Real gases and Vander waal’s equation Questions in English

(B) According to the kinetic molecular theory of gases,one of the assumptions is that there are no intermolecular forces of attraction between gas molecules. However,in real gases,these forces are not negligible,which causes the deviation from ideal behavior.

(B) Using the van der Waals equation: $\left( P + \frac{an^2}{V^2} \right) (V - nb) = nRT$
Given: $a = 3.592 \, atm \, L^2 \, mol^{-2}$,$b = 0.0427 \, L \, mol^{-1}$,$V = 1 \, L$,$n = 5 \, mol$,$T = 273 + 47 = 320 \, K$,$R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$
Substituting the values: $\left( P + \frac{3.592 \times 5^2}{1^2} \right) (1 - 5 \times 0.0427) = 5 \times 0.0821 \times 320$
$\left( P + 89.8 \right) (1 - 0.2135) = 131.36$
$(P + 89.8) (0.7865) = 131.36$
$P + 89.8 = \frac{131.36}{0.7865} \approx 167.02$
$P = 167.02 - 89.8 = 77.22 \, atm$
Thus,the pressure is approximately $77.2 \, atm$.

(B) The compressibility factor,$Z = \frac{PV}{nRT}$.
Substituting the given values: $Z = \frac{40 \times 0.4}{1 \times 0.0821 \times 300} = 0.65$.
Since $Z < 1$,the gas is more compressible than expected for ideal behavior.

(A) The compressibility factor is given by $Z = \frac{PV}{nRT}$.
At $STP$,$P = 1 \, atm$,$T = 273 \, K$,$V = 50 \, L$,and $n = 2 \, mol$.
Using $R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$:
$Z = \frac{1 \times 50}{2 \times 273 \times 0.0821} \approx \frac{50}{44.8266} \approx 1.115$.
Since $Z > 1$,the gas shows positive deviation from ideal gas behavior.

(C) The van der Waals constant $'a'$ is a measure of the magnitude of intermolecular forces of attraction between gas molecules.
$NH_3$ is a polar molecule and exhibits strong intermolecular hydrogen bonding.
Due to these strong attractive forces,the value of $'a'$ is highest for $NH_3$ among the given options.

(C) The van der Waals constant $b$ represents the excluded volume per mole of gas molecules.
For a spherical molecule of radius $r$,the volume of one molecule is $V = \frac{4}{3}\pi r^3$.
The excluded volume for a pair of molecules is the volume of a sphere with radius $2r$,which is $\frac{4}{3}\pi (2r)^3 = 8 \times \frac{4}{3}\pi r^3$.
The excluded volume per molecule is half of this,i.e.,$4 \times \frac{4}{3}\pi r^3$.
Therefore,for $1 \ \text{mole}$ of gas ($N_A$ molecules),the constant $b$ is given by $b = 4 \times \left( \frac{4}{3}\pi r^3 \right) N_A$.

(B) The van der Waals equation for $n$ moles of a real gas is given by $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
For $1$ mole of gas,it is $(P + \frac{a}{V_m^2})(V_m - b) = RT$.
Here,the term $\frac{a}{V_m^2}$ represents the pressure correction factor,which accounts for the intermolecular forces of attraction between gas molecules.
Therefore,the term representing intermolecular forces is $P + \frac{a}{V_m^2}$.

(D) According to the van der Waals equation: $\left( P + \frac{a}{V^2} \right) (V - b) = RT$.
At very low pressure,the volume $V$ is very large.
Therefore,the term $\frac{a}{V^2}$ becomes negligible and can be ignored.
Similarly,the volume correction term $b$ is negligible compared to $V$.
Thus,the equation simplifies to $PV = RT$.
This implies that at very low pressure,a real gas behaves like an ideal gas.
Therefore,the compressibility factor $Z = \frac{PV}{RT} = 1$.

(C) The compressibility factor is defined as $Z = \frac{V_{real}}{V_{ideal}}$.
Since $Z > 1$,it implies that $V_{real} > V_{ideal}$.
At $STP$,the molar volume of an ideal gas is $V_{ideal} = 22.4 \, dm^3$.
Therefore,for a gas with $Z > 1$,the real molar volume $V_m$ must be greater than $22.4 \, dm^3$.

(A) At moderate pressure,the volume $V$ is small enough that the term $\frac{a}{V^2}$ cannot be neglected,but the volume correction '$b$' can be neglected compared to $V$.
The van der Waals equation becomes: $(P + \frac{a}{V^2})V = RT$.
Expanding this: $PV + \frac{a}{V} = RT$,which simplifies to $PV = RT - \frac{a}{V}$.
Dividing by $RT$: $\frac{PV}{RT} = 1 - \frac{a}{RTV}$,so $Z = 1 - \frac{a}{RTV}$.
Since $Z = 1 - \frac{a}{RTV}$,as temperature $T$ increases,the term $\frac{a}{RTV}$ decreases,causing $Z$ to increase towards $1$.

(B) At high pressure,the volume $V$ is very small. The term $\frac{a}{V^2}$ becomes large,but compared to high pressure $P$,it can be neglected. Thus,the van der Waals equation modifies as follows:
$P(V - b) = RT$
$PV = RT + Pb$
$\frac{PV}{RT} = 1 + \frac{Pb}{RT}$
$Z = 1 + \frac{Pb}{RT}$
Since $Z > 1$,the compressibility factor is greater than one and it decreases as the temperature increases.

(C) The van der Waals equation is $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
In this equation,the constant $a$ represents the magnitude of intermolecular forces of attraction,while the constant $b$ represents the excluded volume or the volume occupied by the gas molecules themselves.

(D) The $Van \ der \ Waals$ equation for $n$ moles of a real gas is given by: $\left( P + \frac{n^2a}{V^2} \right) (V - nb) = nRT$.
Here,$P$ is the pressure,$V$ is the volume,$n$ is the number of moles,$R$ is the universal gas constant,$T$ is the temperature,and $a$ and $b$ are $Van \ der \ Waals$ constants.

(B) The van der Waals equation is given by: $\left( P + \frac{n^2 a}{V^2} \right) (V - nb) = nRT$
Given: $n = 2\, \text{mol}$,$V = 5\, L$,$a = 4.17$,$b = 0.03711$,$R = 0.0821\, L\, atm\, K^{-1}\, mol^{-1}$,$T = 27 + 273 = 300\, K$.
Substituting the values: $\left( P + \frac{2^2 \times 4.17}{5^2} \right) (5 - 2 \times 0.03711) = 2 \times 0.0821 \times 300$
$\left( P + \frac{16.68}{25} \right) (5 - 0.07422) = 49.26$
$(P + 0.6672) \times 4.92578 = 49.26$
$P + 0.6672 = \frac{49.26}{4.92578} \approx 10.00$
$P = 10.00 - 0.6672 = 9.3328\, atm \approx 9.333\, atm$.

(D) Real gases deviate from ideal behavior due to intermolecular forces and finite molecular volume. These deviations are minimized at $high \ temperatures$ and $low \ pressures$,where the gas particles are far apart and have high kinetic energy,causing them to behave most like an ideal gas.

(B) The deviation from ideal gas behavior is primarily determined by the magnitude of intermolecular forces of attraction.
$NH_{3}$ is a polar molecule and exhibits strong dipole-dipole interactions,whereas $CH_{4}$,$H_{2}$,and $N_{2}$ are non-polar and exhibit only weak London dispersion forces.
Therefore,$NH_{3}$ shows the maximum deviation from ideal gas behavior.

(A) The constant $b$ represents the excluded volume,which increases with the size of the molecule. The order of $b$ for set-$I$ is: $H_2 (0.0266) < He (0.0237)$ (Note: $He$ is smaller than $H_2$,so $He < H_2$) $< O_2 (0.0318) < CO_2 (0.0427)$. Thus,$I. He < H_2 < O_2 < CO_2$.
The constant $a$ represents the magnitude of intermolecular forces. Larger molecules with more electrons have stronger van der Waals forces. The order of $a$ for set-$II$ is: $CH_4 (2.25) > O_2 (1.36) > H_2 (0.244)$. Thus,$II. CH_4 > O_2 > H_2$.

(D) The value of $a$ is a measure of the magnitude of the attractive forces between the molecules of the gas. Greater the value of $a$,larger is the attractive inter-molecular force between the gas molecules.
The value of $b$ is related to the effective size of the gas molecules. It is also termed as excluded volume.
The gases with higher value of $a$ and lower value of $b$ are more liquefiable. Since $Cl_2$ is more easily liquefied than $C_2H_6$,$a$ for $Cl_2$ must be greater than $a$ for $C_2H_6$,and $b$ for $Cl_2$ must be less than $b$ for $C_2H_6$.

(C) The van der Waals equation for a real gas is $(P + \frac{a}{V^2})(V - b) = RT$.
At high pressure,the term $\frac{a}{V^2}$ is very small compared to $P$ and can be neglected.
Thus,the equation simplifies to $P(V - b) = RT$.
Expanding this,we get $PV - Pb = RT$.
Rearranging the terms,$PV = RT + Pb$.
Dividing both sides by $RT$,we get $\frac{PV}{RT} = 1 + \frac{Pb}{RT}$.
Since the compressibility factor $Z = \frac{PV}{RT}$,we have $Z = 1 + \frac{Pb}{RT}$.
Therefore,at high pressure,$Z > 1$.

(B) The van der Waals equation for $1 \ mol$ of a gas is given by:
$(P + \frac{a}{V^2})(V - b) = RT$
At low pressure,the volume $V$ is very large,so $V >> b$. Therefore,the term $(V - b)$ can be approximated as $V$.
The equation simplifies to:
$(P + \frac{a}{V^2})V = RT$
Expanding the equation:
$PV + \frac{a}{V} = RT$
Rearranging to solve for $PV$:
$PV = RT - \frac{a}{V}$
Dividing both sides by $RT$ to express in terms of the compressibility factor $Z = \frac{PV}{RT}$:
$\frac{PV}{RT} = 1 - \frac{a}{VRT}$
Thus,$Z = 1 - \frac{a}{VRT}$.

(B) The compressibility factor $Z$ for a real gas is given by the van der Waals equation.
Positive deviation $(Z > 1)$ occurs when the repulsive forces or the finite volume of gas molecules become significant.
In the van der Waals equation $(P + \frac{an^2}{V^2})(V - nb) = nRT$,the constant $b$ represents the excluded volume of the gas molecules.
When the pressure is high,the volume of the gas molecules $(nb)$ becomes significant relative to the total volume $V$,leading to $Z > 1$.
Therefore,the molecular volume plays a dominant role in causing positive deviation.

(A) The compressibility factor $Z$ is defined as $Z = \frac{PV_m}{RT}$,where $V_m$ is the molar volume.
Rearranging for $V_m$,we get $V_m = \frac{ZRT}{P}$.
Given values: $Z = 1.5$,$P = 2 \ atm$,$T = 400 \ K$,and $R = 0.08 \ L \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $V_m = \frac{1.5 \times 0.08 \times 400}{2}$.
$V_m = \frac{1.5 \times 32}{2} = 1.5 \times 16 = 24 \ L$.

(C) The van der Waals' constant '$a$' is a measure of the magnitude of intermolecular attractive forces in a gas.
Greater the value of '$a$',stronger are the intermolecular forces.
Stronger intermolecular forces make it easier for a gas to be liquefied.
Comparing the given values: $O_2 (1.360) < N_2 (1.390) < CH_4 (2.253) < NH_3 (4.170)$.
Since $NH_3$ has the highest value of '$a$',it can be most easily liquefied.

(B) The molar volume $V_m$ is calculated as $V_m = \frac{\text{Molar mass}}{\text{density}} = \frac{27}{0.3} = 90 \ L/mol$.
The compressibility factor $Z$ is given by $Z = \frac{P V_m}{R T}$.
Using $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,we get $Z = \frac{1 \times 90}{0.0821 \times 750} \approx 1.46$.
Since $Z > 1$,the gas $A$ shows positive deviation from ideal behavior.

(D) The Van der Waals equation for $n$ moles of a real gas is derived by correcting the ideal gas equation $PV = nRT$ for intermolecular forces and molecular volume.
The pressure correction term is $\frac{n^2 a}{V^2}$ and the volume correction term is $nb$.
Substituting these into the ideal gas equation,we get:
$\left( P + \frac{n^2 a}{V^2} \right) (V - nb) = nRT$
Where:
$P$ = pressure of the gas
$V$ = volume of the gas
$n$ = number of moles
$a$ and $b$ = Van der Waals constants
$R$ = universal gas constant
$T$ = temperature

(A) The compressibility factor $Z$ is defined as $Z = \frac{PV}{nRT}$.
For real gases,the deviation from ideal behavior is represented by the dip in the $Z$ vs $P$ curve.
$A$ greater dip indicates stronger intermolecular forces of attraction.
Stronger intermolecular forces of attraction correspond to a higher critical temperature $(T_c)$,as it is easier to liquefy the gas.
In the given graph,the curve for gas $D$ shows the maximum deviation (the deepest dip) from the ideal behavior $(Z=1)$.
Therefore,gas $D$ has the strongest intermolecular forces and the highest critical temperature.

(B) The compressibility factor $Z$ is given by $Z = \frac{PV_m}{RT}$.
For real gases,at low pressure,the attractive forces dominate,and $Z < 1$. As temperature increases,the gas behaves more ideally,and the dip in the $Z$ vs $P$ curve decreases.
From the graph,the depth of the dip represents the deviation from ideal behavior. The curve with the deepest dip corresponds to the lowest temperature,and the curve with the shallowest dip corresponds to the highest temperature.
Thus,the order of temperatures is $T_1 < T_2 < T_3$.
Since all these curves show a dip below $Z = 1.0$ at low pressures,all these temperatures are below the Boyle temperature $(T_B)$.
Therefore,the correct order is $T_1 < T_2 < T_3 < T_B$.

(A) For a real gas,the critical temperature is given by $T_c = \frac{8a}{27Rb}$.
From the graph,the isotherm at $200 \ K$ shows a horizontal portion,indicating that $200 \ K$ is below the critical temperature $(T_c > 200 \ K)$.
The isotherm at $300 \ K$ does not show a horizontal portion,indicating that $300 \ K$ is above the critical temperature $(T_c < 300 \ K)$.
Thus,$200 < T_c < 300$.
Substituting $T_c = \frac{8a}{27Rb}$,we get $200 < \frac{8a}{27Rb} < 300$.
Rearranging for $\frac{a}{b}$,we have $\frac{200 \times 27 \times R}{8} < \frac{a}{b} < \frac{300 \times 27 \times R}{8}$.
Using $R = 0.08$,we get $\frac{200 \times 27 \times 0.08}{8} < \frac{a}{b} < \frac{300 \times 27 \times 0.08}{8}$.
$25 \times 27 \times 0.08 < \frac{a}{b} < 37.5 \times 27 \times 0.08$.
$54 < \frac{a}{b} < 81$.

(A) For $1$ mol of an ideal gas,the equation is $PV = nRT$. Since $n = 1$,we have $PV = RT$.
At the $PV$ axis,the pressure $P$ approaches $0$. For a real gas,as $P \to 0$,the gas behaves ideally.
Therefore,the intercept on the $PV$ axis is given by $\lim_{P \to 0} PV = nRT$.
Given the intercept is $20$,we have $20 = (1) \times R \times T$.
Solving for $T$,we get $T = \frac{20}{R}$.

(A) The van der Waals constant $a$ represents the magnitude of attractive forces between gas molecules. Larger molecules or those with higher polarizability generally have higher values of $a$.
From the provided table:
For $H_2$,$a = 0.244 \ \text{atm L}^2 \text{mol}^{-2}$
For $O_2$,$a = 1.36 \ \text{atm L}^2 \text{mol}^{-2}$
For $CH_4$,$a = 2.25 \ \text{atm L}^2 \text{mol}^{-2}$
Comparing these values,the increasing order is $0.244 < 1.36 < 2.25$,which corresponds to $H_2 < O_2 < CH_4$.

(C) The van der Waals constant $b$ represents the excluded volume per mole of gas molecules.
The excluded volume for a pair of molecules is $4 \times \text{volume of one molecule}$.
For $1 \ \text{mole}$ of gas,$b = N_{A} \times (2 \times \text{excluded volume per molecule}) = N_{A} \times 2 \times (4 \times \frac{4}{3} \pi r^{3}) / 2 = 4 \times N_{A} \times \text{volume of one molecule}$.
Since the volume of one molecule is $v = \frac{4}{3} \pi r^{3}$,we have $b = 4 N_{A} v$.
Therefore,the excluded volume per molecule is $v = \frac{b}{4 N_{A}}$.

(B) The formula for the critical temperature $(T_c)$ of a gas is given by:
$T_c = \frac{8a}{27Rb}$
Given:
$a = 3.6 \ atm \ L^2 \ mol^{-2}$
$b = 40 \ mL \ mol^{-1} = 0.04 \ L \ mol^{-1}$
$R = 0.08 \ atm \ L \ K^{-1} \ mol^{-1}$
Substituting the values:
$T_c = \frac{8 \times 3.6}{27 \times 0.08 \times 0.04}$
$T_c = \frac{28.8}{27 \times 0.0032}$
$T_c = \frac{28.8}{0.0864}$
$T_c = 333.33 \ K$

(C) The ideal gas law assumes that there are no intermolecular forces of attraction or repulsion between gas molecules.
However,real gases (van der Waals gases) exhibit intermolecular forces of attraction,known as van der Waals forces.
These attractive forces pull the molecules inward,away from the walls of the container.
Consequently,the molecules strike the container walls with less force than they would in the absence of these attractions.
Therefore,the observed pressure of a real gas is lower than the pressure predicted for an ideal gas.

(C) For a gas following van der Waal's equation,the compressibility factor at the critical point is given by $Z_c = \frac{P_c V_c}{R T_c} = \frac{(\frac{a}{27b^2}) (3b)}{R (\frac{8a}{27Rb})} = \frac{3}{8} = 0.375$. This value is a constant for all gases obeying the van der Waal's equation. Thus,Statement $-1$ is true.
At the critical temperature $(T_c)$,the compressibility factor $Z = \frac{P V_m}{R T_c}$ is not constant; it varies with pressure $P$ as the molar volume $V_m$ changes with pressure along the isotherm at $T_c$. Thus,Statement $-2$ is false.

(B) The critical molar volume $V_c$ is calculated as: $V_c = \frac{\text{Molar mass}}{\text{Density}} = \frac{60 \, g/mol}{0.80 \, g/cm^3} = 75 \, cm^3/mol = 0.075 \, L/mol$.
For a van der Waals gas,the critical volume is related to the constant $b$ by: $V_c = 3b$.
Therefore,$b = \frac{V_c}{3} = \frac{0.075}{3} = 0.025 \, L/mol$.
The critical temperature $T_c$ is given by: $T_c = \frac{8a}{27Rb}$.
Given $T_c = \frac{4 \times 10^5}{821} \, K$ and $R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$,we substitute the values:
$\frac{4 \times 10^5}{821} = \frac{8 \times a}{27 \times 0.0821 \times 0.025}$.
Since $821 = 10^4 \times 0.0821$,we simplify:
$\frac{4 \times 10^5}{10^4 \times 0.0821} = \frac{8 \times a}{27 \times 0.0821 \times 0.025}$.
$40 = \frac{8a}{27 \times 0.025} \Rightarrow 40 = \frac{8a}{0.675}$.
$a = \frac{40 \times 0.675}{8} = 5 \times 0.675 = 3.375 \, atm \, L^2 \, mol^{-2}$.

(D) The van der Waals equation for $n \, \text{moles}$ of a gas is $\left(P + \frac{n^2 a}{V^2}\right)(V - nb) = nRT$.
For $1 \, \text{mole}$ $(n=1)$,the equation is $\left(P + \frac{a}{V^2}\right)(V - b) = RT$.
Expanding this,we get $PV - Pb + \frac{a}{V} - \frac{ab}{V^2} = RT$.
At low pressure,the volume $V$ is very large,so the terms $\frac{b}{V}$ and $\frac{ab}{V^2}$ are negligible.
The equation simplifies to $PV + \frac{a}{V} = RT$.
Dividing the entire equation by $RT$,we get $\frac{PV}{RT} + \frac{a}{VRT} = 1$.
Since the compressibility factor $Z = \frac{PV}{RT}$,we can write $Z + \frac{a}{VRT} = 1$.
Therefore,$Z = 1 - \frac{a}{VRT}$.

(B) The Van der Waals equation is $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
From the pressure correction term,we have $P = \frac{an^2}{V^2}$.
Rearranging for $a$,we get $a = \frac{P \cdot V^2}{n^2}$.
Substituting the units: $P$ is in $atm$,$V$ is in $L$,and $n$ is in $mol$.
Therefore,the unit of $a$ is $\frac{atm \cdot L^2}{mol^2}$ or $atm \cdot L^2 \cdot mol^{-2}$.

(C) Based on the provided graph of $PV$ versus $P$:
$1$. Gases like $H_2$ and $He$ show only positive deviation from ideal gas behavior at all pressures.
$2$. Gases like $CO$ and $CH_4$ show negative deviation (where $PV < nRT$) at low pressures and positive deviation (where $PV > nRT$) at high pressures.
$3$. Therefore,$CO$ and $CH_4$ exhibit both types of deviations.

(B) The critical temperature $T_{C}$ is given by $\frac{8a}{27bR}$.
The critical pressure $P_{C}$ is given by $\frac{a}{27b^{2}}$.
Therefore,the ratio $\frac{T_{C}}{P_{C}} = \frac{\frac{8a}{27bR}}{\frac{a}{27b^{2}}}$.
Simplifying this expression: $\frac{8a}{27bR} \times \frac{27b^{2}}{a} = \frac{8b}{R}$.

(B) The Van der Waals equation for a real gas is $(P + \frac{a}{V^2})(V - b) = RT$.
At high pressure,the term $\frac{a}{V^2}$ is very small compared to $P$ and can be neglected.
Thus,the equation simplifies to $P(V - b) = RT$.
Expanding this,we get $PV - Pb = RT$.
Dividing both sides by $RT$,we get $\frac{PV}{RT} - \frac{Pb}{RT} = 1$.
Since the compressibility factor $Z = \frac{PV}{RT}$,we have $Z - \frac{Pb}{RT} = 1$.
Therefore,$Z = 1 + \frac{Pb}{RT}$.
Hence,the correct option is $B$.

(C) According to the van der Waals equation,the constant '$a$' represents the magnitude of intermolecular forces of attraction in a gas.
Greater the value of '$a$',stronger are the intermolecular forces of attraction.
Stronger intermolecular forces make it easier for a gas to be liquefied.
Comparing the given values:
$SO_2 = 6.71$
$Cl_2 = 6.49$
$NH_3 = 4.17$
$CO_2 = 3.59$
Since $SO_2$ has the highest value of '$a$',it can be most easily liquefied.

(B) The compressibility factor $Z$ is given by the formula $Z = \frac{PV}{nRT}$.
Given:
$P = 40 \ atm$
$V = 0.4 \ L$
$n = 1 \ mole$
$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$ (using $0.08$ for calculation as per context)
$T = 300 \ K$
Substituting the values:
$Z = \frac{40 \times 0.4}{1 \times 0.08 \times 300}$
$Z = \frac{16}{24} = 0.666... \approx 0.64$ (based on provided options).
Since $Z < 1$,the gas is more compressible than an ideal gas,indicating that attractive forces are dominant.

(B) Inert gases are monoatomic and non-polar in nature.
Since they lack permanent dipoles,the only attractive forces present between their atoms are weak London dispersion forces,also known as instantaneous dipole-induced dipole forces.
These forces arise due to the temporary fluctuation in electron density,which creates a temporary dipole that induces a dipole in the neighboring atom.
Therefore,these forces are responsible for the liquefaction of inert gases at very low temperatures.

(B) The van der Waals equation for $1 \ mol$ of gas is: $(P + a/V^2)(V - b) = RT$.
At high pressure,the volume $V$ is small,making the term $a/V^2$ negligible compared to $P$.
Thus,$(P + a/V^2) \approx P$.
The equation simplifies to: $P(V - b) = RT$.
Expanding this,we get: $PV - Pb = RT$.
Rearranging gives: $PV = RT + Pb$.

(D) Real gases deviate from ideal behavior due to intermolecular forces and the finite volume of gas molecules.
Depending on the conditions of temperature and pressure,the compressibility factor $Z = \frac{PV}{nRT}$ can be greater than $1$ or less than $1$.
Therefore,a real gas is more or less compressible than an ideal gas.

(A) $(i)$ The van der Waals constant $'a'$ represents the magnitude of intermolecular forces of attraction. Larger molecules with more electrons exhibit stronger van der Waals forces. Comparing the given gases,$CO_2$ has the highest molecular mass and number of electrons,thus it has the highest value of $'a'$.
$(ii)$ The van der Waals constant $'b'$ represents the excluded volume,which is proportional to the size of the gas molecules. Among the given gases,$H_2$ is the smallest molecule,therefore it has the lowest value of $'b'$.
Thus,the correct answer is $(i) CO_2, (ii) H_2$.

(A) The Van der Waals equation for $n$ moles of a gas is given by: $(P + \frac{n^2 a}{V^2})(V - nb) = nRT$.
According to the principle of dimensional homogeneity,the term $\frac{n^2 a}{V^2}$ must have the same units as pressure $(P)$.
Therefore,the unit of $\frac{n^2 a}{V^2} = \text{unit of } P = atm$.
Substituting the units: $\frac{(mol)^2 \cdot a}{(L)^2} = atm$.
Rearranging for $a$: $a = atm \cdot L^2 \cdot mol^{-2}$.
Since $1 \, L = 1 \, dm^3$,then $L^2 = (dm^3)^2 = dm^6$.
Thus,the unit of $a$ is $dm^6 \, atm \, mol^{-2}$.

(B) The compressibility factor is defined as $Z = \frac{PV_m}{RT}$.
For an ideal gas,$Z = 1$ and $V_m = 22.4 \ L$ at $NTP$.
If $Z > 1$,the gas shows positive deviation from ideal behavior,meaning the repulsive forces are dominant.
Consequently,the molar volume $V_m$ is greater than the ideal molar volume,i.e.,$V_m > 22.4 \ L$ at $NTP$.

(C) The van der Waals constant '$a$' represents the magnitude of intermolecular forces of attraction in a gas.
As the intermolecular forces of attraction increase,the value of '$a$' increases.
Among the given options,$NH_3$ is a polar molecule capable of forming hydrogen bonds,which results in significantly stronger intermolecular forces compared to the non-polar gases $H_2$,$Ne$,and $He$.
Therefore,$NH_3$ has the highest value of '$a$'.