Real gases and Vander waal’s equation Questions in English

(A) The critical constants for a gas obeying the van der Waals equation are derived as follows:
$V_C = 3b$
$P_C = \frac{a}{27b^2}$
$T_C = \frac{8a}{27Rb}$

(A) The Boyle temperature $(T_b)$ is the temperature at which a real gas obeys the ideal gas law over an appreciable range of pressure.
It is defined by the relation $T_b = \frac{a}{Rb}$,where $a$ and $b$ are van der Waals constants and $R$ is the universal gas constant.

(C) Real gases deviate from ideal behavior because of intermolecular forces and the finite volume of gas molecules.
These deviations become significant when the pressure is high (making the volume of molecules significant compared to the total volume) and the temperature is low (reducing kinetic energy,making intermolecular forces significant).
Therefore,real gases deviate most from ideal behavior at $High \ Pressure$ and $Low \ Temperature$.

(B) The compressibility factor is defined as $Z = \frac{PV_m}{RT}$.
At $STP$,the molar volume of an ideal gas is $V_{ideal} = 22.4 \ L$.
Since $Z = \frac{V_{real}}{V_{ideal}}$,if $Z < 1$,then $V_{real} < V_{ideal}$.
Therefore,the molar volume will be less than $22.4 \ L$.

(A) The critical temperature $(T_c)$ of a gas is given by the formula: $T_c = \frac{8a}{27Rb}$.
To find the gas with the highest $T_c$,we need to calculate the ratio $(a / b)$ for each gas:
For $Ar$: $1.3 / 3.2 \approx 0.406$
For $Ne$: $0.2 / 1.7 \approx 0.117$
For $Kr$: $5.1 / 1.0 = 5.1$
For $Xe$: $4.1 / 5.0 = 0.82$
Comparing these values,the ratio $(a / b)$ is highest for $Kr$.
Therefore,$Kr$ is expected to have the highest critical temperature.

(D) Below the critical temperature $(T_c)$,the isotherms show a distinct region where the gas and liquid phases coexist. In this region,the pressure remains constant as the volume decreases during the condensation process. This results in a characteristic horizontal or nearly horizontal segment in the $P-V$ isotherm,as shown in the provided figure (option $D$).

(C) The compressibility factor $Z$ is defined as $Z = \frac{PV_{m}}{RT}$.
Given the equation of state: $P(V_{m}-b) = RT$,we can write $PV_{m} - Pb = RT$.
Dividing by $RT$,we get $\frac{PV_{m}}{RT} - \frac{Pb}{RT} = 1$,which simplifies to $Z - \frac{Pb}{RT} = 1$,or $Z = 1 + \frac{Pb}{RT}$.
Now,we differentiate $Z$ with respect to $P$ at constant temperature $T$: $(\frac{\partial Z}{\partial P})_{T} = \frac{\partial}{\partial P}(1 + \frac{Pb}{RT}) = 0 + \frac{b}{RT} = \frac{1 \times b}{RT}$.
Comparing this with the given expression $\frac{xb}{RT}$,we find that $x = 1$.

(D) In the van der Waals equation $(P + \frac{an^{2}}{V^{2}})(V - nb) = nRT$,the term $\frac{an^{2}}{V^{2}}$ is added to pressure $P$.
According to the principle of dimensional homogeneity,only quantities with the same units can be added.
Therefore,the unit of $\frac{an^{2}}{V^{2}}$ must be the same as the unit of pressure ($atm$ or $bar$ or $Pa$).
$\frac{an^{2}}{V^{2}} = \text{Pressure} \Rightarrow a = \text{Pressure} \times \frac{V^{2}}{n^{2}}$.
Substituting the units: $a = atm \times \frac{(dm^{3})^{2}}{(mol)^{2}} = atm \, dm^{6} \, mol^{-2}$.

(A) For a real gas under high pressure,the compressibility factor $Z$ is given by the equation: $Z = 1 + \frac{Pb}{RT}$.
Given $Z = 2$,$P = 99 \ bar$,$T = 25 + 273 = 298 \ K$,and $R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$.
Substituting the values: $2 = 1 + \frac{99 \times b}{0.083 \times 298}$.
$1 = \frac{99 \times b}{24.734}$.
$b = \frac{24.734}{99} \approx 0.2498 \ L \ mol^{-1}$.
$b \approx 0.25 \ L \ mol^{-1} = 25 \times 10^{-2} \ L \ mol^{-1}$.
Thus,the nearest integer value is $25$.

(B) The van der Waals' equation of state for a real gas is given by $(p + \frac{an^2}{V^2})(V - nb) = nRT$.
For a fixed amount of gas at a constant temperature,the $p-V$ isotherm below the critical temperature exhibits a characteristic '$S$' shape,which represents the region of gas-liquid coexistence.
In this region,the pressure remains constant as the volume changes during the phase transition.
The curve shown in option $(b)$ correctly depicts this behavior,showing the characteristic inflection point and the region of phase transition for a van der Waals' gas.

(A) The compressibility factor is defined as $Z = \frac{p V}{n R T}$.
For an ideal gas,$Z = 1$ at all temperatures and pressures.
For a real gas,$Z = \frac{p V_{real}}{n R T} \quad (i)$.
If the gas shows ideal behaviour,then $V_{ideal} = \frac{n R T}{p}$,which implies $\frac{p}{n R T} = \frac{1}{V_{ideal}} \quad (ii)$.
Substituting $(ii)$ into $(i)$,we get $Z = \frac{V_{real}}{V_{ideal}}$.
Considering molar volume,$Z = \frac{(V_{m})_{real}}{(V_{m})_{ideal}}$.
From the given graph,$Z < 1$ at pressure $< 200 \, bar$.
Therefore,$\frac{(V_{m})_{real}}{(V_{m})_{ideal}} < 1$,which implies $(V_{m})_{CH_{4}, real} < (V_{m})_{CH_{4}, ideal}$.

(B) The van der Waals' equation for $n = 1$ mole is $(p + a/V^2)(V - b) = RT$.
Rearranging for the compressibility factor $Z = (pV/RT)$,we get $Z = (V/(V - b)) - (a/(RTV))$.
To decrease $Z$ at a fixed volume $V$,the term $(V/(V - b))$ should decrease and the term $(a/(RTV))$ should increase.
$1$. For $(V/(V - b))$ to decrease,the denominator $(V - b)$ must increase,which means $b$ must decrease.
$2$. For $(a/(RTV))$ to increase,$a$ must increase (at constant $T$ and $V$).
Therefore,$Z$ decreases if $b$ decreases and $a$ increases at constant temperature.

(A) For $1$ mole of a real gas,the van der Waals equation is $(P + \frac{a}{V_m^2})(V_m - b) = RT$.
At low pressure,the attractive forces are dominant,so $Z < 1$.
The compressibility factor $Z$ is given by $Z = \frac{PV_m}{RT}$.
For low pressure,the approximate form of the van der Waals equation is $Z = 1 - \frac{a}{V_mRT}$.
At point $A$,which lies in the region where $Z < 1$ (below the ideal gas line),the compressibility factor is $1 - \frac{a}{V_mRT}$ (or $1 - \frac{a}{RTV}$).

(A) The van der Waals constant '$a$' represents the magnitude of intermolecular forces of attraction in a gas.
It is directly proportional to the molecular size and surface area of the gas molecules.
The values for '$a$' (in $\text{bar L}^2 \text{mol}^{-2}$) are:
$(i)$ $Ar = 1.34$
$(ii)$ $CH_4 = 2.25$
$(iii)$ $H_2O = 5.46$
$(iv)$ $C_6H_6 = 18.57$
Comparing these values,the increasing order is $Ar < CH_4 < H_2O < C_6H_6$,which corresponds to $A < B < C < D$.

(B) The compressibility factor $Z$ is defined as $Z = \frac{PV}{nRT}$,which implies $n = \frac{PV}{ZRT}$.
Since the amount of gas $n$ remains constant,we have $\frac{P_1 V_1}{Z_1 T_1} = \frac{P_2 V_2}{Z_2 T_2}$.
Given values: $P_1 = 100 \ atm$,$V_1 = 0.15 \ dm^3$,$T_1 = 500 \ K$,$Z_1 = 1.07$.
$P_2 = 300 \ atm$,$T_2 = 300 \ K$,$Z_2 = 1.4$.
Substituting these values into the equation:
$\frac{100 \times 0.15}{1.07 \times 500} = \frac{300 \times V_2}{1.4 \times 300}$
$V_2 = \frac{100 \times 0.15 \times 1.4 \times 300}{1.07 \times 500 \times 300}$
$V_2 = \frac{15 \times 1.4}{1.07 \times 500} = \frac{21}{535} \approx 0.03925 \ dm^3$.
Converting to $10^{-4} \ dm^3$ units: $0.03925 \times 10^4 \times 10^{-4} = 392.5 \times 10^{-4} \ dm^3$.
Rounding to the nearest integer,we get $393 \times 10^{-4} \ dm^3$.

(D) The compressibility factor $Z$ for a van der Waals gas is given by $Z = 1 + \frac{(b - a/RT)P}{RT}$ at low pressure.
If $a=0$,$Z = 1 + \frac{bP}{RT}$,which is a linear equation with a positive slope (Gas $A$).
If $b=0$,$Z = 1 - \frac{a}{RT^2}P$,which is a linear equation with a negative slope (Gas $B$).
Gas $C$ represents a typical real gas where both $a$ and $b$ are non-zero,showing a curve with a minimum.
Statement $D$ is incorrect because at high pressure,$Z = 1 + \frac{Pb}{RT}$,which is linear with a positive slope,but the graph for gas $B$ shows a negative slope at all pressures,contradicting the statement that the slope is positive for all real gases at high pressure.

(A) The van der Waal's equation is given by $\left(P + \frac{n^2 a}{V^2}\right)(V - nb) = nRT$.
$A$. At large molar volumes $(V \to \infty)$,the terms $\frac{n^2 a}{V^2}$ and $nb$ become negligible,so the equation reduces to $PV = nRT$,behaving like an ideal gas. This is correct.
$B$. At large pressures,the volume $V$ is small,and the correction terms become significant,so it deviates from ideal behavior. This is incorrect.
$C$. The constants $a$ (intermolecular forces) and $b$ (excluded volume) are characteristic of the specific gas and are independent of temperature. This is correct.
$D$. Since $P = \frac{nRT}{V-nb} - \frac{n^2 a}{V^2}$,the pressure $P$ of a real gas is lower than the ideal pressure $P_{ideal} = \frac{nRT}{V}$ due to the attractive force term $\frac{n^2 a}{V^2}$. This is correct.

(B) The van der Waals equation for $n$ moles of a real gas is given by:
$(P + \frac{an^2}{V^2})(V - nb) = nRT$
In this equation,the term $\frac{an^2}{V^2}$ is added to the measured pressure $P$ to account for the intermolecular attractive forces between gas molecules.
Thus,$\frac{an^2}{V^2}$ is the correction term for attractive forces.

(C) For the given gas: $Z = 0.5$,$V_m = 0.4 \ dm^3 \ mol^{-1}$,$T = 800 \ K$,$P = x \ atm$.
The compressibility factor is given by $Z = \frac{P V_m}{R T}$.
Substituting the values: $0.5 = \frac{x \times 0.4}{0.08 \times 800}$.
$0.5 = \frac{0.4x}{64}$ $\Rightarrow 0.4x = 32$ $\Rightarrow x = 80 \ atm$.
For ideal gas behavior at the same $T$ and $P$,$PV_m = RT$.
$y = V_m = \frac{RT}{P} = \frac{0.08 \times 800}{80} = 0.8 \ dm^3 \ mol^{-1}$.
Therefore,the value of $\frac{x}{y} = \frac{80}{0.8} = 100$.

(C) The van der Waals equation for $1 \ \text{mole}$ of gas is $(P + \frac{a}{V^2})(V - b) = RT$.
Given $b = 0$,the equation simplifies to $(P + \frac{a}{V^2})V = RT$.
Expanding this,we get $PV + \frac{a}{V} = RT$,which can be rearranged as $PV = RT - a(\frac{1}{V})$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = PV$,$x = \frac{1}{V}$,$m = -a$,and $c = RT$.
The slope of the line is $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{20.1 - 21.6}{3.0 - 2.0} = \frac{-1.5}{1.0} = -1.5$.
Since the slope $m = -a$,we have $-a = -1.5$,which gives $a = 1.5 \ \text{atm} \cdot \text{liter}^2 \cdot \text{mol}^{-2}$.

(C) The given equation is $p(V-b) = RT$. This is a simplified form of the van der Waals equation where the attractive force constant $a$ is assumed to be $0$ $(a = 0)$.
This implies that there are no long-range attractive forces between the gas particles.
However,the constant $b$ represents the excluded volume,which accounts for the finite size of the gas particles and the short-range repulsive forces.
For $r < d$ (where $d$ is the hard-sphere diameter),the potential energy $V(r)$ becomes infinite due to strong repulsion.
For $r \geq d$,since $a = 0$,there is no attraction,so the potential energy $V(r)$ remains $0$.
This corresponds to a hard-sphere potential model,which is represented by a graph that is $0$ for $r > d$ and rises vertically to infinity at $r = d$.

(C) The compressibility factor $(Z)$ is defined as the ratio of the molar volume of a real gas $(V_m)$ to the molar volume of an ideal gas $(V_{ideal})$ at the same temperature and pressure.
$Z = \frac{V_m}{V_{ideal}}$
At $STP$,the molar volume of an ideal gas $(V_{ideal})$ is $22.40 \ dm^3 \ mol^{-1}$.
Given $Z = 1.05$,we can calculate the molar volume of the real gas $(V_m)$ as:
$V_m = Z \times V_{ideal}$
$V_m = 1.05 \times 22.40 \ dm^3 \ mol^{-1}$
$V_m = 23.52 \ dm^3 \ mol^{-1}$

(C) The compressibility factor,denoted by $Z$,is defined as the ratio of the actual molar volume of a gas to the molar volume of an ideal gas at the same temperature and pressure.
For $n$ moles of a gas,the ideal gas equation is $PV = nRT$.
Therefore,the compressibility factor is given by the ratio $Z = \frac{PV}{nRT}$.
Thus,option $C$ is the correct formula.

(C) The compressibility factor $Z$ is defined as the ratio of the volume of a real gas $(V_{real})$ to the volume of an ideal gas $(V_{ideal})$ at the same temperature and pressure.
$Z = \frac{V_{real}}{V_{ideal}}$
Given $Z = 1.1$ and $V_{ideal} = 22.4 \ dm^3$ at $STP$.
$V_{real} = Z \times V_{ideal} = 1.1 \times 22.4 \ dm^3 = 24.64 \ dm^3$.

(B) The compressibility factor $(Z)$ is defined as the ratio of the actual molar volume of a gas $(V_m)$ to the molar volume of an ideal gas $(V_{ideal})$ at the same temperature and pressure.
For an ideal gas,the equation of state is $PV = nRT$. Thus,$V_{ideal} = \frac{nRT}{P}$.
The compressibility factor is given by the formula $Z = \frac{PV}{nRT}$.
For an ideal gas,$Z = 1$. For real gases,$Z \neq 1$,which indicates the deviation from ideal behavior.

(A) The force of attraction in real gases tends to show slightly lower pressure as compared to ideal gases.
$p_{ideal} = p_{real} + \frac{a n^2}{V^2}$ or $p_{real} = p_{ideal} - \frac{a n^2}{V^2}$
Here,'$a$' is the constant which measures the magnitude of attractive forces among the molecules of gas,'$n$' is the number of moles present,and '$V$' is the volume of the gas.

(A) The wrong relation for real gases is given in option $A$.
The compressibility factor $Z$ is defined as the ratio of the molar volume of a real gas to the molar volume of an ideal gas at the same temperature and pressure.
Therefore,the correct form is $Z = \frac{V_{real}}{V_{ideal}}$.
Option $B$ represents the pressure correction term in the van der Waals equation.
Option $C$ represents the volume correction term where $V_{real} = V_{ideal} + nb$ is generally used for the excluded volume correction,but in the context of the van der Waals equation $(p + \frac{an^2}{V^2})(V - nb) = nRT$,the relation $V_{real} = V_{ideal} + nb$ is standard.
Option $D$ is the van der Waals equation for $1 \text{ mole}$ of gas.

(A) At high pressure,the volume of the gas is decreased significantly,which brings the molecules closer to each other.
Due to this proximity,the intermolecular attractive forces become significant.
Consequently,the gas deviates from ideal behaviour as the assumption of negligible intermolecular forces in the kinetic molecular theory of gases no longer holds true.

(B) The compressibility factor $Z$ is defined as $Z = \frac{PV}{nRT}$.
Given values are $Z = 1.1$,$n = 1 \ mol$,$T = 300 \ K$,$P = 2.706 \ atm$,and $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$.
Rearranging the formula for volume $V$: $V = \frac{ZnRT}{P}$.
Substituting the values: $V = \frac{1.1 \times 1 \times 0.082 \times 300}{2.706}$.
$V = \frac{27.06}{2.706} = 10 \ L$.

(C) For an ideal gas,the compressibility factor $Z = \frac{PV}{RT} = 1$.
From the given data table,we observe that for pressure values $P = 1, 2, \text{ and } 3 \text{ bar}$,the value of $Z = 1$,which indicates ideal behavior.
However,for pressure values ranging from $3.1 \text{ bar}$ to $4.0 \text{ bar}$,the value of $Z$ is greater than $1$ $(1.2, 1.4, 1.5)$,which indicates a deviation from ideal behavior.
Therefore,the gas deviates from ideal behavior in the pressure range $3.1 \text{ to } 4.0 \text{ bar}$.

(C) The van der Waals equation is given by $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
For $1 \ mol$ of gas,this becomes $(P + \frac{a}{V^2})(V - b) = RT$.
At high temperature,the kinetic energy of gas molecules is very high,making the intermolecular forces of attraction (represented by the constant $a$) negligible.
At low pressure,the volume $V$ is very large,making the volume occupied by gas molecules (represented by the constant $b$) negligible compared to the total volume.
Under these conditions,$(P + 0)(V - 0) = RT$,which simplifies to $PV = RT$,the ideal gas equation.

(A) 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 $1 \text{ mole}$ of a real gas,we substitute $n = 1$ into the equation.
This results in $(p + \frac{a}{V^2})(V - b) = RT$.

(D) For an ideal gas,the compressibility factor $Z$ is defined as $Z = \frac{pV}{nRT}$.
For an ideal gas,$pV = nRT$,therefore $Z = 1$ for all values of pressure $p$.
Thus,the graph of $Z$ versus $p$ for an ideal gas is a horizontal straight line at $Z = 1.0$.

(D) The compressibility factor $(Z)$ is defined as $Z = \frac{PV}{nRT}$.
Given values are $P = 2.706 \ atm$,$V = 10 \ L$,$n = 1 \ mol$,$T = 300 \ K$,and $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$.
Substituting these values into the formula:
$Z = \frac{2.706 \times 10}{1 \times 0.082 \times 300}$
$Z = \frac{27.06}{24.6}$
$Z = 1.1$.

(A) In the graph of compressibility factor $Z$ versus pressure $P$,a positive deviation from ideal behavior occurs when $Z > 1$.
This indicates that the gas is less compressible than an ideal gas.
From the given graph,the curves for gases $a$,$b$,and $c$ lie entirely above the line $Z = 1$ for all pressures.
Therefore,gases $a$,$b$,and $c$ show only positive deviation from ideal behavior.

(C) The ideal gas equation is $pV = nRT$. At constant temperature,$p = \frac{nRT}{V}$,which represents a rectangular hyperbola $(p \propto \frac{1}{V})$.
For a real gas,the behavior deviates from the ideal gas law due to intermolecular forces and finite molecular volume.
At high pressures,the volume of a real gas is greater than that of an ideal gas because of the excluded volume effect ($V_{real} > V_{ideal}$ for a given $p$).
Therefore,for a fixed pressure,the curve for the real gas $(A)$ lies above the curve for the ideal gas $(B)$ in the $p$ vs $V$ plot.
This corresponds to the plot shown in option $C$.

(C) The deviation from ideal gas behavior is primarily determined by the magnitude of intermolecular forces and the molecular size,represented by the van der Waals constants $a$ and $b$.
$NH_3$ molecules exhibit strong intermolecular forces due to hydrogen bonding,which is significantly stronger than the London dispersion forces present in $He$,$CH_4$,and $H_2$.
Because $NH_3$ has the highest attractive forces (highest $a$ value) among the given options,it deviates the most from ideal gas behavior under the same conditions of temperature and pressure.

(D) Statement $(i)$ is incorrect because for real gases,$Z$ can be equal to $1$ at specific conditions (Boyle temperature).
Statement $(ii)$ is correct because real gases behave ideally at low pressure and high temperature.
Statement $(iii)$ is incorrect because real gases possess intermolecular forces of attraction.
Statement $(iv)$ is incorrect because real gases obey the Van der Waals equation,$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$,not the ideal gas equation $PV = nRT$.
Therefore,statements $(i), (iii),$ and $(iv)$ are incorrect. However,based on the provided options,$(iii)$ and $(iv)$ are clearly incorrect.

(A) Real gases deviate from ideal behavior because of the intermolecular forces and the finite volume of gas molecules.
At $high$ $pressure$,the volume of gas molecules becomes significant compared to the total volume.
At $low$ $temperature$,the kinetic energy of molecules decreases,allowing intermolecular forces (van der Waals forces) to become significant.
Therefore,a gas deviates most from ideal behavior under conditions of $low$ $temperature$ and $high$ $pressure$.

(B) Real gases exhibit ideal gas behavior at high temperature and low pressure.
Under these conditions,the volume $V$ is very large,and the correction terms in the van der Waals equation,specifically $\frac{a}{V^2}$ and $b$,become negligible.
Consequently,the van der Waals equation,$(p + \frac{an^2}{V^2})(V - nb) = nRT$,simplifies to the ideal gas equation,$pV = nRT$.

(D) The kinetic molecular theory of gases assumes that there are no intermolecular forces of attraction between gas molecules and that the volume of the gas molecules is negligible compared to the total volume of the gas.
Real gases deviate from ideal behavior primarily because these two assumptions are not strictly true.
Specifically,real gas molecules possess a finite volume and exert intermolecular forces of attraction on each other,which causes them to deviate from the ideal gas law $(PV = nRT)$.

(A) The ease of liquefaction of a gas depends on the magnitude of intermolecular forces of attraction,which is represented by the van der Waals' constant '$a$'.
Greater the value of '$a$',stronger are the intermolecular forces,and easier it is to liquefy the gas.
Among the given gases,$SO_2$ has the highest value of '$a$' due to its polar nature and strong dipole-dipole interactions.
Additionally,$SO_2$ has a high critical temperature,making it the easiest to liquefy under moderate conditions.

(D) The van der Waals' constant $a$ represents the magnitude of intermolecular attractive forces in a gas.
Greater intermolecular attraction leads to a higher value of $a$.
Among the given options,$NH_3$ molecules exhibit strong hydrogen bonding in addition to dipole-dipole interactions.
$H_2$,$O_2$,and $CH_4$ only exhibit weaker London dispersion forces.
Therefore,$NH_3$ has the strongest intermolecular forces and the highest value of $a$.

(C) The Joule-Thomson effect describes the temperature change of a real gas as it is forced through a valve or porous plug while kept insulated so that no heat is exchanged with the environment.
For most gases,expansion results in cooling because work is done to overcome intermolecular forces.
However,for gases like $H_2$ and $He$,the inversion temperature is very low (for $He$,it is approximately $40 \ K$).
If the gas temperature is above its inversion temperature during expansion,it exhibits a heating effect instead of cooling.
Since the inversion temperature of $He$ is very low,it shows a heating effect under standard conditions.

(A) Real gases deviate from ideal behaviour because they possess intermolecular forces of attraction and have a finite volume.
At high pressures,the molecules are closer to each other,making the intermolecular forces of attraction significant.
These attractive forces pull the molecules inward,reducing the frequency and force of collisions with the container walls,which causes the observed pressure to be lower than the ideal pressure.

(B) The compressibility factor $z$ is defined as $z = \frac{PV}{nRT}$.
For real gases,the deviation from ideal behavior depends on the magnitude of intermolecular attractive forces.
$CO_2$ is a gas with relatively strong intermolecular forces compared to $H_2, He,$ and $N_2$,leading to a more significant negative deviation $(z < 1)$ at low pressures.
Among the given curves,the curve that shows the deepest dip below the ideal line $(z = 1)$ corresponds to the gas with the strongest attractive forces.
Therefore,curve $A$ represents $SO_2$,curve $B$ represents $CO_2$,curve $C$ represents $N_2$,and curve $D$ represents $H_2$ (which shows $z > 1$ due to dominant repulsive forces).
Thus,the plot for $CO_2$ gas is $B$.

(A) The compressibility factor $(z)$ is defined as $z = PV/nRT$.
For real gases,the deviation from ideal behavior is primarily due to intermolecular forces of attraction,represented by the van der Waals constant $'a'$.
$A$ higher value of $'a'$ indicates stronger intermolecular forces of attraction.
Stronger attractive forces lead to a greater negative deviation from ideal behavior,resulting in a lower compressibility factor $(z < 1)$.
Since $NH_3$ and $CO_2$ are polar or more easily liquefiable than $N_2$,they have larger values of the constant $'a'$ compared to $N_2$.
Therefore,the compressibility factor is lower for $NH_3$ and $CO_2$ than for $N_2$.

(D) For a real gas,the van der Waals equation is $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
For $1 \ mol$ of gas,this becomes $(P + \frac{a}{V^2})(V - b) = RT$.
At high pressure,the volume $V$ is small,so the term $\frac{a}{V^2}$ is negligible compared to $P$.
The equation simplifies to $P(V - b) = RT$.
$PV - Pb = RT$.
Dividing 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,$Z = 1 + \frac{Pb}{RT}$.

(D) The van der Waal's constant '$a$' is a measure of the magnitude of attractive forces between gas molecules.
Larger molecules or those with stronger intermolecular forces (like dipole-dipole interactions or hydrogen bonding) exhibit higher values of '$a$'.
Among the given options,$NH_3$ is a polar molecule capable of forming hydrogen bonds,which results in stronger intermolecular attractions compared to $H_2$,$He$,and $CO_2$.
Therefore,$NH_3$ has the maximum value of '$a$'.

(C) For a real gas at high pressure,the van der Waals equation is $P(V - nb) = nRT$.
Given: $n = 1 \ mol$,$T = 300 \ K$,$P = 100 \ bar$,$b = 0.005 \ L \ mol^{-1}$,$R = 0.08314 \ bar \ L \ K^{-1} \ mol^{-1}$.
Substituting the values: $100(V - 0.005) = 1 \times 0.08314 \times 300$.
$100(V - 0.005) = 24.942$.
$V - 0.005 = 0.24942$.
$V_{real} = 0.25442 \ L$.
The ideal volume is $V_{ideal} = \frac{nRT}{P} = \frac{1 \times 0.08314 \times 300}{100} = 0.24942 \ L$.
Compressibility factor $Z = \frac{PV_{real}}{nRT} = \frac{100 \times 0.25442}{1 \times 0.08314 \times 300} = \frac{25.442}{24.942} \approx 1.02$.
$\%$ Deviation of volume $= \frac{V_{real} - V_{ideal}}{V_{real}} \times 100 = \frac{0.25442 - 0.24942}{0.25442} \times 100 = \frac{0.005}{0.25442} \times 100 \approx 1.96 \% \approx 2 \%$.
Thus,the values are $Z = 1.02$ and $\%$ deviation $= 2$.