States of Matter

From ideal gas law, $p V=n R T$

For isobaric process, $p$ is constant

$ \Rightarrow \quad V=\frac{n R T}{p} $

The above equation is in the form of $y=k x$ where $y=V, x=T$ and $k=\frac{n R}{p}$.

It is given that, $p_1 < p_2 < p_3$

And from the obtained relation $m \propto \frac{1}{p}$

Thus, $m_1>m_2>m_3$

The force $(F)$ required to maintain the steady flow of liquid layers is defined by Newton's law of viscosity. This force is directly proportional to the area of contact $(A)$ and the velocity gradient $\left(\frac{d u}{d z}\right)$, with the proportionality constant being the coefficient of viscosity ( $\eta$ ). Mathematically, this relationship is expressed as

$ F=\eta A \frac{d u}{d z} $

Using the ideal gas law, $n=\frac{p V}{R T}$.

Since the temperature and $R$ are constant, they can be treated as a single term.

• Moles of oxygen in flask

$ A: n_{\mathrm{O}_2}=\frac{5 \times V}{R T} $

• Moles of helium in flask

$ B: n_{\mathrm{Hc}}=\frac{3 \times 2}{R T}=\frac{6}{R T} $

• Total moles after mixing:

$ n_{\text {total }}=n_{\mathrm{O}_2}+n_{\mathrm{He}}=\frac{5 V+6}{R T} $

The mole fraction of oxygen in the mixture is given as 0.2 .

$ x_{\mathrm{O}_2}=\frac{n_{\mathrm{O}_2}}{n_{\text {total }}}=\frac{5 V / R T}{(5 V+6) / R T}=0.2 $

The $R T$ terms cancel out, leaving

$ \begin{aligned} & \frac{5 V}{5 V+6}=0.2 \\ & 5 V=0.2(5 V+6) \\ & 5 V=V+1.2 \Rightarrow 4 V=1.2 \\ & V=0.3 \mathrm{~L} \end{aligned} $

The root mean square (rms) speed of argon gas is given as $20~\mathrm{ms}^{-1}$.

The molar mass of argon is $40~\mathrm{g~mol}^{-1}$. This can also be written as $0.040~\mathrm{kg~mol}^{-1}$ (because $1000~\mathrm{g} = 1~\mathrm{kg}$).

The formula for average kinetic energy per mole of a gas is: $E = \frac{1}{2} M v_{\mathrm{rms}}^2$, where $M$ is the molar mass in kg/mol and $v_{\mathrm{rms}}$ is the rms speed.

Substitute the values:
$E = \frac{1}{2} \times 0.040 \times (20)^2$

Calculate $ (20)^2 = 400 $, so:

$E = \frac{1}{2} \times 0.040 \times 400$

$\frac{1}{2} \times 0.040 = 0.020$, so:
$E = 0.020 \times 400 = 8$

So, the average kinetic energy per mole of argon is $8~\mathrm{J~mol}^{-1}$ at this temperature.

$ \begin{aligned} &\text { Using formula for partial pressure }\\ &\begin{aligned} p_i & =\frac{n_i}{n_{\text {total }}} \times p_{\text {total }} \\ p_{\text {total }} & =\frac{n_{\text {total }} \cdot R \cdot T}{V}=\frac{136}{20}=6.8 \mathrm{~atm} \\ p_A & =\frac{96}{136} \times 6.8=4.8 \mathrm{~atm} \\ p_B & =\frac{40}{136} \times 6.8=2 \mathrm{~atm} \end{aligned} \end{aligned} $

$v_{\mathrm{rms}}$ is given by, $v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$

Also, it is given that

$ \begin{aligned} v_{\mathrm{rms}\left(\mathrm{H}_2\right)} & =\sqrt{7} v_{\mathrm{rms}\left(\mathrm{~N}_2\right)} \\ \sqrt{\frac{3 R T_{\mathrm{H}_2}}{M_{\mathrm{H}_2}}} & =\sqrt{7} \times \sqrt{\frac{3 R T_{\mathrm{N}_2}}{M_{\mathrm{N}_2}}} \\ \sqrt{\frac{3 T_{\mathrm{H}_2}}{2}} & =\sqrt{7} \times \sqrt{\frac{3 T_{\mathrm{N}_2}}{28}} \\ \frac{T_{\mathrm{H}_2}}{2} & =\frac{T_{\mathrm{N}_2}}{4} \Rightarrow T_{\mathrm{H}_2}=\frac{T_{\mathrm{N}_2}}{2} \end{aligned} $

The RMS velocity is given by

$ v_{\mathrm{ms}}=\sqrt{\frac{3 R T}{M}} $

According to question, $ v_{\mathrm{mms}} $ of $ \mathrm{SO}_{2}= $ Most probable speed of $ \mathrm{O}_{2} $

$ \begin{array}{l} \sqrt{\frac{3 R T_{\mathrm{SO}_{2}}}{M_{\mathrm{SO}_{2}}}}=\sqrt{\frac{2 R T_{\mathrm{O}_{2}}}{M_{\mathrm{O}_{2}}}} \\\\ \sqrt{\frac{3 T_{\mathrm{SO}_{2}}}{M_{\mathrm{SO}_{2}}}}=\sqrt{\frac{2 T_{\mathrm{O}_{2}}}{M_{\mathrm{O}_{2}}}} \end{array} $

Squaring both side and substituting the values,

$ \frac{3 \times 400}{64}=\frac{T_{\mathrm{O}_{2}} \times 2}{32} \Rightarrow T_{\mathrm{O}_{2}}=300 \mathrm{~K} $

From ideal gas equation,

$ p V=n R T $

The plot is between volume and number of moles

Slope $ =\frac{V}{n} $

From Eq. (i) $ \frac{V}{n} $ is $ \frac{R T}{p} $

where, $ R=0.0821 \mathrm{~L} \mathrm{~atm} / \mathrm{mol} / \mathrm{K} $

$ T=300 \mathrm{~K}, p=1 \mathrm{~atm} $

Slope $ =\frac{V}{n}=\frac{0.082 \times 300}{1} $

$ \frac{V}{n}=24.6 \mathrm{~L} / \mathrm{mol} $

Given:

Initial temperature, $ T_{1} = 400 \, \mathrm{K} $

Initial volume, $ V_{1} = 0.5 \, \mathrm{m}^{3} $

Initial pressure, $ p_{1} = 203 \, \mathrm{kPa} $

For the final state:

Final volume, $ V_{2} = 0.2 \, \mathrm{m}^{3} $

Final pressure, $ p_{2} = 304 \, \mathrm{kPa} $

To find the change in temperature using the ideal gas law, we can use the relation:

$ pV = nRT \quad \text{which implies} \quad \frac{pV}{T} = \text{constant} $

By comparing the initial and final states, we have:

$ \frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}} $

Solving for $ T_{2} $:

$ T_{2} = \frac{p_{2} V_{2} T_{1}}{p_{1} V_{1}} $

Substituting the given values:

$ T_{2} = \frac{304 \times 0.2 \times 400}{203 \times 0.5} \approx 240 \, \mathrm{K} $

Therefore, the change in temperature is:

$ \text{Change in temperature} = T_{1} - T_{2} = 400 - 240 = 160 \, \mathrm{K} $

Given,

$ \begin{aligned} & d=0.920 \mathrm{~g} / \mathrm{L} \\\\ & T=400 \mathrm{~K}, p=1 \mathrm{~atm} \\\\ & R=0.082 \mathrm{~L} \mathrm{~atm} / \mathrm{mol} / \mathrm{K} \end{aligned} $

Using the formula,

$ p M=d R T \text { or } M=\frac{d R T}{p} \quad \ldots \text { (i) } $

Substituting the value in Eq. (i),

$ \begin{aligned} & M=\frac{0.920 \times 0.082 \times 400}{1} \\\\ & \Rightarrow \quad M=30.176 \quad \ldots \text { (ii) } \\\\ & 30.176=\frac{28 n_{\mathrm{N}_2}+32 n_{\mathrm{O}_2}}{n_{\mathrm{N}_2}+n_{\mathrm{O}_2}} \\\\ & 30.176 n_{\mathrm{N}_2}+30.176 n_{\mathrm{O}_2}=28 n_{\mathrm{N}_2}+32 n_{\mathrm{O}_2} \\\\ & 2.176 n_{\mathrm{N}_2}=1.824 n_{\mathrm{O}_2} \\\\ & n_{\mathrm{N}_2}=\frac{1.824}{2.176} n_{\mathrm{O}_2} \quad \ldots \text { (iii) } \\\\ & \text { Now, } \begin{aligned} \chi_{\mathrm{N}_2} & =\frac{n_{\mathrm{N}_2}}{n_{\mathrm{N}_2}+n_{\mathrm{O}_2}} \\\\ & =\frac{\frac{1.824}{2.176} n_{\mathrm{O}_2}}{\frac{1.824}{2.176} n_{\mathrm{O}_2}+n_{\mathrm{O}_2}}=0.456 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Given, } \\ & T_1=27^{\circ} \mathrm{C} \text { or } 300 \mathrm{~K}, T_2=727^{\circ} \mathrm{C} \text { or } 1000 \mathrm{~K} \\ & \end{aligned} $

According to ideal gas equation

$ p V=n R T $

or $p_1 V_1=n_1 R T_1$ and $p_2 V_2=n_2 R T_2$

$ \frac{n_1}{n_2}=\frac{T_2}{T_1} \quad \ldots \text { (i) } $

or $\quad n_2=\frac{T_1}{T_2} \times n_1 \quad\left[n_1=1\right.$ unit $]$

$ n_2=\frac{300}{1000} $

or $\quad n_2=\frac{3}{10}$

Fraction remains in vessel,

$ 1-n_2=1-\frac{3}{10}=\frac{7}{10} $

Both vessels contain same gas.

∴ Molar mass of gas in both the vessels is same.

Mass of the gas in $B=x \mathrm{~g}$

and mass of gas in $A=y \mathrm{~g}$

In vessel $B, p V=n R T$

$ \begin{aligned} & \Rightarrow \quad \frac{p V}{T}= \frac{w}{M} R \\ & {\left[\because n=\frac{\operatorname{Mass}(w)}{\operatorname{Molecular~mass~}(M)}\right] } \\ & \frac{p V_B}{T_B}=\frac{x}{M} R \\ & {[\because \text { Mass of gas in } B(w)=x] } \\ & \frac{M}{R}=\frac{x T_B}{p_B V_B}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

In vessel $A$, pressure, $p_A=4 p_B$

Volume, $V_A=4 V_B$

and temperature $T_A=4 T_B$

$ \begin{aligned} \therefore \quad \frac{M}{R} & =\frac{y \times 4 T_B}{4 p_B \times 4 V_B} \\ \frac{M}{R} & =\frac{y T_B}{4 p_B V_B} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i,i)\end{aligned} $

From (i) and (ii),

$ \begin{aligned} \frac{x T}{p V} & =\frac{y T}{4 p V} \Rightarrow x=\frac{y}{4} \\ y & =4 x \end{aligned} $

Thus, the mass of the gas in $A$ will be $4 x \mathrm{~g}$.

Let the rate of diffusion of gas,

$ A=r_A=\sqrt{5} $

and of gas, $B=r_B$

We know, $\frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}}$

$ \begin{aligned} \frac{\sqrt{5 r}}{r} & =\sqrt{\frac{M_B}{x}} \\ or\,\,\,\,\,\,5 & =\frac{M_B}{x} \text { or }\left[M_B=5 x\right] \end{aligned} $

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 $80\%$.

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.

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.