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} $

Graham's Law of Diffusion:

The rate at which a gas diffuses is inversely proportional to the square root of its molar mass. This means, lighter gases diffuse faster than heavier gases.

Setting up the ratio:

The formula is:

$\frac{\text{Rate of } \mathrm{H}_2}{\text{Rate of } \mathrm{O}_2} = \sqrt{\frac{\text{Molar mass of } \mathrm{O}_2}{\text{Molar mass of } \mathrm{H}_2}}$

The same amount of time is used for both gases. This means we can compare the number of moles of each gas that can diffuse:

$\frac{n_{\mathrm{H}_2}}{n_{\mathrm{O}_2}} = \sqrt{\frac{M_{\mathrm{O}_2}}{M_{\mathrm{H}_2}}}$

Finding the values:

Molar mass of $\mathrm{H}_2 = 2\,\mathrm{g/mol}$ and molar mass of $\mathrm{O}_2 = 32\,\mathrm{g/mol}$.

$n_{\mathrm{H}_2}$ is the number of moles of hydrogen:

$n_{\mathrm{H}_2} = \frac{2.0\,\mathrm{g}}{2\,\mathrm{g/mol}} = 1\,\mathrm{mol}$

$\frac{1}{n_{\mathrm{O}_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$

So,

$n_{\mathrm{O}_2} = \frac{1}{4} = 0.25\,\mathrm{mol}$

Calculate the mass of $\mathrm{O}_2$:

$\text{Mass of } \mathrm{O}_2 = n_{\mathrm{O}_2} \times \text{Molar mass of } \mathrm{O}_2$

$\text{Mass of } \mathrm{O}_2 = 0.25 \times 32 = 8\,\mathrm{g}$

Final answer:

8 grams of $\mathrm{O}_2$ would diffuse from the container in the same time under similar conditions.

$v_{\mathrm{rms}}=412 \mathrm{~m} / \mathrm{s}$

Molar mass of $\mathrm{CO}_2=M=0.044 \mathrm{~kg} / \mathrm{mol}$

$ \begin{aligned} \mathrm{KE} & =\frac{1}{2} M v_{\mathrm{rms}}^2 \\ \Rightarrow \quad \mathrm{KE} & =\frac{1}{2} \times 0.044 \times(412)^2=3.7343 \mathrm{~kJ} \end{aligned} $

The correct graph for compressibility factor vs pressure for an ideal gas is

$ \begin{aligned} &\begin{aligned} p V & =n R T \\ Z & =\frac{n R T}{p V} \end{aligned}\\ &\text { for ideal gas } Z=1 \end{aligned} $

Statements I and II are correct, while III is incorrect. The correct form of III is,

Compressibility factor of ideal gas is one.

The correct equation (van der Waal's equation) for 1 mole of a real gas is

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

Step 1: Use Boyle's Law for Gas A

Boyle's Law says that for a fixed temperature, pressure times volume stays the same when the gas changes containers.

The formula is: $ p_A V_A = p_A' V_{\text{final}} $

So for gas A: $ p_A' = \frac{1 \times 2}{100} = 0.02 $ bar

Step 2: Use Dalton’s Law of Partial Pressures

Dalton’s Law says that the total pressure from a mixture of gases is the sum of the pressures each gas would have by itself in the same container.

Total pressure after mixing is given in the question as 0.1 bar. So, $ p_{\text{total}} = p_A' + p_B' $ $ 0.1 = 0.02 + p_B' $ This means $ p_B' = 0.08 $ bar

Step 3: Find the Original Pressure of Gas B with Boyle's Law

Again, use Boyle's Law for gas B as we did for gas A.

$ p_B \times V_B = p_B' \times V_{\text{final}} $

Put the numbers in: $ p_B \times 4 = 0.08 \times 100 $

So, $ p_B = \frac{0.08 \times 100}{4} = 2 $ bar

Statement I

If thermal energy is stronger than intermolecular forces, the substance prefers to be in gaseous state.

• When thermal (kinetic) energy of molecules exceeds the intermolecular attractive forces, the molecules move freely, far apart — this is the gaseous state.

✅ Statement I is correct.

Statement II

At constant temperature, the density of an ideal gas is proportional to its pressure.

From the ideal gas law:

$ PV = nRT $

or

$ P = \frac{n}{V}RT $

Now, $ \frac{n}{V} = \frac{\rho}{M} $, where $ \rho $ is density and $ M $ is molar mass.

Thus,

$ P = \frac{\rho}{M} RT \Rightarrow \rho = \frac{PM}{RT} $

At constant $ T $ and $ M $:

$ \rho \propto P $

✅ Statement II is also correct.

✅ Final Answer:

Option C — Both Statement-I and Statement-II are correct.

Calculating the partial pressure of $\mathrm{H}_2$ in 10 L flask

$p_{\mathrm{H}_2 \text { initial }} V_{\mathrm{H}_2 \text { initial }}=p_{\mathrm{H}_2 \text { final }} V_{\text {total }}$

$ p_{\mathrm{H}_2 \text { final }}=\frac{1 \times 1}{10}=0.1 \mathrm{bar} $

Using Boyle's law for partial pressure of $\mathrm{O}_2$

$ \begin{aligned} p_{\mathrm{O}_2 \text { final }} & =\frac{p_{\mathrm{O}_2 \text { initial }} \times V_{\mathrm{O}_2 \text { initial }}}{V_{\text {total }}} \\ & =\frac{2 \times 2}{10}=0.4 \mathrm{bar} \end{aligned} $

Use Dalton's law of partial pressure.

$ \begin{aligned} p_{\text {total }} & =p_{\mathrm{H} \text { final }}+p_{\mathrm{O}_2 \text { final }} \\ & =0.1+0.4=0.5 \text { bar } \end{aligned} $

Statement given in option (b) is incorrect. The correct form of option (b) is critical temperature of $\mathrm{CO}_2$ is $31^{\circ} \mathrm{C}$. Critical temperature is the temperature above which a substance cannot be

liquified regardless of the pressure applied.

$ \begin{aligned} &\text { Molar mass of } \mathrm{H}_2=2 \mathrm{~g} / \mathrm{mol}\\ &\text { Number of moles }=n_{\mathrm{H}_2}=\frac{\text { Given mass }}{\text { Molar mass }} \end{aligned} $

$ =\frac{4}{2}=2 \mathrm{~g} / \mathrm{mol} $

Moles of $\mathrm{O}_2=\frac{6.4}{32} \approx 0.20 \mathrm{~mol}$

Now, $\frac{\mathrm{KE}_{\mathrm{O}_2}}{\mathrm{KE}_{\mathrm{H}_2}}=\frac{n_{\mathrm{O}_2} \times T_{\mathrm{O}_2}}{n_{\mathrm{H}_2} \times T_{\mathrm{H}_2}}$

$ \Rightarrow \quad \mathrm{KE}_{\mathrm{O}_2}=\frac{x \times 0.200 \times 400}{2 \times 300}=\frac{2 x}{15} \mathrm{~J} $

Assume a total mass : 1 g of $\mathrm{H}_2$ and 2 g of $\mathrm{O}_2$

$ n_{\mathrm{H}_2}=\frac{\text { Given mass }}{\text { Molar mass }}=\frac{1}{2}=0.5 \mathrm{~mol} $

Similarly, $n_{\mathrm{O}_2}=\frac{2}{32}=0.0625 \mathrm{~mol}$

$ \begin{aligned} \text { Total moles } & =n_{\text {Total }}=n_{\mathrm{H}_2}+n_{\mathrm{o}_2} \\ & =0.5625 \mathrm{~mol} \end{aligned} $

Mole fraction of $\mathrm{O}_2$

$ \chi_{\mathrm{o}_2}=\frac{n_{\mathrm{o}_2}}{n_{\text {total }}}=\frac{0.0625}{0.5625}=\frac{1}{9} $

Partial pressure of $\mathrm{O}_2$

$ p_{\mathrm{O}_2}=\chi_{\mathrm{O}_2} \cdot p_{\text {total }}=\frac{p}{9} \mathrm{~atm} $

We want to find the temperature where the rms (root mean square) speed of SO₂ molecules is equal to the rms speed of O₂ molecules at 27°C.

The formula for rms velocity is: $ V_{\rm rms} = \sqrt{\frac{3RT}{M}} $ where $R$ is the gas constant, $T$ is temperature in Kelvin, and $M$ is molar mass.

We set the rms speeds equal: $ V_{\rm rms(SO_2)} = V_{\rm rms(O_2)} $ So, $ \sqrt{\frac{3RT_{\rm SO_2}}{M_{\rm SO_2}}} = \sqrt{\frac{3RT_{\rm O_2}}{M_{\rm O_2}}} $

Squaring both sides removes the square root: $ \frac{T_{\rm SO_2}}{M_{\rm SO_2}} = \frac{T_{\rm O_2}}{M_{\rm O_2}} $

Rearrange for $T_{\rm SO_2}$: $ T_{\rm SO_2} = T_{\rm O_2} \times \frac{M_{\rm SO_2}}{M_{\rm O_2}} $

The temperature given for O₂ is 27°C. Convert this to Kelvin: 27 + 273 = 300 K

The molar mass of SO₂ is 64, and of O₂ is 32.

Plug in the values: $ T_{\rm SO_2} = 300 \times \frac{64}{32} = 300 \times 2 = 600~K $

The slope of an isochore (a line where volume does not change) for an ideal gas is given by $\frac{d p}{d T} = \frac{nR}{V}$, where:
$n$ = number of moles, $R$ = gas constant, $V$ = volume.

Here, the slope is $0.082~\mathrm{atm~K}^{-1}$, $n = 1$, and $R = 0.082~\mathrm{L~atm~mol}^{-1}~\mathrm{K}^{-1}$.

Substitute the known values into the formula:
$0.082 = \frac{1 \times 0.082}{V}$

Solve for $V$:
$V = 1~\mathrm{L}$

Now, use the ideal gas equation: $pV = nRT$.

Substitute the values: $p \times 1 = 1 \times 0.082 \times 12.2$

Calculate the multiplication: $0.082 \times 12.2 = 1.0004$

So, $p = 1.0004~\mathrm{atm}$. Round to nearest whole number: $p = 1~\mathrm{atm}$

$p_{\text {Total }}=p_{\mathrm{O}_2}+p_{\mathrm{H}_2}$

$ \Rightarrow \quad p_{\mathrm{O}_2}=2-1.778=0.22 \mathrm{bar} $

Mole fraction of $\mathrm{H}_2$ and $\mathrm{O}_2$

$ \begin{aligned} X_{\mathrm{H}_2} & =\frac{p_{\mathrm{H}_2}}{p_{\text {Total }}}=\frac{1.778}{2}=0.889 \\ X_{\mathrm{O}_2} & =1-X_{\mathrm{H}_2}=0.111 \\ m_{\mathrm{H}_2} & =X_{\mathrm{H}_2} \cdot M_{\mathrm{H}_2}=1.778 \\ m_{\mathrm{O}_2} & =X_{\mathrm{O}_2} \cdot M_{\mathrm{O}_2}=3.552 \end{aligned} $

$ \begin{aligned} \frac{W}{W} \% \mathrm{H}_2 & =\frac{m_{\mathrm{H}_2}}{m_{\mathrm{H}_2}+m_{\mathrm{O}_2}} \times 100 \\ & =\frac{1.778}{1.778+3.552} \times 100=33.33 \% \end{aligned} $

Most probable speed is given by

$ U_{m p}=\sqrt{\frac{2 R T}{M}} $

On rearranging to obtain the value of $R T$.

$ \frac{U_{m p}^2 \times M}{2}=R T $

Substituting the values

$ =\frac{(2 \times 10)^2 \times(2) \times 10^{-3}}{2} $

$ R T=40 $

Moles of $\mathrm{H}_2=\frac{8}{2}=4$

$ \mathrm{KE}=\frac{3}{2} n R T \Rightarrow \frac{3}{2} \times 4 \times 40=240 \mathrm{~J} $

Ideal gas equations, $p V=n R T$

$ p \propto n, T \Rightarrow p \propto \frac{1}{V} $

So correct graph is

Root mean square speed is,

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

Most probable speed is, $v_{m p}=\sqrt{\frac{2 R T}{M}}$

$ \begin{aligned} & \frac{v_{\mathrm{rms}}}{v_{\mathrm{mp}}}=\sqrt{\frac{3}{2}} \Rightarrow v_{\mathrm{MP}}=v_{\mathrm{rms}} \sqrt{\frac{2}{3}} \\ \Rightarrow & 3.16 \times 10^2 \times \sqrt{\frac{2}{3}} \\ & v_{\mathrm{mp}}=2.58 \times 10^2 \mathrm{~m} / \mathrm{s} \end{aligned} $

Both statements I and II are not correct.

The incorrect forms are

If intermolecular forces are stronger than thermal energy, the substance will prefer to be in solid or liquid state.

The total number of elements that are gases at room temperature is II.

At high temperature and low pressure, van der Waals' equation of state changes to ideal gas equation.

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} $

The root mean square (rms) velocity, mean velocity, and most probable velocity of a gas differ in value at a given temperature. Here is the detailed explanation for each:

Root Mean Square Velocity ($u_{\mathrm{rms}}$):

$ u_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}} $

Mean Velocity ($u_{\text{average}}$):

$ u_{\text{average}} = \sqrt{\frac{8RT}{\pi M}} $

Most Probable Velocity ($u_{\text{mp}}$):

$ u_{\text{mp}} = \sqrt{\frac{2RT}{M}} $

These formulas indicate the velocities in terms of the universal gas constant $ R $, absolute temperature $ T $, and molar mass $ M $.

Considering the ratios:

Ratios Derived:

$ u_{\mathrm{rms}} : u_{\mathrm{av}} : u_{\mathrm{mp}} = \sqrt{3} : \sqrt{\frac{8}{\pi}} : \sqrt{2} $

Evaluating these expressions approximately gives:

$ 1.732 : 1.596 : 1.414 $

Calculated Ratios:

$ \frac{u_{\mathrm{rms}}}{u_{\mathrm{av}}} = 1.08, \quad \frac{u_{\mathrm{av}}}{u_{\mathrm{mp}}} = 1.12, $

$ \frac{u_{\mathrm{rms}}}{u_{\mathrm{mp}}} = 1.22, \quad \frac{u_{\mathrm{av}}}{u_{\mathrm{rms}}} = 0.92 $

Based on these calculations, the correct option is the one that accurately reflects the derived ratios, specifically noting that the ratio $\frac{u_{\mathrm{av}}}{u_{\mathrm{mp}}}$ equals 1.12, hence confirming the accuracy of this comparison.

Given the data:

Volume of $\mathrm{SO}_2$ ($V_{\mathrm{SO}_2}$) = $60 \mathrm{~cm}^3$

Time for $\mathrm{SO}_2$ ($t_{\mathrm{SO}_2}$) = $x$ minutes

Molar mass of $\mathrm{SO}_2$ ($M_{\mathrm{SO}_2}$) = $64 \mathrm{~g/mol}$

For the other gas:

Volume ($V_g$) = $360 \mathrm{~cm}^3$

Time ($t_g$) = $y$ minutes

Molar mass ($M_g$) = $4 \mathrm{~g/mol}$

We apply Graham's law of diffusion, which states that the rate of diffusion ($r$) is inversely proportional to the square root of the molar mass, as shown below:

$ r \propto \frac{1}{\sqrt{M}} $

This can be expressed as:

$ \frac{r_{\mathrm{SO}_2}}{r_g} = \sqrt{\frac{M_g}{M_{\mathrm{SO}_2}}} = \frac{V_{\mathrm{SO}_2} \times t_g}{V_g \times t_{\mathrm{SO}_2}} $

Substituting the known values:

$ \sqrt{\frac{4}{64}} = \frac{60 \times y}{360 \times x} $

Simplifying the above equation gives:

$ \frac{x}{y} = \frac{2}{3} $

Thus, the ratio of $x$ to $y$ is $2:3$.

When a 10 L vessel containing 1 mole of an ideal gas is divided into two equal parts, each section now holds a volume of 5 L and 0.5 moles of gas.

To understand the effect on pressure and temperature, we can use the ideal gas law:

$ PV = nRT $

Since the ratio $\frac{n}{V}$ remains constant for each section of the divided vessel, the pressure remains unchanged. This is because:

The number of moles $n$ is halved ($0.5$ moles) and the volume $V$ is also halved (5 L), thus the ratio $\frac{n}{V}$ stays the same.

Both sections maintain the same conditions because dividing the vessel doesn't add or remove any energy from the system.

Therefore, both the pressure $p$ and the temperature $T$ remain the same in each part.

Given, Slope $=32.8 \mathrm{~L} \mathrm{~atm} / \mathrm{mol}$

From the graph,

$ \begin{aligned} \text { slope } & =\frac{p}{\frac{1}{V}} \text { or } p V \\ p V & =32.8 \mathrm{~L} \mathrm{~atm} / \mathrm{mol} \,\,\,\,\,...(i)\end{aligned} $

Now, using ideal gas equation for 1 mo of ideal gas,

$ p V=R T \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Compare Eqs. (i) and (ii),

$ \begin{aligned} 32.8 & =R T \\ T & =\frac{32.8}{0.082}=400 \mathrm{~K} \end{aligned} $

When a vessel containing an ideal gas is divided into two equal parts, the characteristics of the gas in each part are determined by the properties of volume and temperature:

Volume is an extensive property, meaning it depends on the size or extent of the system. Therefore, when the vessel is partitioned into two equal halves, the volume in each part will be half of the original, i.e., $\frac{V}{2}$ liters.

Temperature is an intensive property, indicating that it does not depend on the size or amount of the system. As a result, the temperature remains the same in each part after the partitioning, i.e., $T$ Kelvin.

Thus, for each part of the partitioned vessel, the volume is $\frac{V}{2}$ liters and the temperature is $T$ Kelvin.

$ \begin{aligned} &\text { Given, } 1 \text { mole of ideal gas, } n=1\\ &\begin{aligned} & \text { slope }=2000 \mathrm{~J} / \mathrm{mol} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { slope }=\frac{y}{x}=\frac{p}{1 / V}=p V \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} \end{aligned} $

Using, ideal gas equation, $p V=R T$

$ R T=2000 \mathrm{~J} / \mathrm{mol}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) $

Kinetic energy of ideal gas is given by

$ \begin{aligned} \mathrm{KE} & =\frac{3}{2} R T=\frac{3}{2} \times 2000 \mathrm{~J} / \mathrm{mol} \\ & =3000 \mathrm{~J} / \mathrm{mol} \end{aligned} $

Assuming He and CH₄ behave as ideal gases, note that at STP, 1 mole occupies 22.4 liters.

Calculations:

Helium:

Moles of He that escaped = $ \frac{2}{22.4} $

Remaining moles of He = $ 1 - \frac{2}{22.4} = 0.91 $

Methane:

Moles of CH₄ that escaped = $ \frac{1}{22.4} $

Remaining moles of CH₄ = $ 1 - \frac{1}{22.4} = 0.95 $

Mole Fractions:

Mole fraction of He:

$ \frac{0.91}{0.91 + 0.95} = 0.489 \approx 0.488 $

Mole fraction of CH₄:

$ 1 - 0.488 = 0.512 $

The mole fractions of the remaining gases are approximately 0.488 for helium and 0.512 for methane.

The RMS velocity of ideal gas is given by

$ v_{\mathrm{ms}}=\sqrt{\frac{3 R T}{M}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Squaring on both side,

$ v_{\mathrm{ms}}^2=\frac{3 R T}{M} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, using ideal gas equation for 1 mole of gas.

$ p V=R T \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Substitute the value of $R T$ in Eq. (ii),

$ v_{\mathrm{ms}}^2=\frac{3}{M} p V \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

Compare it with straight line equation passing from origin.

$ \begin{array}{ll} y =m x+c \quad(\text { as } c=0) \\ \because\ m =\frac{3}{M} \end{array} $

In the system of three layers of liquid flowing over a fixed solid surface, the velocities of these layers are ordered as follows:

$ v_3 > v_2 > v_1 $

This sequence indicates that the velocity of the liquid is highest at the topmost layer and decreases as you move to the layers beneath it. The reduced frictional resistance experienced by the top layer, compared to the layers closer to the solid surface, allows it to achieve the highest velocity. Conversely, the bottom layer, in direct contact with the solid surface, encounters the greatest viscosity and thus has the lowest velocity.

RMS velocity of ideal gas is given by

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

Squaring both sides

$ u_{\mathrm{rms}}^2=\frac{3 R T}{M} $

Now plotting graph between $\mu_{\text {rms }}^2$ on $Y$-axis and $T$ on $X$-axis

Slope $=m$

Compare Eq. (i) with straight line equation $\Rightarrow y=m x+C$

$ C=0, m=\frac{3 R}{M} $

Statement I: "Viscosity of liquid decreases with increase in temperature."

This is correct in general. As the temperature rises, the intermolecular forces in a liquid weaken, causing the viscosity to decrease.

Statement II: "The units of viscosity coefficient are pascal?"

The SI unit of (dynamic) viscosity is actually Pascal-second ($\text{Pa}\cdot\text{s}$), not just Pascal (Pa). So, stating that the units are only "pascal" is incorrect.

Hence,

Statement I is correct.

Statement II is incorrect.

Therefore, the correct choice is:

Option C: Statement I is correct but statement II is not correct.

To find the kinetic energy of one mole of an ideal gas, we'll use the following given data:

The root mean square velocity, $v_{\mathrm{rms}}$, is $4 \times 10^2 \, \mathrm{m/s}$.

The molar mass is $0.1 \, \mathrm{kg/mol}$.

First, we need to calculate the mass of one mole of the gas.

$ \text{Mass} = \text{molar mass} \times \text{number of moles} = 0.1 \, \mathrm{kg/mol} \times 1 \, \text{mol} = 0.1 \, \mathrm{kg} $

The relationship between kinetic energy ($E$) and $v_{\mathrm{rms}}$ is given by the following formula:

$ v_{\mathrm{rms}} = \sqrt{\frac{2E}{m}} $

Squaring both sides of the equation, we have:

$ v_{\mathrm{rms}}^2 = \frac{2E}{m} $

From this, we can solve for $E$:

$ E = \frac{v_{\mathrm{rms}}^2 \times m}{2} $

Now, substitute the known values into the equation:

$ E = \frac{(4 \times 10^2)^2 \times 0.1}{2} = \frac{16 \times 10^4 \times 0.1}{2} $

Simplifying, we find:

$ E = \frac{1.6 \times 10^4}{2} = 8 \times 10^3 \, \mathrm{J/mol} $

Therefore, the kinetic energy of one mole of the ideal gas is $8 \times 10^3 \, \mathrm{J/mol}$.

Let's analyze each statement step by step:

$\textbf{Statement I:}$

The compressibility factor $Z$ for a gas is defined as the ratio of its molar volume $V$ to that of an ideal gas $V_{\text{ideal}}$ at the same temperature and pressure.

For an ideal gas, under a given temperature $T$ and pressure $p$, the molar volume is given by $V_{\text{ideal}} = \frac{RT}{p}.$

Therefore, the compressibility factor is:

$Z = \frac{V}{V_{\text{ideal}}} = \frac{V}{RT/p} = \frac{pV}{RT}.$

This matches the statement provided. Hence, Statement I is correct.

$\textbf{Statement II:}$

The root-mean-square (rms) velocity of gas molecules is given by:

$v_{\text{rms}} = \sqrt{\frac{3RT}{M}},$

where $M$ is the molar mass of the gas.

From this expression, it is clear that $v_{\text{rms}}$ is proportional to the square root of the temperature $T$ (when $M$ is constant), i.e.,

$v_{\text{rms}} \propto \sqrt{T}.$

Therefore, Statement II is also correct.

Since both statements are correct, the correct answer is:

Option A: Both statement I and statement II are correct.

Given:

Temperature, $ \mathrm{T} = 133.33 \, \mathrm{K} $

Molar mass, $ M = 0.083 \, \mathrm{kg/mol} $

Universal gas constant, $ R = 8.3 \, \mathrm{J/(mol \cdot K)} $

The root mean square (RMS) velocity for an ideal gas is calculated using the formula:

$ v_{\text{RMS}} = \sqrt{\frac{3RT}{M}} $

Substituting the given values into the formula:

$ v_{\text{RMS}} = \sqrt{\frac{3 \times 8.3 \times 133.33}{0.083}} $

$ v_{\text{RMS}} = \sqrt{39,999} $

Therefore:

$ v_{\text{RMS}} \approx 200 \, \mathrm{m/s} $

Let's analyze each statement step by step:

$\textbf{Statement I:}$

For an ideal gas, the equation of state is

$pV = nRT.$

The compressibility factor $Z$ is defined as

$Z = \frac{pV}{nRT}.$

For an ideal gas, this gives:

$Z = 1.$

Therefore, Statement I is correct.

$\textbf{Statement II:}$

The average translational kinetic energy per mole for any ideal gas is given by

$\langle E_k \rangle = \frac{3}{2}RT,$

regardless of the molecular mass. Although the average speed of the molecules depends on the mass (with lighter molecules moving faster), the kinetic energy per mole at a given temperature is the same for all gases. Hence, if the kinetic energy per mole of NO(g) is $x$ J, then that of $\mathrm{N}_2\mathrm{O}_4(g)$ must also be $x$ J – not $2x$ J.

Therefore, Statement II is incorrect.

$\textbf{Statement III:}$

From kinetic theory, the rate of diffusion (or effusion) of a gas is proportional to the average speed, which is given by

$\bar{v} \propto \frac{1}{\sqrt{M}},$

where $M$ is the molar mass. For gases under the same conditions, the density ($\rho$) is proportional to the molar mass. Thus, the diffusion rate is inversely proportional to the square root of the density:

$\text{Rate} \propto \frac{1}{\sqrt{\rho}}.$

Hence, Statement III is correct.

In summary:

Statement I is correct.

Statement II is incorrect.

Statement III is correct.

The correct option is therefore:

Option C: I, III only.

Velocity of gas is given by

$ v_{\mathrm{MF}}=\sqrt{\frac{2 R T}{M}} \quad \text { or } \quad v_{\mathrm{MP}} \propto \sqrt{\frac{T}{M}} $

As the temperature is increased, the value of most probable speed increases but fraction of particles passing most probable speed decreases.

Hence, $T_2>T_1>T_3$

Slope from the above graph is

$ \frac{\left(u_{\mathrm{rms}}\right)^2}{T}=249 \mathrm{~m}^2 \mathrm{~s}^2 \mathrm{~K}^{-1} $

From, RMS velocity of gas molecule

$ u_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}} \text { or } \frac{\left(u_{\mathrm{mss}}\right)^2}{T}=\frac{3 R}{M} $

Comparing Eq. (i) and (ii),

$ \begin{aligned} & 249=\frac{3 R}{M} \text { or } M=\frac{3 R}{249} \Rightarrow \frac{3 \times 8.3}{249} \\ & \Rightarrow M=0.1 \mathrm{~kg} / \mathrm{mol} \text { or } 1 \times 10^{-1} \mathrm{~kg} \mathrm{~mol}^{-1} \end{aligned} $

Statement I is correct but Statement II is incorrect. The correct form of incorrect statement is:

SI unit of visocity is $\mathrm{kg}-\mathrm{m}^{-1} \mathrm{~s}^{-1}$.

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} $