Real gases and Vander waal’s equation Questions in English

(A) For real gases,the deviation from ideal behavior is described by the compressibility factor $Z = PV/nRT$.
For gases with weak intermolecular forces like $H_2$ and $He$,the repulsive forces dominate even at low pressures,causing $PV$ to increase continuously with pressure.
Among $H_2$ and $He$,$He$ has weaker intermolecular forces and a smaller molecular size,leading to a steeper slope in the $PV$ vs $P$ plot.
In the given graph,curves $1$ and $2$ show a continuous increase in $PV$ with $P$. Curve $1$ has a steeper slope than curve $2$.
Therefore,curve $1$ represents $He$ and curve $2$ represents $H_2$.

(C) The compressibility factor $Z$ is defined as $Z = \frac{pV}{nRT}$.
For real gases,$Z$ deviates from $1$.
Helium $(He)$ is a light,noble gas with very weak intermolecular forces of attraction. For such gases,the volume correction term $(nb)$ dominates over the attractive force term $(a/V^2)$ even at moderate pressures.
Therefore,for helium,$Z$ is always greater than $1$ $(Z > 1)$ and increases linearly with an increase in pressure $p$.
This corresponds to a graph that starts at $Z=1$ and shows a positive slope as $p$ increases.

(B) The compressibility factor $Z$ is defined as $Z = \frac{pV}{nRT}$.
For an ideal gas,$Z = 1$ at all pressures,which corresponds to curve $1$.
Real gases deviate from ideal behavior.
At low pressures,attractive forces dominate,leading to $Z < 1$ (negative deviation),as seen in curve $3$.
At high pressures,repulsive forces dominate,leading to $Z > 1$ (positive deviation),as seen in curves $2$ and $3$.
Therefore,curves $2$ and $3$ represent the behavior of real gases at different conditions of temperature and pressure.

(C) At $NTP$ condition $(p = 1 \ atm, T = 273 \ K)$:
For $1 \ mol$ of $H_2$ gas,the molar volume is slightly greater than the ideal gas volume of $22.4 \ L \ mol^{-1}$ due to repulsive forces.
Using the formula $Z = \frac{pV}{nRT}$,where $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
For hydrogen gas at $273 \ K$ and $1 \ atm$,the value of $Z$ is approximately $1.0006$.
Since $Z > 1$,the compressibility factor for hydrogen gas under these conditions is greater than one.

(C) According to the van der Waals equation for real gases,the pressure correction term is added to the observed pressure to account for intermolecular forces.
The corrected pressure is given by $P_{ideal} = P_{observed} + \frac{a n^2}{V^2}$.
Therefore,the correction term for pressure is $\frac{a n^2}{V^2}$.
Thus,option $(C)$ is the correct answer.

(B) The compressibility factor $(Z)$ is defined as $Z = \frac{PV_m}{RT}$.
For an ideal gas,$Z = 1$ at all pressures,which corresponds to the horizontal line $(A)$.
Real gases show deviations from ideal behavior.
For gases like $H_2$ and $He$,the intermolecular forces are weak,and the volume effect dominates,leading to $Z > 1$ at all pressures,as shown by curve $(B)$.
For easily liquefiable gases like $CO_2$,the attractive forces dominate at low pressures $(Z < 1)$,while the volume effect dominates at high pressures $(Z > 1)$,resulting in the characteristic dip shown by curve $(C)$.
Therefore,$(A)$ is an ideal gas,$(B)$ is $H_2$,and $(C)$ is $CO_2$.

(B) The van der Waals' equation for $n$ moles of a gas is given by $\left(p+\frac{a n^2}{V^2}\right)(V-n b)=n R T$.
Given $n = 0.5 \ mol$,we substitute this into the equation:
$\left(p+\frac{a(0.5)^2}{V^2}\right)(V-0.5 b) = 0.5 R T$
$\left(p+\frac{a}{4 V^2}\right)(V-0.5 b) = 0.5 R T$
Multiplying both sides by $2$,we get:
$\left(p+\frac{a}{4 V^2}\right)(2V-b) = R T$.
Thus,option $B$ is correct.

(D) The Van der Waals equation is a modified form of the ideal gas equation that accounts for the non-ideal behavior of real gases.
For $n$ moles of a real gas,the equation is given by:
$\left(P+\frac{an^2}{V^2}\right)(V-nb)=nRT$
Expanding this expression:
$P(V-nb) + \left(\frac{an^2}{V^2}\right)(V-nb) = nRT$
$PV - Pnb + \frac{an^2}{V} - \frac{an^2b}{V^2} = nRT$
Rearranging the terms,we get:
$PV + \frac{an^2}{V} - \frac{an^2b}{V^2} - Pnb = nRT$
Thus,option $D$ represents the expanded form of the Van der Waals equation.

(C) Real gases deviate from ideal behaviour due to intermolecular forces and the finite volume of gas molecules.
At high temperature,the kinetic energy of gas molecules is high,making intermolecular forces negligible.
At low pressure,the volume occupied by the gas molecules becomes negligible compared to the total volume of the container.
Therefore,real gases approach ideal gas behaviour under conditions of high temperature and low pressure.

(C) Boyle's temperature $(T_B)$ is defined as the temperature at which a real gas behaves like an ideal gas over an appreciable range of pressure.
At this temperature,the second virial coefficient of the real gas becomes zero,and the gas follows the ideal gas equation $PV = nRT$ for a significant pressure range.

(B) For $1 \ mol$ of a van der Waal gas,the equation is $(P+\frac{a}{V^2})(V-b)=RT$.
At high pressure,the term $\frac{a}{V^2}$ is negligible compared to $P$,so $P+\frac{a}{V^2} \approx P$.
Substituting this into the equation,we get $P(V-b)=RT$.
Expanding this gives $PV-Pb=RT$,which rearranges to $PV=RT+Pb$.
The compressibility factor $z$ is defined as $z=\frac{PV}{RT}$.
Dividing the equation $PV=RT+Pb$ by $RT$,we get $z=\frac{RT}{RT}+\frac{Pb}{RT}$.
Therefore,$z=1+\frac{Pb}{RT}$.

(B) The dimensions of the term $\left(\frac{a b}{V^{2}}\right)$ can be analyzed as follows:
In the van der Waals' equation,the term $\left(\frac{a}{V^{2}}\right)$ has the dimensions of pressure $(P)$.
Therefore,$\left(\frac{a}{V^{2}}\right) \times b = \text{Pressure} \times \text{Volume}$.
Since $\text{Pressure} = \frac{\text{Force}}{\text{Area}}$,we have $\text{Pressure} \times \text{Volume} = \frac{\text{Force}}{\text{Area}} \times \text{Volume} = \frac{MLT^{-2}}{L^{2}} \times L^{3} = ML^{2} T^{-2}$.
This dimension $(ML^{2} T^{-2})$ corresponds to energy (joule).
Thus,the term $\left(\frac{a b}{V^{2}}\right)$ represents energy.

(C) The van der Waals' equation for $1$ mole of gas is $(p+\frac{a}{V^2})(V-b) = RT$.
Since '$a$' is negligible,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 $Z = \frac{pV}{RT}$,the equation becomes $Z - \frac{pb}{RT} = 1$.
Therefore,$Z = 1 + \frac{pb}{RT}$.

(C) The van der Waals' equation for $n$ moles of a gas is $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
At very low pressure,the volume $V$ is very large,making the term $\frac{a}{V^2}$ negligible.
Also,the volume $V$ is much larger than the excluded volume $b$,so $(V - b) \approx V$.
Thus,the equation reduces to $PV = RT$,which is the ideal gas equation.
Therefore,a van der Waals' gas behaves ideally at very low pressure.

(A) The van der Waals constant $a$ represents the magnitude of intermolecular attractive forces in a gas.
Higher the value of $a$,stronger are the intermolecular forces,and consequently,the gas is more easily liquefiable.
Given the order of $a$ values is $Q < R < S < P$,the gas $P$ has the highest value of $a$.
Therefore,$P$ is the most liquefiable gas among the given options.