States of Matter

For ideal gas $A, w_A=4 \mathrm{~g}$

Molar mass $=M_A$

Volume $=V \mathrm{~L}$

Pressure, $p_A=5 \mathrm{~atm}$

Temperature $=300 \mathrm{~K}$

For ideal gas $B, w_B=16 \mathrm{~g}$

Molar mass $=M_B$

Volume and temperature are same as of ideal gas $A$.

$ \begin{aligned} \because \quad p_{\text {Total }} & =p_A+p_B=10 \mathrm{~atm} \\ \therefore \quad p_B & =10-5=5 \mathrm{~atm} \\ p V & =n R T \text { (for ideal gas) } \\ & =\frac{w}{M} R T \end{aligned} $

For gas $A, p_A=\frac{4}{M_A} \cdot \frac{R T}{V}$

$ \Rightarrow \quad 5=\frac{4}{M_A} \cdot \frac{R T}{V} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For gas $B, p_B=\frac{16}{M_B} \cdot \frac{R T}{V}$

$ \begin{aligned} &5=\frac{16}{M_B} \cdot \frac{R T}{V}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ &\text { From Eq. (i) and Eq. (ii), we get }\\ &\begin{aligned} & \frac{5}{5}=\frac{\frac{4}{M_A} \cdot \frac{R T}{V}}{\frac{16}{M_B} \cdot \frac{R T}{V}} \Rightarrow 1=\frac{\frac{4}{M_A}}{\frac{16}{M_B}} \\ & \Rightarrow \quad \frac{1}{1}=\frac{\frac{1}{M_A}}{\frac{4}{M_B}} \Rightarrow \frac{1}{1}=\frac{M_B}{4 M_A} \\ & \Rightarrow 4 M_A=M_B \end{aligned} \end{aligned} $

Rate of diffusion

$ \propto \sqrt{\frac{1}{\text { Molecular mass }}} $

So, for $X$, rate of diffusion will be

$ r_X \propto \sqrt{\frac{1}{36}} \Rightarrow r_X \propto \frac{1}{6} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Similarly for $Y$, rate of diffusion, will be

$ r_Y \propto \sqrt{\frac{1}{64}} \Rightarrow r_Y \propto \frac{1}{8}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

Divide Eq. (i) and Eq. (ii),

So, $\frac{r_X}{r_Y}=\frac{1}{6} / \frac{1}{8}=\frac{8}{6}=\frac{4}{3}=4: 3$

According to Graham's law, the rate of diffusion of gases is inversely proportional to the square root of their molar masses. This relationship is represented as:

$ \frac{r_0}{r_X} = \sqrt{\frac{M_X}{M_0}} $

where:

$ r_0 $ and $ r_X $ represent the rate of diffusion of oxygen and another gas $ X $, respectively.

$ M_X $ and $ M_0 $ are the molar masses of gas $ X $ and oxygen, respectively.

The rate of diffusion is defined as the volume diffused divided by the time taken, which gives us:

$ \text{Rate of diffusion} = \frac{\text{Volume diffused}}{\text{Time taken}} $

By applying this to the context of diffusion for gases, we reformulate equation (i) as:

$ \frac{\left(\frac{V_0}{20}\right)}{\left(\frac{V_X}{Y}\right)} = \sqrt{\frac{M_X}{16}} $

Given that the volumes $ V_0 $ (oxygen) and $ V_X $ (gas $ X $) are equal, the equation simplifies to:

$ \frac{Y}{20} = \frac{\sqrt{M_X}}{4} $

Solving for $ Y $ and $\sqrt{M_X}$:

$\frac{Y}{\sqrt{M_X}} = 5$

Therefore, $ Y = 5 $ seconds.

$\sqrt{M_X} = 1$, which implies $ M_X = 1^2 = 1$.

The molar mass of 1 indicates that the gas $ X $ is hydrogen ($ \mathrm{H}_2 $).

To find the mass percent of oxygen in a gas mixture at 300 K and 760 torr with a density of $0.543 \, \text{g/L}$, we can use the ideal gas law in its density form:

$ p \times M = d \times R \times T $

where:

$ p = $ pressure in atm (convert 760 torr to 1 atm),

$ M = $ molar mass of the gas mixture,

$ d = $ density ($0.543 \, \text{g/L}$),

$ R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} $,

$ T = $ temperature ($300 \, \text{K}$).

First, calculate the molar mass $ M $:

$ M = \frac{d \times R \times T}{p} = \frac{0.543 \times 0.0821 \times 300}{1} = 13.37 \, \text{g/mol} $

Assume $ y $ is the mole fraction of oxygen ($\text{O}_2$) in the mixture and $ (1-y) $ is the mole fraction of helium ($\text{He}$).

The average molecular weight of the mixture is given by:

$ M = y \times M_{\text{O}_2} + (1-y) \times M_{\text{He}} $

Substitute the known values:

$ 13.37 = y \times 32 + (1-y) \times 4 $

$ 13.37 = 32y + 4 - 4y $

$ 13.37 = 28y + 4 $

$ 9.37 = 28y $

$ y = \frac{9.37}{28} \approx 0.335 $

The mass percent of oxygen is calculated as follows:

$ \text{Mass percent of } \text{O}_2 = \frac{y \times M_{\text{O}_2} \times 100}{(1-y) \times M_{\text{He}} + y \times M_{\text{O}_2}} $

$ = \frac{0.335 \times 32 \times 100}{(1-0.335) \times 4 + 0.335 \times 32} $

Simplifying:

$ = \frac{10.72}{10.72 + 2.66} \times 100 $

$ = \frac{10.72}{13.38} \times 100 \approx 80\% $

Thus, the mass percent of oxygen in the mixture is approximately $80\%$.

To determine the pressure ratio of the two gases, we can use the ideal gas law given by

$PV = nRT$

where:

• $ P $ = Pressure
• $ V $ = Volume
• $ n $ = Number of moles
• $ R $ = Universal gas constant
• $ T $ = Temperature

Since the volumes and temperatures of both gases are the same, we can simplify the relationship to

$P \propto n$

Therefore, the pressure ratio will be equal to the ratio of the number of moles of the two gases. First, we need to find the number of moles of each gas.

The number of moles is calculated as:

$n = \frac{m}{M}$

where $ m $ is the mass and $ M $ is the molar mass.

For methane, $ \text{CH}_4 $:

$M_{\mathrm{CH}_4} = 12 + 4 \times 1 = 16 \mathrm{\ g/mol}$

$n_{\mathrm{CH}_4} = \frac{10 \mathrm{\ g}}{16 \mathrm{\ g/mol}} = 0.625 \mathrm{\ mol}$

For oxygen, $ \mathrm{O}_2 $:

$M_{\mathrm{O}_2} = 2 \times 16 = 32 \mathrm{\ g/mol}$

$n_{\mathrm{O}_2} = \frac{10 \mathrm{\ g}}{32 \mathrm{\ g/mol}} = 0.3125 \mathrm{\ mol}$

Now, we calculate the pressure ratio:

$\frac{P_{\mathrm{CH}_4}}{P_{\mathrm{O}_2}} = \frac{n_{\mathrm{CH}_4}}{n_{\mathrm{O}_2}} = \frac{0.625}{0.3125} = 2$

Therefore, the pressure ratio of $ \mathrm{CH}_4 $ to $ \mathrm{O}_2 $ is 2:1.

The correct option is Option D.

The correct order of pressure is $p_1>p_3>p_2$.

Given, $T=300 \mathrm{~K}$

$ \begin{aligned} & n=1 \text { mole } \\ & p=100 \text { bar } \\ & b=0.005 \mathrm{~L} / \mathrm{mol} \\ & Z=? \end{aligned} $

% deviation = ?

As pressure is very high $a$ will be negligible.

$ \therefore\left(p+\frac{a n^2}{V^2}\right)(V-n b)=n R T $

becomes

$ \begin{aligned} p(V-n b) & =n R T \\ 100(V-1 \times 0.005) & =1 \times 0.0821 \times 300 \\ 100 V-0.5 & =24.63 \\ 100 V=25.13 & \\ \Rightarrow \quad V=\frac{25.13}{100} & =0.2513 \end{aligned} $

Now, $Z=\frac{p V}{R T}$

$ =\frac{100 \times 0.2513}{0.0821 \times 300}=1.02 $

% deviation in volume

$ \begin{aligned} & =\frac{0.2513-0.2463}{0.2513} \times 100 \\ & \cong 2 \% \end{aligned} $

Given, $V=1 \mathrm{~L}$

$ \begin{aligned} \text { Methane } & =4 \mathrm{~g} \\ \mathrm{CO}_2 & =4.4 \mathrm{~g} \\ p & =? \\ T & =27^{\circ} \mathrm{C}, \text { i.e. } 300 \mathrm{~K} \end{aligned} $

Number of moles of methane $=\frac{4}{16}$

$ =\frac{1}{4}=0.25 $

Number of moles of

$ \mathrm{CO}_2=\frac{4.4}{44}=\frac{1}{10}=0.1 $

Total number of moles of gases $=0.35$

$ \begin{aligned} p V & =n R T(\text { ideal gas equation }) \\ p \times 1 & =0.35 \times 0.0821 \times 300 \\ p & =8.62 \mathrm{~atm} \end{aligned} $

Rate of diffusion of gases is directly proportional to partial pressure and inversely proportional to square root of molecular mass.

$ \frac{R_1}{R_X}=\sqrt{\frac{M_X}{M_1}} $

(where $R_1=$ rate of diffusion of methane gas,

$R_X=$ rate of diffusion of ' $X^{\prime}$ gas,

$M_X=$ molecular mass of ' $X^{\prime}$ gas,

$M_1=$ molecular mass of methane gas)

As, $R_1=2 R_X$

$ \begin{aligned} \frac{2 R_X}{R_X} & =\sqrt{\frac{M_X}{16}} \\ 2 & =\sqrt{\frac{M_X}{16}} \end{aligned} $

Squaring both sides,

$ \begin{aligned} 4 & =\frac{M_X}{16} \\ \therefore \quad M_X & =64 \end{aligned} $

$\mathrm{NH}_3$ has largest value of ' $a$ ' van der Waals' constant due to higher force of attraction in ammonia, i.e. hydrogen bonding.

Real gas equation is

$ \left(p+\frac{a}{V^2}\right)(V-b)=R T $

At high pressure, $p \ggg \frac{a}{V^2}$

So, $\frac{a}{V^2}$ can be neglected

$ \begin{array}{rlrl} \therefore & p(V-b) =R T \\ \Rightarrow & p V-p b =R T \\ \frac{p V}{R T}-\frac{p b}{R T} & =1 \\ Z & =p V / R T \\ Z & =1+\frac{p b}{R T} \end{array} $

The compressibility factor $(Z)$ is lower for $\mathrm{NH}_3$ and $\mathrm{CO}_2$ gases than that of $\mathrm{N}_2$ gas because van der Waals' constant ' $a$ ' of $\mathrm{CO}_2$ and $\mathrm{NH}_3$ are greater than $\mathrm{N}_2 \cdot \mathrm{NH}_3$ has ability to form strong hydrogen bonding and $\mathrm{CO}_2$ has stronger van der Waals' attractive interactions.

Compressibility factor $(z)$ becomes less than unity when the size of the gas becomes larger. $\mathrm{CO}_2$ has large size which means strong van der waals' attraction. Hence, $\mathrm{CO}_2$ shows lowest sip in graph.

Plot $D$ represents the $\mathrm{CO}_2$ gas.

Dipole-induced dipole interactions are present between a polar and a non-polar molecule.

In $\mathrm{NH}_3$ and $\mathrm{H}_2, \mathrm{NH}_3$ is polar and $\mathrm{H}_2$ is non-polar, so they have dipole-induced dipole interactions.

In $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$, both are polar, so they do not have dipole-induced dipole interactions.

In $\mathrm{Cl}_2$ and $\mathrm{CCl}_4$, and $\mathrm{SiF}_4$ and $\mathrm{BF}_3$ both are non-polar, so they do not have dipole-induced dipole interactions.

Given, Moles of $A=2$;

Moles of $B=3$; Moles of $C=5$;

Moles of $D=10$

Partial pressure of $C, p_C=1.5 \mathrm{~atm}$

$ p_c=p_{\text {Total }} \times \text { mole fraction of gas } C $

$ \begin{aligned} & \Rightarrow 1.5=p_{\text {Total }} \times \frac{5}{2+3+5+10} \\ & \Rightarrow 1.5=p_{\text {Total }} \times \frac{5}{20} \Rightarrow p_{\text {Total }}=1.5 \times 4=6 \mathrm{~atm} \end{aligned} $

At high pressure, molecules tend to be more crowded together, if they are doser together, the intermolecular forces are stronger, and cause more deviations from ideal gas behaviour. The ideal gases are assumed to occupy no space and have no attractions for one another.

The kinetic-molecular theory of gases assumes that

A. The particles are much smaller than the distance between particles. Most of the volume of a gas is therefore empty space.

B. Collisions between gas particles or collisions with the walls of the container are perfectly elastic. None of the energy of a gas particle is lost when it collides with another particle or with the walls of the container.

C. Different particles have different speed.

D. The pressure of a gas results from collisions between the gas particles and the walls of the container.

So, option (A) and (D) are only correct.

Ideal gas equation $=p V=n R T$

On rearranging $p=\frac{n R T}{V}$ or $V=\frac{n R T}{p}$

i.e. $p \propto \frac{1}{V}$ or $V \propto \frac{1}{p}$

At low pressures, the real gas shows very small deviation from ideal behaviour and the curve for ideal gas and real gas almost coincides. There is a large deviation from ideal gas behaviour at high pressure, hence the curve are fai apart.

Mixture of acetylene and oxygen is used as a fuel for oxy-welding. As this mixture forms flames hot enough, cutting and welding of metals becomes easier.

A gas shows a positive deviation from ideal behaviour, if there is a force of attraction between the molecules. For such gases, the compressibility factor, $Z$ is more than 1 or

$Z=\frac{p V}{n R T}>1$

Hence, only gases (b) and (c) shows positive deviation from ideal behaviour at all pressures.

Option C In case of liquids, decreases and in case of gases, increases.

Explanation :

Viscosity, in general, refers to a fluid's resistance to flow. It can be thought of as the internal friction between molecules in a fluid (liquids or gases).

For liquids, as temperature increases, the viscosity usually decreases. This is because at higher temperatures, the energy in the liquid is greater, allowing particles to move more easily and reducing the internal friction (or viscosity).

In contrast, for gases, as temperature increases, viscosity tends to increase. This is due to the fact that, in gases, increased temperatures lead to higher molecular speeds and more frequent collisions, effectively increasing the internal friction or viscosity.

From the graph it is clear that, the value of 'Z' decreases with increase of pressure. We can explain as follows on the basis of van der Waals' equation.

At high pressure, when 'p' is large, V will be small and one cannot ignore 'b' in comparison to V. However, the team a/V² may be considered negligible in comparison to 'p' in van der Waals' equation.

$\left(p+\frac{a}{V^2}\right)$ (V $-$ b) = nRT

p(V $-$ b) = nRT

$\Rightarrow$ pV $-$ pb = nRT

or $\frac{pV}{nRT}=1+\frac{pb}{nRT}$

or Z = 1 + $\frac{pb}{nRT}$

Thus, Z is greater than 1. As pressure is increased (at constant T), the factor $\frac{pb}{nRT}$ increases. This explains why after minima in the curves, Z increase continuously with pressure. Hence, the only incorrect statement is (b).

The temperature is changed from $30^{\circ} \mathrm{C}$ to $930^{\circ} \mathrm{C}$.

We need to convert these temperatures to Kelvin:
$T_1 = 30^{\circ} \mathrm{C} = 303\, \mathrm{K}$
$T_2 = 930^{\circ} \mathrm{C} = 1203\, \mathrm{K}$

The root mean square (rms) speed of a gas is given by:
$v_{\mathrm{rms}} = \sqrt{\frac{3 K T}{m}}$
This means that $v_{\mathrm{rms}}$ is proportional to $\sqrt{T}$.
So, when temperature changes:
$\frac{(v_{\mathrm{rms}})_1}{(v_{\mathrm{rms}})_2} = \sqrt{\frac{T_1}{T_2}}$
Put the values in:
$\sqrt{\frac{303}{1203}}$

By calculating, we find:
$(v_{\mathrm{rms}})_2 = 2(v_{\mathrm{rms}})_1$

Therefore, the root mean square speed becomes double.

Three flasks of equal volumes contain $\mathrm{CH}_4, \mathrm{CO}_2$ and $\mathrm{Cl}_2$ gases respectively. They will contain equal number of molecules if temperature and pressure of all flasks are same.

According to Avogadro law, at same condition of temperature and pressure, equal volumes of different gases contain equal number of molecules.

(i) Compressibility factor (Z) of real gas cannot be unity, while compressibility factor for an ideal gas is one.

(ii) At low pressure and high temperature a gas behaves more like an ideals gas, so the deviation is very less.

(iii) Intermolecular forces among gas molecules are weak as compare to liquid and solid but, these forces cannot be zero.

(iv) Ideal gas follows van der Waals’ equation

$pV=nRT$

Gases which do not follow this equation are real gases.

Therefore, statements (iii) and (iv) are incorrect.

The deviations from ideal gas $(p V=n R T)$ depend on the temperature pressure and nature of gas. Among all options, $\mathrm{NH}_3$ is the one which show maximum deviation under the same conditions. $\mathrm{NH}_3$ will show maximum deviation from ideal gas due to more dipole-dipole attractions which leads to more attractive forces between molecules of $\mathrm{NH}_3$. In $\mathrm{NH}_3$ there are strong intermolecular force of attraction, so the van der Waals' constant value is high.

Boyle’s law states that pressure of given mass of an ideal gas is inversely proportional to its volume at a constant temperature.

For ideal gas, $pV=nRT$

where, T = constant

$pV=k$ .... (i)

For Eq. (i), i.e. $xy=c^2$ (i.e. hyperbola)

$p=kT$ ..... (ii)

i.e. $y=mx$ (i.e. straight line)

$pV$ = constant ...... (iii)

For an ideal gas, $pV=nRT$ .... (i)

Here, p = pressure, V = volume of gas, n = moles, R = rate constant, T = Temperature

$pV=RT$ (For 1 mole)

$V=RT/p$ .... (ii)

Now, density is given by $D=M/V$

Put value of (ii) in density,

we get, $D = {M \over {RT/p}} \Rightarrow D = {{pM} \over {RT}}$

Suppose the equal mass of methane and oxygen, w = 1 g

Mole fraction of oxygen = $ = {{w/32} \over {{w \over {32}} + {w \over {16}}}} = {{1/32} \over {{3 \over {32}}}} = {1 \over 3}$

Let, the total pressure = p

Pressure exerted by oxygen (partial pressure)

$\therefore$ The fraction of total pressure exerted by oxygen is ${1 \over 3}p$.

Given, pressure of mixture $\left(p_0\right) =1.1 \mathrm{~atm}$

Molecular weight of two gases $=4$ and 40.

∴ Number of moles of lighter gas $=m / 4$ where, $m=$ mass of gases and, number of moles of heavier gas $=\frac{m}{40}$

Thus, total number of moles $=\frac{m}{4}+\frac{m}{40}$

$ =\frac{11 m}{40} $

Hence, mole fraction of lighter gas

$ \frac{m / 4}{11 m / 40}=\frac{10}{11} $

∴ Partial pressure of lighter gas

$ =p_0 \times \frac{10}{11}=1.1 \times \frac{10}{11}=1 \mathrm{~atm} $

Hence, option (d) is the correct answer.

$\because Z=\frac{p V}{n R T}$

where,

$Z=$ compressibility factor, $p=$ pressure

$V=$ volume,$n=$ number of moles

$R=$ gas constant, $T=$ temperature,

Also, curves for ( $Z$ vs $p$ )

(i) If $Z>1$ the gas show positive deviation from ideal behaviour and is less compressible.

(ii) If $Z<1$, gas show negative deviation.

(iii) No, real gas give a straight line parallel to $(p)$.

Hence, option (b) is the correct answer, i.e. path way (2) and (3) only.

The expression of average kinetic energy of a gas molecule is,

$ \mathrm{KE}=\frac{3}{2} \times n R T $

When, temperature ( $T$ in K ) is constant, $\mathrm{KE} \propto n$ (mole)

$ \Rightarrow \quad \frac{(\mathrm{KE})_{\mathrm{N}_2}}{(\mathrm{KE})_{\mathrm{O}_2}}=\frac{n_{\mathrm{N}_2}}{n_{\mathrm{O}_2}}=\frac{7 / 28}{4 / 32}=\frac{1}{4} \times \frac{8}{1}=2: 1 $

For a gas, density $(d)=\frac{p M}{R T}$

For equal amount of the same gas, $d \propto \frac{p}{T}$

$ \Rightarrow \quad \frac{d_2}{d_1}=\frac{p_2}{p_1} \times \frac{T_1}{T_2} \Rightarrow \frac{p_2}{p_1}=\frac{d_2}{d_1} \times \frac{T_2}{T_1}=\frac{2}{1} \times \frac{1}{2}=1: 1 $

It should be mentioned in the question to match with option (answer).

In the given plot,

$ T_2>T_1 \text { and } p_1>p_2 $

Because, according to Boyle's law,

$ V \propto \frac{1}{p} $

(at constant temperature) and according to Charles' law,

$ V \propto T $

(at constant pressure)

Also,

∵ At constant temperature, product of $p V$ is constant.

$ \therefore \quad T_2>T_1 $

and for a given temperature ( $T$ ), as $V$ is more, $p$ is less, i.e. as $V_1$ is more, $p_2$ will be smaller, i.e. $p_1>p_2$.

Hence, option (b) is the correct answer.

Given, initial pressure $=2 \mathrm{~atm}$

Final volume $=1 / 4$ th of initial volume ( $V$ )

Let, initial volume $=V$

then, final volume $=\frac{V}{4}$ (at constant temperature)

∵ Final pressure $=p_2$ (to find)

$ \begin{aligned} & V_1 p_1 =V_2 p_2 \\ \therefore & \,\,p_2 =\frac{V_1 p_1}{V_2}=\frac{V}{V / 4} \times 2=8 \mathrm{~atm} \end{aligned} $

Hence, increase in pressure (i.e. final pressure) $=8 \mathrm{~atm}$

Given, root mean square (rms) speed of $\mathrm{O}_2$ is $500 \mathrm{~m} / \mathrm{s}$ at a constant temperature.

Root mean square speed in given by the following expression

$ u_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M w}} $

For $\mathrm{H}_2$ gas, $u_{\text {rms }\left(\mathrm{H}_2\right)}=\sqrt{\frac{3 R T}{M w_{\mathrm{H}_2}}}$

$ \left(M w \text { of } \mathrm{H}_2=2 \mathrm{~g} / \mathrm{mol}\right) $

For $\mathrm{O}_2$ gas, $u_{\text {rms }\left(\mathrm{O}_2\right)}=\sqrt{\frac{3 R T}{M w_{\mathrm{O}_2}}}$

$ \left(M w \text { of } \mathrm{O}_2=32 \mathrm{~g} / \mathrm{mol}\right) $

$ \frac{u_{\mathrm{rms}\left(\mathrm{H}_2\right)}}{u_{\mathrm{rms}\left(\mathrm{O}_2\right)}}=\sqrt{\frac{3 R T}{M w_{\mathrm{H}_2}}} \times \sqrt{\frac{M w_{\mathrm{O}_2}}{3 R T}} $

(Temperature constant, $R=8.33 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )

$ \begin{aligned} u_{\mathrm{rms}\left(\mathrm{H}_2\right)} & =\sqrt{\frac{M w_{\mathrm{O}_2}}{M w_{\mathrm{H}_2}}} \times u_{\mathrm{rms}\left(\mathrm{O}_2\right)} \\ & =\sqrt{\frac{32}{2}} \times 500=2000 \mathrm{~m} / \mathrm{s} \end{aligned} $

Average kinetic energy of $\mathrm{H}_2$ can be calculate as

$ \begin{aligned} & =\frac{1}{2} m u^2 \\ {[u=2000 \mathrm{~m} / \mathrm{s}, m=} & \mathrm{H}_2=2 \mathrm{~g} / \mathrm{mol}= 0.002 \mathrm{~kg} / \mathrm{mol}] \end{aligned} $

$ \begin{aligned} & =\frac{1}{2} \times 0.002 \times(2000)^2 \\ & =\frac{1}{2} \times \frac{2}{1000} \times 2000 \times 2000 \\ & =4000 \mathrm{~J} \mathrm{~mol}^{-1} \text { or } 4 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $

(i), (ii), (iii)

All gases are made up of a very large number of extremely small particles called molecules. All the molecules of a particular gas are identical in mass and size. The molecules are separated from one another by large space. Hence, the actual volume occupied by the molecules is negligible as compared to the total volume of the gas.

The collisions of gas particles with one another and with the walls of the container are perfectly elastic.

Statement I and III are correct and statement II incorrect.

(I) ${{{T_c}} \over {{p_c}}} = {{8a} \over {27\,Rb}}.\,{{27{b^2}} \over a} = {{8b} \over R}$ $\therefore$ ${{{T_c}} \over {{p_c}}} \propto b$

(II) ${T_c} = {{8a} \over {27\,Rb}}$ and ${T_b} = {a \over {Rb}}$

Thus, T_b > T_c

(III) When, T_c is attained.

$\therefore$ $\left( {{{\partial p} \over {\partial V}}} \right)$ = 0