ \frac{-RT _{C}}{(V _{C}-b)^{2}} + \frac{1a}{V _{C}^{3}}$
$\displaystyle\ \frac{RT _{C}}{(V _{C}-b)^{2}} = \displaystyle\ \frac{2a}{V _{C}^{3}}$ .... (ii)
Also at critical point,
$\displaystyle\ \frac{d^{2}P _{C}}{dV _{C}^{2}} = 0$
$0 = \displaystyle\ \frac{2RT}{(V _{C}-b)^{3}}- \frac{6a}{V _{C}^{4}}$
$\displaystyle\ \frac{2RT _{C}}{(V _{C-b})^{3}}$ = $\displaystyle\ \frac{5a}{V _{C}^{4}}$ ....(iii)
Dividing (ii) by (iii) we get
$\displaystyle\ \frac{1}{2}(V _{C}-b)$ = $\displaystyle\ \frac{1}{3} V _{C}$
$V _{C} = 3b$ ..... (iv)
Putting this value in (ii), we get 
$\displaystyle\ \frac{RT _{C}}{4b^{2}} = \displaystyle\ \frac{2a}{27{b}^{3}}$
$T _{C} = \displaystyle\ \frac{8a}{27bR}$ .......(v)
Putting the value of $V _{C}$ and $T _{C}$ in (i), we get 
$P _{C}$ = $\displaystyle\ \frac{R}{2b}\left(\frac{8a}{27bR}\right)$ = $\displaystyle\ \frac{a}{9b^{2}}$
$= \displaystyle\ \frac{a}{27b^{2}}$

$\begin{array}{l} \dfrac { { { P _{ 1 } }{ V _{ 1 } } } }{ { { T _{ 1 } } } } =\dfrac { { { P _{ 2 } }{ V _{ 2 } } } }{ { { T _{ 2 } } } }  \ \Rightarrow \dfrac { { { { 10 }^{ 5 } }\times \left( { 1{ m^{ 3 } } } \right)  } }{ { 300K } } =\dfrac { { P\left( 2 \right)  } }{ { 600 } }  \ \Rightarrow P={ 10^{ 5 } }N/{ m^{ 2 } } \ Hence, \ option\, \, A\, \, is\, correct\, \, answer. \end{array}$

The ideal gas law treats the molecules of a gas as point particles with  perfectly elastic collisions. This works well for dilute gases in many experimental circumstances. But gas molecules are not point masses, and there are circumstances where the properties of the molecules have an experimentally measurable effect.

The Bhopal gas tragedy was an industrial catastrophe that occurred in 1984 at the Methyl isocynate gas ($CH _3NCO$) was leaked from the plant. Union Carbide India Limited (UCIL) pesticide plant in Bhopal, Madhya Pradesh.

Real gas can be approximated as ideal gas when the pressure is low and the temperature is high
This means that per unit volume, there are less number of gas molecules because there is less force (pressure) and there is more energy (temperature), so the molecules will tend to move apart
So, density will be low.

Generally, a gas behaves more like an ideal gas at higher temperature and lower pressure as the forces against intermolecular forces becomes less significant compared to the particles' kinetic energy, and the size of the molecules becomes less significant compared to the empty space between them.

For  a reversible adiabatic process, we have $dS = 0$ (Entropy change = 0)

${  \quad \quad \quad \quad \quad \quad \quad PCl } _{ 5 }\rightleftharpoons { PCl } _{ 3(9) }+{ Cl } _{ 2(9) }\\ Initial\quad mole\quad \quad \quad 1\quad  \quad 0\quad\quad\quad 0\\ After\quad mole\quad \quad 1-\alpha \quad \quad  \alpha \quad\quad  \alpha \\ decomposition$

Total mole$=1-\alpha+\alpha+\alpha\\=1+\alpha$

Total pressure$=P$

Partial pressure of $PCl _5=P(\cfrac{1-\alpha}{1+\alpha})$

PArtial pressure of $PCl _3=P(\cfrac{\alpha}{1+\alpha})$

Partial pressure of $PCl _2=P(\cfrac{\alpha}{1+\alpha})$

Then $K _p=\cfrac{(PCl _3)(Cl _2)}{(PCl _5)}\\ \quad=\cfrac{P(\cfrac{\alpha}{1+\alpha})P(\cfrac{\alpha}{1+\alpha})}{P(\cfrac{1-\alpha}{1+\alpha})}\\ \quad=\cfrac{P^2\alpha^2}{(1+\alpha)^2}\times\cfrac{(1+\alpha)}{P(1-\alpha)}\\K _p=\cfrac{P\alpha^2}{1-\alpha^2}$

now, $1-\alpha^2<<1$

so that $K _p=P\alpha^2\\ \alpha^2=\cfrac{K _p}{P}\\ \alpha=\sqrt{\cfrac{K _p}{P}}$

According to Van Der waal's equation for $n$ mole of real gas