Characteristics and Measurable properties of gases Questions in English

(C) According to Dalton's law of partial pressure:
$p_i = x_i \times P_T$
Where $p_i$ is the partial pressure of the $i^{th}$ component,$x_i$ is the mole fraction of the $i^{th}$ component,and $P_T$ is the total pressure of the mixture.
Given: $n_{H_2} = 1 \ mol$,$n_{He} = 1 \ mol$,$n_{O_2} = 1 \ mol$.
Total moles $n_{total} = 1 + 1 + 1 = 3 \ mol$.
Mole fraction of $H_2$ is $x_{H_2} = \frac{n_{H_2}}{n_{total}} = \frac{1}{3}$.
Given partial pressure of $H_2$ is $p_{H_2} = 2 \ atm$.
Using the formula: $2 \ atm = \frac{1}{3} \times P_T$.
Therefore,$P_T = 2 \ atm \times 3 = 6 \ atm$.

(D) Step $1$: Calculate the number of moles of each gas.
$n_{N_2} = \frac{7 \ g}{28 \ g \ mol^{-1}} = 0.25 \ mol = \frac{1}{4} \ mol$
$n_{Ar} = \frac{8 \ g}{40 \ g \ mol^{-1}} = 0.2 \ mol = \frac{1}{5} \ mol$
Step $2$: Calculate the mole fraction of $N_2$ $(x_{N_2})$.
$x_{N_2} = \frac{n_{N_2}}{n_{N_2} + n_{Ar}} = \frac{1/4}{1/4 + 1/5} = \frac{0.25}{0.25 + 0.20} = \frac{0.25}{0.45} = \frac{5}{9}$
Step $3$: Calculate the partial pressure of $N_2$ $(p_{N_2})$.
$p_{N_2} = x_{N_2} \times P_{total} = \frac{5}{9} \times 27 \ bar = 5 \times 3 \ bar = 15 \ bar$.

(B) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
First,calculate the molar masses of the given compounds:
$M(CO_{2}) = 12 + 2 \times 16 = 44 \ g/mol$
$M(SO_{2}) = 32 + 2 \times 16 = 64 \ g/mol$
$M(SO_{3}) = 32 + 3 \times 16 = 80 \ g/mol$
$M(PCl_{3}) = 31 + 3 \times 35.5 = 31 + 106.5 = 137.5 \ g/mol$
Since the rate of diffusion is inversely proportional to the square root of the molar mass,the order of the rate of diffusion will be the reverse of the order of their molar masses:
$M(CO_{2}) < M(SO_{2}) < M(SO_{3}) < M(PCl_{3})$
Therefore,the rate of diffusion order is: $CO_{2} > SO_{2} > SO_{3} > PCl_{3}$.

(C) For $London$ dispersion forces,the interaction energy $E$ is inversely proportional to the sixth power of the distance $r$ between the particles.
Mathematically,$E \propto \frac{1}{r^{6}}$,which can be written as $E \propto r^{-6}$.
Therefore,comparing this to $E \propto r^{x}$,we get $x = -6$.

(D) The solubility of gases in liquids decreases with an increase in temperature.
Therefore,as the temperature of water increases,the concentration of dissolved $O_{2}$ decreases.
Among the given options,$4^{\circ} C$ is the lowest temperature.
Thus,water at $4^{\circ} C$ will have the maximum concentration of dissolved $O_{2}$.

(C) Dalton's law of partial pressure states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases: $p = p_{1} + p_{2} + p_{3}$.
The partial pressure of a gas $i$ is given by $p_{i} = \chi_{i} p$,where $\chi_{i}$ is the mole fraction of the gas in the mixture.
Option $A$ is correct because $p = \sum n_{i} \frac{RT}{V} = (\sum n_{i}) \frac{RT}{V} = n_{total} \frac{RT}{V}$,which is the ideal gas law for the mixture.
Option $C$ represents Raoult's Law for solutions,not Dalton's Law for gaseous mixtures. Therefore,it is the incorrect equation.

(B) Let the total mass of the mixture be $100 \, g$.
Mass of $H_2 = 40 \, g$,so moles of $H_2$ $(n_{H_2})$ $= \frac{40}{2} = 20 \, mol$.
Mass of $O_2 = 60 \, g$,so moles of $O_2$ $(n_{O_2})$ $= \frac{60}{32} = 1.875 \, mol$.
Mole fraction of $H_2$ $(x_{H_2})$ $= \frac{n_{H_2}}{n_{H_2} + n_{O_2}} = \frac{20}{20 + 1.875} = \frac{20}{21.875} \approx 0.9143$.
Partial pressure of $H_2$ $(P_{H_2})$ $= x_{H_2} \times P_{total} = 0.9143 \times 2.2 \, bar = 2.01146 \, bar$.
The nearest integer is $2 \, bar$.

(B) According to Dalton's Law of Partial Pressures,$P_{total} = P_{O_2} + P_{Ne}$.
Given $P_{total} = 25 \ bar$ and $P_{Ne} = 20 \ bar$,so $P_{O_2} = 25 - 20 = 5 \ bar$.
Partial pressure is related to mole fraction $(X)$ by $P_i = X_i \times P_{total}$.
Moles of $Ne = \frac{200 \ g}{20 \ g \ mol^{-1}} = 10 \ mol$.
Moles of $O_2 = \frac{x \ g}{32 \ g \ mol^{-1}} = \frac{x}{32} \ mol$.
Mole fraction of $O_2$ is $X_{O_2} = \frac{n_{O_2}}{n_{O_2} + n_{Ne}} = \frac{x/32}{x/32 + 10}$.
Using $P_{O_2} = X_{O_2} \times P_{total}$,we get $5 = \left( \frac{x/32}{x/32 + 10} \right) \times 25$.
Dividing by $5$,we get $1 = \left( \frac{x/32}{x/32 + 10} \right) \times 5$.
$1 = \frac{5x/32}{(x + 320)/32} = \frac{5x}{x + 320}$.
$x + 320 = 5x \implies 4x = 320 \implies x = 80 \ g$.

(D) Step $1$: Calculate the number of moles of each gas.
$n_X = \frac{0.6 \ g}{20 \ g \ mol^{-1}} = 0.03 \ mol$
$n_Y = \frac{0.45 \ g}{45 \ g \ mol^{-1}} = 0.01 \ mol$
Step $2$: Calculate the mole fraction of gas $X$ $(\chi_X)$.
$\chi_X = \frac{n_X}{n_X + n_Y} = \frac{0.03}{0.03 + 0.01} = \frac{0.03}{0.04} = 0.75$
Step $3$: Calculate the partial pressure of gas $X$ $(P_X)$.
$P_X = \chi_X \times P_{total} = 0.75 \times 740 \ mm \ Hg = 555 \ mm \ Hg$

(C) The density of water vapour depends on temperature and pressure.
At the boiling point of water $(100^{\circ}C)$ and standard atmospheric pressure $(1 \ atm)$,the density of water vapour is approximately $6 \times 10^{-4} \ g \ cm^{-3}$.

(A) The molar volume of any ideal gas at standard temperature and pressure $(STP)$ is $22.414 \ L$.
Since $1 \ m^3 = 1000 \ L$,we have $1 \ L = 10^{-3} \ m^3$.
Therefore,$22.414 \ L = 22.414 \times 10^{-3} \ m^3 = 0.022414 \ m^3$.

(B) According to Dalton's Law of Partial Pressures,the partial pressure of a gas $(P_1)$ in a mixture is the product of its mole fraction $(x_1)$ and the total pressure $(P_{total})$ of the mixture.
$P_1 = x_1 \cdot P_{total}$
Rearranging this equation to solve for the mole fraction $(x_1)$ gives:
$x_1 = \frac{P_1}{P_{total}}$
Therefore,the correct relation is $x_1 = \frac{P_1}{P_{total}}$.

(B) The $SI$ unit of viscosity is the Pascal-second $(Pa \ s)$,which is equivalent to $N \ s \ m^{-2}$ or $Kg \ m^{-1} \ s^{-1}$.
Therefore,the correct option is $(B)$.

(B) $1$. Calculate the number of moles for each gas:
$n(N_2) = \frac{28 \ g}{28 \ g/mol} = 1 \ mol$
$n(He) = \frac{8 \ g}{4 \ g/mol} = 2 \ mol$
$n(Ne) = \frac{40 \ g}{20 \ g/mol} = 2 \ mol$
$2$. Calculate the total number of moles:
$n_{total} = 1 + 2 + 2 = 5 \ mol$
$3$. Calculate the mole fraction of $N_2$:
$x(N_2) = \frac{n(N_2)}{n_{total}} = \frac{1}{5} = 0.2$
$4$. Calculate the partial pressure of $N_2$ using Dalton's Law:
$P(N_2) = x(N_2) \times P_{total} = 0.2 \times 20 \ bar = 4 \ bar$

(C) The rate of diffusion is defined as the amount of gas (in moles) that diffuses per unit time.
Mathematically,$\text{Rate} = \frac{\text{Amount of substance}}{\text{Time}}$.
The $SI$ unit for the amount of substance is $mol$ and the $SI$ unit for time is $s$.
Therefore,the $SI$ unit for the rate of diffusion is $mol \cdot s^{-1}$.

(A) According to Dalton's Law of partial pressure,the partial pressure of a gas is directly proportional to its mole fraction $(P_i = X_i P_{total})$.
Since the temperature and volume are constant,the mole fraction $X_i$ is proportional to the number of moles $n_i$.
Let the mass of each gas be $m$.
The number of moles $n_i$ is given by $n_i = \frac{m}{M_i}$,where $M_i$ is the molar mass.
For $H_2$: $n = \frac{m}{2}$
For $He$: $n = \frac{m}{4}$
For $CO_2$: $n = \frac{m}{44}$
For $Ne$: $n = \frac{m}{20}$
Comparing the number of moles,the gas with the smallest molar mass will have the highest number of moles.
Since $H_2$ has the smallest molar mass $(2 \ g/mol)$,it will have the maximum number of moles and therefore exert the maximum partial pressure.

(C) According to Dalton's Law of Partial Pressure,the partial pressure of a gas is directly proportional to its mole fraction $(x_i)$ in the mixture,i.e.,$P_i = x_i \times P_{total}$.
Since $x_i = \frac{n_i}{n_{total}}$,the partial pressure is directly proportional to the number of moles $(n_i)$ of the gas.
Given that the masses $(m)$ of all gases are equal,the number of moles is calculated as $n = \frac{m}{M}$,where $M$ is the molar mass.
Thus,$n \propto \frac{1}{M}$.
The molar masses are: $M(H_2) = 2 \ g/mol$,$M(He) = 4 \ g/mol$,$M(Ne) = 20 \ g/mol$,and $M(CO_2) = 44 \ g/mol$.
Since $CO_2$ has the highest molar mass $(44 \ g/mol)$,it will have the minimum number of moles $(n = \frac{m}{44})$ for a fixed mass $m$.
Therefore,$CO_2$ exerts the minimum partial pressure.

(C) Let the mass of both helium $(He)$ and oxygen $(O_2)$ be $m \ g$.
The molar mass of $He$ is $4 \ g/mol$ and the molar mass of $O_2$ is $32 \ g/mol$.
The number of moles of $He$ $(n_{He})$ = $\frac{m}{4}$.
The number of moles of $O_2$ $(n_{O_2})$ = $\frac{m}{32}$.
The total number of moles $(n_{total})$ = $\frac{m}{4} + \frac{m}{32} = \frac{8m + m}{32} = \frac{9m}{32}$.
According to Dalton's law of partial pressures,the fraction of total pressure exerted by a gas is equal to its mole fraction.
Mole fraction of $He$ $(x_{He})$ = $\frac{n_{He}}{n_{total}} = \frac{m/4}{9m/32} = \frac{m}{4} \times \frac{32}{9m} = \frac{8}{9}$.

(D) The partial pressure $(P_i)$ of a gas in a mixture is directly proportional to its number of moles $(n_i)$ according to Dalton's Law of Partial Pressures ($P_i = x_i P_{total}$,where $x_i = \frac{n_i}{n_{total}}$).
Number of moles $(n)$ is calculated as $n = \frac{\text{mass}}{\text{molar mass}}$.
For $H_2$: $n = \frac{5 \text{ g}}{2 \text{ g/mol}} = 2.5 \text{ mol}$.
For $He$: $n = \frac{8 \text{ g}}{4 \text{ g/mol}} = 2.0 \text{ mol}$.
For $CO_2$: $n = \frac{50 \text{ g}}{44 \text{ g/mol}} \approx 1.136 \text{ mol}$.
For $Ne$: $n = \frac{20 \text{ g}}{20 \text{ g/mol}} = 1.0 \text{ mol}$.
Since $Ne$ has the minimum number of moles $(1.0 \text{ mol})$,it exerts the minimum partial pressure.

(B) Step $1$: Calculate the number of moles of each gas.
$n_{O_2} = \frac{32 \ g}{32 \ g/mol} = 1 \ mol$
$n_{Ar} = \frac{80 \ g}{40 \ g/mol} = 2 \ mol$
$n_{H_2} = \frac{4 \ g}{2 \ g/mol} = 2 \ mol$
Step $2$: Calculate the total number of moles.
$n_{\text{total}} = n_{O_2} + n_{Ar} + n_{H_2} = 1 + 2 + 2 = 5 \ mol$
Step $3$: Calculate the mole fraction of dioxygen $(x_{O_2})$.
$x_{O_2} = \frac{n_{O_2}}{n_{\text{total}}} = \frac{1}{5} = 0.2$
Step $4$: Calculate the partial pressure of dioxygen $(P_{O_2})$.
$P_{O_2} = x_{O_2} \times P_{\text{total}} = 0.2 \times 10 \ bar = 2 \ bar$

(C) According to Dalton's Law of Partial Pressures,the total pressure $P_{Total}$ is the sum of partial pressures of individual gases: $P_{Total} = P_A + P_B = 4.5 \ bar + 5.5 \ bar = 10.0 \ bar$.
The mole fraction $x_i$ of a gas is given by the ratio of its partial pressure to the total pressure: $x_A = \frac{P_A}{P_{Total}} = \frac{4.5 \ bar}{10.0 \ bar} = 0.45$.
Similarly,for gas $B$: $x_B = \frac{P_B}{P_{Total}} = \frac{5.5 \ bar}{10.0 \ bar} = 0.55$.
Therefore,the mole fractions of $A$ and $B$ are $0.45$ and $0.55$ respectively.

(A) First,calculate the number of moles for each gas:
$n_{O_2} = \frac{64 \ g}{32 \ g/mol} = 2 \ mol$
$n_{Ne} = \frac{160 \ g}{20 \ g/mol} = 8 \ mol$
Next,calculate the mole fraction of dioxygen $(x_{O_2})$:
$x_{O_2} = \frac{n_{O_2}}{n_{O_2} + n_{Ne}} = \frac{2}{2 + 8} = \frac{2}{10} = 0.2$
Finally,calculate the partial pressure of dioxygen $(P_{O_2})$ using Dalton's Law:
$P_{O_2} = x_{O_2} \times P_{Total} = 0.2 \times 25 \ bar = 5 \ bar$

(B) In a pressure cooker,the pressure inside is significantly higher than the atmospheric pressure. According to the relationship between vapor pressure and boiling point,as the pressure increases,the boiling point of water also increases. Consequently,water boils at a higher temperature (above $100 \ ^{\circ}C$),which provides more thermal energy to the food,leading to faster cooking.

(D) Dalton's law of partial pressure is applicable only to a mixture of non-reacting gases.
In the given options,$NH_{3} + HCl$ react to form $NH_{4}Cl$,$NO + O_{2}$ react to form $NO_{2}$,and $H_{2} + Cl_{2}$ react to form $HCl$.
However,$CO$ and $H_{2}$ do not react with each other under normal conditions.
Therefore,Dalton's law of partial pressure is applicable to the system $CO + H_{2}$.

(D) Assertion: The statement is incorrect because the standard boiling point (boiling point at $1 \ bar$ pressure) of a liquid is slightly lower than the normal boiling point (boiling point at $1 \ atm$ pressure).
Reason: The statement is correct because $1 \ atm = 1.01325 \ bar$,which means $1 \ bar$ pressure is slightly lower than $1 \ atm$ pressure.
Since the Assertion is incorrect and the Reason is correct,the correct option is $D$.

(B) The temperature at which the vapor pressure of a liquid becomes equal to the external pressure is defined as the boiling point of the liquid.
At a standard atmospheric pressure of $1.013 \ bar$ $(1 \ atm)$,this temperature is specifically known as the normal boiling point.

(B) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$ of the gas: $r \propto \frac{1}{\sqrt{M}}$.
The molar mass of $SO_2$ is $32 + (16 \times 2) = 64 \ g/mol$.
The molar mass of $CH_4$ is $12 + (1 \times 4) = 16 \ g/mol$.
Therefore,the ratio of the rate of diffusion of $SO_2$ to $CH_4$ is:
$\frac{r_{SO_2}}{r_{CH_4}} = \sqrt{\frac{M_{CH_4}}{M_{SO_2}}} = \sqrt{\frac{16}{64}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the ratio is $1: 2$.

(B) The diffusion rate is given as $120 \ mL \ min^{-1}$.
To find the time required for the diffusion of $3000 \ mL$ of the gas,we use the formula:
$\text{Time} = \frac{\text{Volume}}{\text{Rate}}$
$t = \frac{3000 \ mL}{120 \ mL \ min^{-1}} = 25 \ min$.
Since the question asks for the time in seconds:
$t = 25 \ min \times 60 \ s \ min^{-1} = 1500 \ s$.

(C) Given: $P_{\text{total}} = 2 \ bar$.
For $100 \ g$ of the mixture,mass of $H_2 = 20 \ g$ and mass of $O_2 = 80 \ g$.
Moles of $H_2$ $(n_{H_2})$ = $\frac{20 \ g}{2 \ g/mol} = 10 \ mol$.
Moles of $O_2$ $(n_{O_2})$ = $\frac{80 \ g}{32 \ g/mol} = 2.5 \ mol$.
Mole fraction of $O_2$ $(x_{O_2})$ = $\frac{n_{O_2}}{n_{H_2} + n_{O_2}} = \frac{2.5}{10 + 2.5} = \frac{2.5}{12.5} = 0.2$.
Partial pressure of $O_2$ $(P_{O_2})$ = $x_{O_2} \times P_{\text{total}} = 0.2 \times 2 \ bar = 0.4 \ bar$.

(C) At higher altitudes,the atmospheric pressure is lower than at sea level.
Because the boiling point of water is dependent on external pressure,a decrease in atmospheric pressure leads to a decrease in the boiling point of water.
Since the water boils at a lower temperature,it cannot provide enough heat to cook the food efficiently,thus taking more time.

(A) Isochores are drawn at constant volume,$\Delta V = 0$.
We get a plot of pressure $(p)$ $vs$ temperature $(T$,in $K)$ as shown in the graph.
For $1 \ mol$ of an ideal gas,$pV = nRT$.
At isochoric condition,$V$ is constant,so $p \propto T$.

(D) As we go higher in altitude,the atmospheric pressure decreases.
Since the boiling point of a liquid is the temperature at which its vapor pressure equals the external atmospheric pressure,a decrease in external pressure leads to a decrease in the boiling point.
Therefore,water boils at a lower temperature on the top of a mountain.

(A) Let the mass of $H_2$ be $w \ g$ and the mass of $O_2$ be $2w \ g$.
Number of moles of $H_2$ $(n_{H_2})$ = $\frac{w}{2}$.
Number of moles of $O_2$ $(n_{O_2})$ = $\frac{2w}{32} = \frac{w}{16}$.
Total moles $(n_{total})$ = $\frac{w}{2} + \frac{w}{16} = \frac{8w + w}{16} = \frac{9w}{16}$.
Mole fraction of $O_2$ $(x_{O_2})$ = $\frac{n_{O_2}}{n_{total}} = \frac{w/16}{9w/16} = \frac{1}{9}$.
Partial pressure of $O_2$ = $x_{O_2} \times P_{total} = \frac{1}{9} \times p = \frac{p}{9}$ atm.

(D) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ of the gas: $r \propto \frac{1}{\sqrt{M}}$.
Since the time $t$ is the same,the rate of diffusion is directly proportional to the mass $w$ diffused: $r = \frac{w}{t}$.
Therefore,$\frac{w_1}{w_2} = \sqrt{\frac{M_2}{M_1}}$.
Given: $w_{H_2} = 2.0 \ g$,$M_{H_2} = 2 \ g/mol$,$M_{O_2} = 32 \ g/mol$.
Substituting the values: $\frac{2.0}{w_{O_2}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
$w_{O_2} = 2.0 \times 4 = 8.0 \ g$.

(B) Let the equal weight of each gas be $w \ g$. The molar masses are $M(H_2) = 2 \ g/mol$,$M(D_2) = 4 \ g/mol$,and $M(T_2) = 6 \ g/mol$.
Number of moles: $n(H_2) = \frac{w}{2}$,$n(D_2) = \frac{w}{4}$,$n(T_2) = \frac{w}{6}$.
Total moles $n_{\text{total}} = w(\frac{1}{2} + \frac{1}{4} + \frac{1}{6}) = w(\frac{6+3+2}{12}) = \frac{11w}{12}$.
Mole fractions: $x(H_2) = \frac{w/2}{11w/12} = \frac{6}{11} \approx 0.545$,$x(D_2) = \frac{w/4}{11w/12} = \frac{3}{11} \approx 0.273$,$x(T_2) = \frac{w/6}{11w/12} = \frac{2}{11} \approx 0.182$.
Since partial pressure $P_i = x_i P_{\text{total}}$,the ratio of partial pressures $P(T_2) : P(D_2) : P(H_2)$ is equal to the ratio of their mole fractions $x(T_2) : x(D_2) : x(H_2) = 0.18 : 0.27 : 0.54$.

(A) $P_{total} = 1 \ bar$. Let total mass $= 100 \ g$.
$W(H_2) = 20 \ g, W(He) = 16 \ g, W(O_2) = 100 - (20 + 16) = 64 \ g$.
$n(H_2) = \frac{20}{2} = 10 \ mol$.
$n(He) = \frac{16}{4} = 4 \ mol$.
$n(O_2) = \frac{64}{32} = 2 \ mol$.
Total moles $= 10 + 4 + 2 = 16 \ mol$.
$x(H_2) = \frac{10}{16} = 0.625$.
$x(He) = \frac{4}{16} = 0.250$.
$x(O_2) = \frac{2}{16} = 0.125$.
Partial pressure $P_i = x_i \times P_{total}$.
$P(H_2) = 0.625 \times 1 = 0.625 \ bar$.
$P(He) = 0.250 \times 1 = 0.250 \ bar$.
$P(O_2) = 0.125 \times 1 = 0.125 \ bar$.

(A) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
Molecular mass of $H_2$ $(M_{H_2})$ = $2 \ g/mol$.
Molecular mass of $He$ $(M_{He})$ = $4 \ g/mol$.
$\frac{r_{He}}{r_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{He}}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Therefore,the ratio is $1 : \sqrt{2}$.

(D) According to Graham's Law of diffusion,the rate of effusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
Since the time taken is the same for both gases,the ratio of the rates of effusion is equal to the ratio of the volumes effused $(V)$: $\frac{r_x}{r_{CH_4}} = \frac{V_x}{V_{CH_4}}$.
Therefore,$\frac{V_x}{V_{CH_4}} = \sqrt{\frac{M_{CH_4}}{M_x}}$.
Given: $V_x = 300 \ mL$,$M_x = 32 \ g \ mol^{-1}$,$M_{CH_4} = 16 \ g \ mol^{-1}$.
Substituting the values: $\frac{300}{V_{CH_4}} = \sqrt{\frac{16}{32}} = \sqrt{\frac{1}{2}} = \frac{1}{1.414}$.
$V_{CH_4} = 300 \times 1.414 = 424.26 \ mL \approx 424 \ mL$.

(A) According to Graham's law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$:
$\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}}$
Given:
$M_A = 100 \ g/mol$
$M_B = 64 \ g/mol$
$r_A = 12 \times 10^{-3} \ mol/s$
Substituting the values:
$\frac{12 \times 10^{-3}}{r_B} = \sqrt{\frac{64}{100}}$
$\frac{12 \times 10^{-3}}{r_B} = \frac{8}{10} = 0.8$
$r_B = \frac{12 \times 10^{-3}}{0.8} = 15 \times 10^{-3} \ mol/s$

(A) $(I)$ According to Dalton's Law of Partial Pressures: $P_{total} = P_A + P_B + P_C$
$\Rightarrow P_C = P_{total} - (P_A + P_B) = 10 - (3 + 1) = 6 \,atm$
$(II)$ The partial pressure of a gas is given by $P_i = \chi_i \times P_{total} = (n_i / n_{total}) \times P_{total}$
$\Rightarrow P_C = (n_C / 10) \times 10 = n_C$
$\Rightarrow n_C = 6 \,mol$
$(III)$ Weight of $C = n_C \times \text{Molar Mass of } C = 6 \,mol \times 2 \,g/mol = 12 \,g$

(B) According to Dalton's law of partial pressures,the partial pressure of a gas in a mixture is equal to the product of its mole fraction and the total pressure of the mixture.
Mathematically,this is expressed as: $p_i = \chi_i \times p_{\text{total}}$.

(B) According to Graham's law of effusion,the rate of effusion $(r)$ of a gas is inversely proportional to the square root of its molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
For two gases $CO$ and $N_2$ at the same temperature and pressure,the ratio of their rates of effusion is given by: $\frac{r_{CO}}{r_{N_2}} = \sqrt{\frac{M_{N_2}}{M_{CO}}}$.
The molar mass of $CO$ is $12 + 16 = 28 \ g/mol$.
The molar mass of $N_2$ is $14 \times 2 = 28 \ g/mol$.
Substituting these values: $\frac{r_{CO}}{r_{N_2}} = \sqrt{\frac{28}{28}} = \sqrt{1} = 1$.
Thus,the ratio is $1: 1$.

(C) According to Graham's Law of diffusion,the rate of diffusion $(R)$ is inversely proportional to the square root of the molecular weight $(M)$: $R \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the rates of diffusion for gases $A$ and $B$ is given by: $\frac{R_A}{R_B} = \sqrt{\frac{M_B}{M_A}}$.
Given $\frac{R_A}{R_B} = \frac{1}{0.707}$ and $M_B = 32$,we substitute these values into the equation:
$\frac{1}{0.707} = \sqrt{\frac{32}{M_A}}$.
Squaring both sides,we get: $(\frac{1}{0.707})^2 = \frac{32}{M_A}$.
Since $0.707 \approx \frac{1}{\sqrt{2}}$,then $(\frac{1}{0.707})^2 \approx 2$.
So,$2 = \frac{32}{M_A}$,which gives $M_A = \frac{32}{2} = 16$.

(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.
Therefore,$\frac{r_X}{r_{\text{gas}}} = \sqrt{\frac{M_{\text{gas}}}{M_X}}$.
Given that $r_X = 3 \sqrt{3} \times r_{\text{gas}}$,we have $\frac{r_X}{r_{\text{gas}}} = 3 \sqrt{3}$.
Substituting the values: $3 \sqrt{3} = \sqrt{\frac{54}{M_X}}$.
Squaring both sides: $(3 \sqrt{3})^2 = \frac{54}{M_X}$.
$27 = \frac{54}{M_X}$.
$M_X = \frac{54}{27} = 2 \ g \ mol^{-1}$.

(C) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.
Let $d_1$ be the distance traveled by $CH_4$ and $d_2$ be the distance traveled by $SO_2$.
Since they start at the same time and meet at time $t$,the ratio of distances is equal to the ratio of their rates of diffusion: $\frac{d_1}{d_2} = \frac{r_{CH_4}}{r_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{CH_4}}}$.
Given $M_{CH_4} = 16 \ g/mol$ and $M_{SO_2} = 64 \ g/mol$.
$\frac{d_1}{d_2} = \sqrt{\frac{64}{16}} = \sqrt{4} = 2$.
So,$d_1 = 2d_2$.
Since the total length is $1 \ km = 1000 \ m$,$d_1 + d_2 = 1000$.
Substituting $d_2 = \frac{d_1}{2}$,we get $d_1 + \frac{d_1}{2} = 1000$.
$\frac{3d_1}{2} = 1000 \implies d_1 = \frac{2000}{3} \approx 667 \ m$.

(C) $(I)$ Gases mix evenly without the help of any mechanical change. The large intermolecular spaces allow molecules of different gases to mix easily.
Thus,statement $(I)$ is correct.
$(II)$ Due to the negligible intermolecular forces of attraction,gas molecules move randomly in all directions and collide with the walls of the container,exerting pressure in all directions.
Thus,statement $(II)$ is correct.
$(III)$ Gases are highly compressible because their molecules are separated by large distances compared to solids and liquids,and they have lower density.
Thus,statement $(III)$ is correct.
$(IV)$ Gases have neither a fixed shape nor a fixed volume; they occupy the entire space available to them.
Thus,statement $(IV)$ is incorrect.
Therefore,statements $(I)$,$(II)$,and $(III)$ are correct,making option $(C)$ the correct choice.

(C) Viscosity of liquids depends strongly on temperature. It usually decreases with increasing temperature because an increase in temperature increases the kinetic energy of the molecules,allowing them to overcome the intermolecular attractive forces.
The $SI$ unit of viscosity is Pascal-second $(Pa \ s)$,which is equivalent to $kg \ m^{-1} \ s^{-1}$.
Statement $I$ is correct,but Statement $II$ is incorrect because the unit provided is $kg \ m^{-1} \ s^{-2}$ instead of $kg \ m^{-1} \ s^{-1}$.

(B) Volume is an extensive property,meaning it depends on the quantity of matter. When the vessel is divided into two equal parts,the volume of each part becomes $\frac{V}{2} \ L$.
Temperature is an intensive property,meaning it is independent of the quantity of matter. Therefore,the temperature in each part remains $T \ K$.

(C) Pressure and temperature are intensive properties,which means they are independent of the amount of matter present in the system.
When a vessel is divided into two equal parts,the volume of each part becomes half,but the pressure and temperature remain the same as the original system.
Therefore,the pressure in each part is $P \ atm$ and the temperature is $T \ K$.

(D) In the given $V-T$ graph:
$1$. Process $C \rightarrow A$: The temperature remains constant while the volume changes. This is an isothermal process.
$2$. Process $A \rightarrow B$: The volume remains constant while the temperature changes. This is an isochoric process.
$3$. Process $B \rightarrow C$: The graph is a straight line passing through the origin in a $V-T$ plot,which implies $V \propto T$. According to Charles's Law,this represents an isobaric process.
Therefore,the processes $C \rightarrow A$,$B \rightarrow C$,and $A \rightarrow B$ are isothermal,isobaric,and isochoric respectively.