Mix Examples- States of Matter Questions in English

(C) Option $(C)$ is correct because both liquids and gases are fluids,meaning they can flow and exhibit the property of viscosity.
Option $(A)$ is incorrect because molecules in solids do not possess random translational motion; they only vibrate about fixed positions.
Option $(B)$ is incorrect because gases can undergo deposition to form solids directly.
Option $(D)$ is incorrect because Boyle's law states that $PV = \text{constant}$,not $V/P = \text{constant}$.

(A) From the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$ and $d = \frac{m}{V}$,we get $P = \frac{dRT}{M}$.
Assuming the molar masses $(M)$ of the two gases are the same or comparing the pressure-density-temperature relationship for a given gas,the ratio is given by $\frac{P_1}{P_2} = \frac{d_1 T_1}{d_2 T_2}$.
Given $\frac{d_1}{d_2} = \frac{1}{2}$ and $\frac{T_1}{T_2} = \frac{2}{1}$.
Substituting these values: $\frac{P_1}{P_2} = \frac{1}{2} \times \frac{2}{1} = \frac{1}{1}$.
Thus,the ratio of their pressures is $1:1$.

(C) In a closed flask,the volume $(V)$ and the mass of the gas remain constant.
According to Gay-Lussac's Law,at constant volume,$P \propto T$.
Since the temperature increases from $300 \ K$ to $600 \ K$,the pressure $(P)$ must increase.
As temperature increases,the average kinetic energy of the gas molecules increases,which leads to an increase in the frequency and rate of collisions.
Since the flask is closed,no gas is added or removed,so the number of moles $(n)$ remains constant.
Therefore,the statement that the number of moles of gas increases is incorrect.

(A) The Avogadro number is defined as $6.022 \times 10^{23} \ mol^{-1}$.
Option $(A)$ states the value as $6.02 \times 10^{21}$,which is incorrect.
Option $(B)$ correctly relates average velocity $(\bar{v})$ and root mean square velocity $(u)$ as $\bar{v} = \sqrt{8RT/\pi M}$ and $u = \sqrt{3RT/M}$,leading to $\bar{v} = 0.9213 \ u$.
Option $(C)$ is correct because the mean kinetic energy of an ideal gas depends only on temperature $(KE = 3/2 RT)$.
Option $(D)$ is the correct formula for root mean square velocity.

(C) Let the mass of each gas be $m \, g$.
Number of moles of lighter gas $(n_1)$ $= \frac{m}{4}$.
Number of moles of heavier gas $(n_2)$ $= \frac{m}{40}$.
Total number of moles $(n_{total})$ $= \frac{m}{4} + \frac{m}{40} = \frac{10m + m}{40} = \frac{11m}{40}$.
Mole fraction of lighter gas $(x_1)$ $= \frac{n_1}{n_{total}} = \frac{m/4}{11m/40} = \frac{10}{11}$.
Partial pressure of lighter gas $= x_1 \times P_{total} = \frac{10}{11} \times 1.1 \, atm = 1 \, atm$.

(D) 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,$\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}}$.
Given that $r_A = 2r_B$,we have $\frac{r_A}{r_B} = 2$.
Substituting this into the equation: $2 = \sqrt{\frac{M_B}{M_A}}$.
Squaring both sides: $4 = \frac{M_B}{M_A}$.
Thus,the ratio of molecular weights of $A$ to $B$ is $\frac{M_A}{M_B} = \frac{1}{4} = 0.25$.

(A) Dalton's law of partial pressure is applicable only to a mixture of non-reacting gases.
$A$. The mixture of $SO_2$ and $Cl_2$ reacts chemically to form sulfuryl chloride $(SO_2Cl_2)$: $SO_2(g) + Cl_2(g) \rightarrow SO_2Cl_2(g)$.
Since these gases react with each other,they do not follow Dalton's law of partial pressure.

(A) The average kinetic energy of a gas molecule is given by the formula $KE_{avg} = \frac{3}{2}RT$.
Since the temperature $(T)$ is the same for both gases at $STP$,the average kinetic energy per mole of both gases will be equal.
Therefore,$0.50 \ mol$ of $H_2$ and $1.0 \ mol$ of $He$ will have equal average kinetic energies.

(D) The average translational kinetic energy $(KE)$ per molecule is given by $\frac{3}{2}kT$,which depends only on temperature. Since both gases are at $298 \ K$,their average translational $KE$ per molecule is the same.
According to Avogadro's Law,at the same temperature and pressure,equal volumes of gases contain the same number of molecules $(n)$. Since $KE_{total} = n \times (\text{average } KE \text{ per molecule})$,the total translational $KE$ is also the same.
Density $(d)$ is given by $d = \frac{PM}{RT}$. Since $P$,$T$,and $R$ are constant,$d \propto M$. The molar mass of $N_2$ $(28 \ g/mol)$ is less than that of $CO_2$ $(44 \ g/mol)$,so the density of $N_2$ is less than that of $CO_2$.
Therefore,all statements are correct.

(C) Volume of steam = $1\ L = 10^3\ cm^3$.
Mass of $10^3\ cm^3$ steam = $\text{density} \times \text{Volume} = 0.0006\ g\ cm^{-3} \times 10^3\ cm^3 = 0.6\ g$.
The actual volume occupied by $H_2O$ molecules is equal to the volume of liquid water of the same mass.
Actual volume of $H_2O$ molecules = $\frac{\text{mass of steam}}{\text{density of liquid water}} = \frac{0.6\ g}{1.0\ g\ cm^{-3}} = 0.6\ cm^3$.

(B) Adiabatic demagnetisation is a technique used to achieve extremely low temperatures,often near absolute zero,by removing the magnetic field from a paramagnetic substance that has been pre-cooled.

(D) For a monoatomic gas,${C_v} = \frac{3}{2}R$ and ${C_p} = \frac{5}{2}R$.
For a diatomic gas,${C_v} = \frac{5}{2}R$ and ${C_p} = \frac{7}{2}R$.
Since the gases are at the same temperature and pressure,equal volumes contain an equal number of moles $(n_1 = n_2 = 1)$.
The molar heat capacity at constant volume for the mixture is ${C_{v,mix}} = \frac{n_1{C_{v,1}} + n_2{C_{v,2}}}{n_1 + n_2} = \frac{\frac{3}{2}R + \frac{5}{2}R}{2} = 2R$.
The molar heat capacity at constant pressure for the mixture is ${C_{p,mix}} = \frac{n_1{C_{p,1}} + n_2{C_{p,2}}}{n_1 + n_2} = \frac{\frac{5}{2}R + \frac{7}{2}R}{2} = 3R$.
Therefore,the ratio of specific heats is $\gamma = \frac{{C_{p,mix}}}{{C_{v,mix}}} = \frac{3R}{2R} = 1.5$.

(C) . Solid $CO_2$ is known as dry ice because it undergoes sublimation,meaning it evaporates directly from the solid phase to the gas phase at $-78 \ ^\circ C$ at atmospheric pressure without passing through the liquid state.

(A) is the correct statement.
$1$. Both gases and liquids exhibit fluidity,which is characterized by viscosity.
$2$. Molecules in the solid state are fixed in position and do not possess translational motion; they only exhibit vibrational motion.
$3$. Gases can be converted directly into solids through the process of deposition.
$4$. While liquids have significant vapour pressure,solids generally have negligible vapour pressure at room temperature.

(B) When the temperature is raised,the viscosity of liquid decreases. This is because an increase in temperature increases the average kinetic energy of the molecules,which allows them to overcome the intermolecular attractive forces that hinder flow.

(A) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$:
$r \propto \frac{1}{\sqrt{M}}$
Let $r_A$ and $r_B$ be the rates of diffusion of gas $A$ and gas $B$,and $M_A$ and $M_B$ be their respective molar masses.
$\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}}$
Given that $r_A = 3r_B$ and $M_A = 4$:
$\frac{3r_B}{r_B} = \sqrt{\frac{M_B}{4}}$
$3 = \sqrt{\frac{M_B}{4}}$
Squaring both sides:
$9 = \frac{M_B}{4}$
$M_B = 9 \times 4 = 36$

(B) $1$. Initial state: Each part has a volume $V = 0.5 \, L$.
$2$. For $H_2$: Moles $n_1 = \frac{2 \, g}{2 \, g/mol} = 1 \, mol$.
$3$. For $CH_4$: Moles $n_2 = \frac{16 \, g}{16 \, g/mol} = 1 \, mol$.
$4$. Using ideal gas law $PV = nRT$,since $n_1 = n_2 = 1 \, mol$ and $V$ and $T$ are the same for both,the pressure in each part is $P = \frac{nRT}{V}$.
$5$. After removing the partition,the total volume becomes $V_{total} = 1 \, L$ and total moles $n_{total} = n_1 + n_2 = 1 + 1 = 2 \, mol$.
$6$. The new total pressure $P_{total} = \frac{n_{total}RT}{V_{total}} = \frac{2RT}{1} = 2 \times \frac{RT}{V_{part}} = 2P$.

(D) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
Given the molar ratio $n_{CH_4} : n_{O_2} = 1:2$.
The rate of diffusion is also proportional to the number of moles $(n)$: $r \propto n$.
Therefore,the ratio of rates is $\frac{r_{O_2}}{r_{CH_4}} = \frac{n_{O_2}}{n_{CH_4}} \times \sqrt{\frac{M_{CH_4}}{M_{O_2}}}$.
Substituting the values: $M_{CH_4} = 16 \ g/mol$ and $M_{O_2} = 32 \ g/mol$.
$\frac{r_{O_2}}{r_{CH_4}} = \frac{2}{1} \times \sqrt{\frac{16}{32}} = 2 \times \sqrt{\frac{1}{2}} = 2 \times \frac{1}{1.414} = \frac{1.414}{1}$.
Thus,the ratio is $1.414:1$.

(D) The universal gas constant $R$ has different values depending on the units used:
$1$. In $L \, atm \, K^{-1} \, mol^{-1}$,$R \approx 0.0821$.
$2$. In $CGS$ units $(erg \, K^{-1} \, mol^{-1})$,$R \approx 8.314 \times 10^7$.
$3$. In calories $(cal \, K^{-1} \, mol^{-1})$,$R \approx 1.987 \approx 2$.
Since all provided options represent correct values of $R$ in different unit systems,the correct answer is $D$.

(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: $r_1 / r_2 = \sqrt{M_2 / M_1}$.
Here,$r_1$ is the rate of helium $(He)$ and $r_2$ is the rate of methane $(CH_4)$.
The molar mass of helium $(M_1)$ is $4 \ g/mol$.
The molar mass of methane $(M_2)$ is $16 \ g/mol$.
Substituting the values: $r_{He} / r_{CH_4} = \sqrt{16 / 4} = \sqrt{4} = 2$.
Therefore,the rate of diffusion of helium is $2$ times that of methane.

(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M_w$ for a given amount of gas.
$\frac{r_{O_2}}{r_{H_2}} = \frac{n_{O_2}}{n_{H_2}} = \sqrt{\frac{M_{w(H_2)}}{M_{w(O_2)}}}$
Given: $n_{O_2} = \frac{4 \ g}{32 \ g/mol} = 0.125 \ mol$.
Let $w$ be the mass of $H_2$. Then $n_{H_2} = \frac{w}{2 \ g/mol}$.
Substituting the values: $\frac{0.125}{w/2} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
$\frac{0.25}{w} = \frac{1}{4} \implies w = 1 \ g$.

(D) Using Boyle's Law,$P_1V_1 = P_2V_2$ at constant temperature.
For $N_2$: $P_{N_2} = (720 \ mm \times 200 \ cm^3) / 1000 \ cm^3 = 144 \ mm$.
For $O_2$: $P_{O_2} = (750 \ mm \times 400 \ cm^3) / 1000 \ cm^3 = 300 \ mm$.
Total pressure $P_{total} = P_{N_2} + P_{O_2} = 144 \ mm + 300 \ mm = 444 \ mm$.

(A) The Avogadro number is defined as $6.022 \times 10^{23} \text{ mol}^{-1}$.
Option $A$ states it as $6.02 \times 10^{21}$,which is incorrect.
Option $B$ is correct as $\bar{\upsilon} = \sqrt{\frac{8RT}{\pi M}}$ and $u = \sqrt{\frac{3RT}{M}}$,so $\frac{\bar{\upsilon}}{u} = \sqrt{\frac{8}{3\pi}} \approx 0.9213$.
Option $C$ is correct because the average kinetic energy of an ideal gas depends only on temperature ($KE = \frac{3}{2}RT$ for $1 \text{ mole}$).
Option $D$ is the standard formula for root mean square velocity.

(C) For one mole of an ideal gas at $P = 1 \ atm$,the ideal gas equation is $PV = nRT$.
Since $n = 1$ and $P = 1 \ atm$,the equation becomes $V = RT$.
Using the gas constant $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,we have $V = 0.0821 \times T$.
At $T_1 = 273 \ K$,$V_1 = 0.0821 \times 273 \approx 22.4 \ L$.
At $T_2 = 373 \ K$,$V_2 = 0.0821 \times 373 \approx 30.6 \ L$.
Comparing these values with the given options,option $C$ matches these coordinates.

(B) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
For $H_2$,$M_1 = 2 \ g/mol$. For $He$,$M_2 = 4 \ g/mol$.
$\frac{r_{H_2}}{r_{He}} = \sqrt{\frac{M_{He}}{M_{H_2}}} = \sqrt{\frac{4}{2}} = \sqrt{2} \approx 1.414$.
Therefore,the rate of diffusion of $H_2$ is $1.4$ times that of $He$.

(B) The pressure of the dry gas is calculated by subtracting the vapor pressure of water from the total pressure: $P_{gas} = P_{total} - P_{H_2O} = 744 \ mm \ Hg - 31.8 \ mm \ Hg = 712.2 \ mm \ Hg$.
Converting pressure to atm: $P = \frac{712.2}{760} \ atm \approx 0.937 \ atm$.
Using the ideal gas equation $PV = \frac{w}{M_w}RT$:
$0.937 \times 1.47 = \frac{1.98}{M_w} \times 0.0821 \times 303$.
$M_w = \frac{1.98 \times 0.0821 \times 303}{0.937 \times 1.47} \approx 35.76 \ g/mol$.

(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: $r \propto \frac{1}{\sqrt{M}}$.
For two gases to diffuse at the same rate,they must have the same molar mass.
Calculate the molar masses of the given pairs:
$A: NO = 14 + 16 = 30 \ g/mol$; $CO = 12 + 16 = 28 \ g/mol$ (Not equal)
$B: NO = 30 \ g/mol$; $CO_2 = 12 + 32 = 44 \ g/mol$ (Not equal)
$C: NH_3 = 14 + 3 = 17 \ g/mol$; $PH_3 = 31 + 3 = 34 \ g/mol$ (Not equal)
$D: NO = 30 \ g/mol$; $C_2H_6 = (2 \times 12) + (6 \times 1) = 24 + 6 = 30 \ g/mol$ (Equal)
Since $NO$ and $C_2H_6$ have the same molar mass,they will diffuse at the same rate.

(C) The density of liquid water is $\rho_l = 1.0 \, \text{g cm}^{-3}$.
The density of water vapor is $\rho_v = 0.0006 \, \text{g cm}^{-3} = 6 \times 10^{-4} \, \text{g cm}^{-3}$.
In $1 \, \text{L}$ $(1000 \, \text{cm}^3)$ of steam,the mass of water vapor is $m = \rho_v \times V = (6 \times 10^{-4} \, \text{g cm}^{-3}) \times (1000 \, \text{cm}^3) = 0.6 \, \text{g}$.
Since the mass of water remains constant during phase change,this $0.6 \, \text{g}$ of water occupies a volume in the liquid state given by $V_l = \frac{m}{\rho_l} = \frac{0.6 \, \text{g}}{1.0 \, \text{g cm}^{-3}} = 0.6 \, \text{cm}^3$.

(B) During the process of evaporation,the molecules with higher kinetic energy escape from the surface of the liquid.
As a result,the average kinetic energy of the remaining molecules decreases,which leads to a decrease in the temperature of the liquid.

(C) Let the volume of each flask be $V$. The total volume of the system is $2V$.
Initially,both flasks are at $T_1 = 100 \, ^\circ C = 373 \, K$. The total moles of gas $n = \frac{P_1 V_{total}}{R T_1} = \frac{0.5 \times 2V}{R \times 373}$.
When one flask is at $T_{ice} = 0 \, ^\circ C = 273 \, K$ and the other is at $T_{boil} = 100 \, ^\circ C = 373 \, K$,the total moles remain constant:
$n = \frac{P_2 V}{R \times 273} + \frac{P_2 V}{R \times 373} = \frac{P_2 V}{R} \left( \frac{1}{273} + \frac{1}{373} \right)$.
Equating the moles: $\frac{0.5 \times 2V}{373} = P_2 V \left( \frac{373 + 273}{273 \times 373} \right)$.
$P_2 = \frac{0.5 \times 2 \times 273}{646} = \frac{273}{646} \approx 0.4226 \, atm$.
However,using the simplified average temperature approach: $T_{avg} = \frac{273 + 373}{2} = 323 \, K$.
Using $\frac{P_1}{T_1} = \frac{P_2}{T_2}$:
$P_2 = \frac{0.5 \times 323}{373} \approx 0.433 \, atm$.

(C) According to Graham's law of diffusion for gases at different pressures,the rate of diffusion $r$ is given by $r \propto \frac{P}{\sqrt{M}}$.
Thus,$\frac{r_{HCl}}{r_{NH_3}} = \frac{P_{HCl}}{P_{NH_3}} \sqrt{\frac{M_{NH_3}}{M_{HCl}}}$.
Given that $NH_4Cl$ forms at $60 \, cm$ from the $HCl$ end,the distance traveled by $HCl$ is $60 \, cm$ and by $NH_3$ is $40 \, cm$. Since the rate is proportional to the distance traveled in the same time,$\frac{r_{HCl}}{r_{NH_3}} = \frac{60}{40} = 1.5$.
Substituting the values: $1.5 = \frac{P}{1} \sqrt{\frac{17}{36.5}}$.
$P = 1.5 \times \sqrt{\frac{36.5}{17}} = 1.5 \times \sqrt{2.147} \approx 1.5 \times 1.465 = 2.198 \, atm$.

(A) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$: $\frac{r_{He}}{r_{CH_4}} = \sqrt{\frac{M_{CH_4}}{M_{He}}}$.
Substituting the molar masses ($M_{CH_4} = 16 \ g/mol$,$M_{He} = 4 \ g/mol$): $\frac{r_{He}}{r_{CH_4}} = \sqrt{\frac{16}{4}} = \sqrt{4} = 2$.
This means the rate of effusion of $He$ is $2$ times that of $CH_4$.
The composition of the effusing mixture is given by the ratio of the rates of effusion multiplied by their initial molar ratio in the container: $\frac{\text{Rate}_{He}}{\text{Rate}_{CH_4}} = \frac{r_{He}}{r_{CH_4}} \times \frac{n_{He}}{n_{CH_4}} = 2 \times \frac{4}{1} = \frac{8}{1}$.
Thus,the initial composition of the effusing gas mixture is $8:1$.

(A) The molar mass of water $(H_2O)$ is $18 \, \text{g/mol}$.
Since the density of water is approximately $1 \, \text{g/cm}^3$,the volume of $1 \, \text{mol}$ of liquid water is $18 \, \text{cm}^3$.
At $STP$,the volume of $1 \, \text{mol}$ of an ideal gas is $22.4 \, \text{L} = 22400 \, \text{cm}^3$.
The actual volume occupied by the molecules is the volume of the liquid state.
The percentage of volume occupied by the molecules is calculated as: $\frac{18 \, \text{cm}^3}{22400 \, \text{cm}^3} \times 100 \approx 0.08\%$.
This value is less than $1\%$ of the total volume of $22.4 \, \text{L}$.

(A) According to Boyle's Law,at constant temperature,$V \propto \frac{1}{P}$. Therefore,the graph of $V$ versus $P$ should be a rectangular hyperbola,not a straight line passing through the origin. Thus,the graph in option $A$ is incorrect.

(C) According to the ideal gas equation,$PV = nRT$. For $n = 1 \ mol$ and $P = 1 \ atm$,the equation becomes $V = \frac{RT}{P} = \frac{0.0821 \ L \ atm \ K^{-1} \ mol^{-1} \times T}{1 \ atm} = 0.0821 \times T$.
At $T_1 = 273 \ K$,$V_1 = 0.0821 \times 273 \approx 22.4 \ L$.
At $T_2 = 373 \ K$,$V_2 = 0.0821 \times 373 \approx 30.6 \ L$.
Thus,the graph must pass through the points $(22.4 \ L, 273 \ K)$ and $(30.6 \ L, 373 \ K)$.

(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ for a given time $t$: $\frac{n_1}{n_2} = \sqrt{\frac{M_2}{M_1}}$,where $n$ is the number of moles.
Since $n = \frac{w}{M}$,the ratio of weights $w$ is given by: $\frac{w_1}{w_2} = \sqrt{\frac{M_1}{M_2}}$.
Here,$w_{O_2} = 4 \ g$,$M_{O_2} = 32 \ g/mol$,and $M_{H_2} = 2 \ g/mol$.
Substituting the values: $\frac{4}{w_{H_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
Therefore,$w_{H_2} = \frac{4}{4} = 1 \ g$.

(A) According to Graham's Law of Diffusion: $\frac{r_1}{r_2} = \frac{V_1/t}{V_2/t} = \frac{V_1}{V_2} = \sqrt{\frac{M_2}{M_1}}$.
At $STP$,the volume of $0.48 \, g$ of $O_2$ is calculated as: $V_{O_2} = \frac{22400 \, mL}{32 \, g} \times 0.48 \, g = 336 \, mL$.
Using the diffusion ratio: $\frac{V_{O_2}}{V_{CO_2}} = \sqrt{\frac{M_{CO_2}}{M_{O_2}}} = \sqrt{\frac{44}{32}} = \sqrt{1.375} \approx 1.1726$.
Therefore,$V_{CO_2} = \frac{V_{O_2}}{1.1726} = \frac{336}{1.1726} \approx 286.5 \, mL$.

(B) $P_{\text{total}} = P_{\text{dry gas}} + P_{\text{water vapor}}$
At $T_1$,$P_{\text{dry gas}} = 830 - 30 = 800 \, mm$
Using Gay-Lussac's Law for the dry gas: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$
Given $T_2 = T_1 - 0.01 T_1 = 0.99 T_1$
$P_2 = P_1 \times \frac{T_2}{T_1} = 800 \times 0.99 = 792 \, mm$
New total pressure = $P_2 + P_{\text{water vapor at } T_2} = 792 + 25 = 817 \, mm$

(D) Using the ideal gas law $PV = nRT$ for dry $O_2$ gas:
$n = 0.0168 \, mol$,$V = 428 \, mL = 0.428 \, L$,$T = 298 \, K$,$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.
$P_{dry} = \frac{nRT}{V} = \frac{0.0168 \times 0.0821 \times 298}{0.428} \approx 0.9607 \, atm$.
Convert $P_{dry}$ to $mm \, Hg$: $0.9607 \, atm \times 760 \, mm \, atm^{-1} \approx 730.1 \, mm \, Hg$.
Total pressure $P_{total} = P_{dry} + P_{water \, vapor}$.
$P_{water \, vapor} = P_{total} - P_{dry} = 754 \, mm - 730.1 \, mm \approx 23.9 \, mm \approx 24 \, mm$.

(A) According to Graham's Law of Effusion,the rate of effusion $r \propto \frac{1}{\sqrt{M}}$.
Calculating molar masses: $M(H_2) = 2 \ g/mol$,$M(CH_4) = 16 \ g/mol$,$M(SO_2) = 64 \ g/mol$.
Since the rate of effusion is inversely proportional to the square root of molar mass,the order of effusion rates is $r_{H_2} > r_{CH_4} > r_{SO_2}$.
This means that $H_2$ effuses out the fastest,leaving the least amount behind,while $SO_2$ effuses the slowest,leaving the most amount behind.
Since partial pressure is directly proportional to the number of moles remaining,the order of partial pressures will be $P_{SO_2} > P_{CH_4} > P_{H_2}$.

(C) According to Charles's Law,for a fixed amount of gas at constant pressure,the volume is directly proportional to the absolute temperature: $V \propto T$ or $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 22.4 \ L$,$T_1 = 273 \ K$,$T_2 = 373 \ K$.
We need to find $V_2$:
$V_2 = V_1 \times \frac{T_2}{T_1} = 22.4 \ L \times \frac{373 \ K}{273 \ K} \approx 30.6 \ L$.
Thus,the plot showing the point $(30.6 \ L, 373 \ K)$ is correct.

(C) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.
The number of moles effused $n$ in time $t$ is given by $n = r \times t$.
Therefore,the ratio of moles effused is $\frac{n_{H}}{n_{O}} = \frac{r_{H} \times t}{r_{O} \times t} = \frac{\sqrt{M_{O}}}{\sqrt{M_{H}}}$.
Given $M_{H} = 2 \ g/mol$ and $M_{O} = 32 \ g/mol$.
$\frac{n_{H}}{n_{O}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
So,$n_{H} = 4 \times n_{O}$.
We are given that $n_{H} = 1/2 \ mole$ (half of the initial $1 \ mole$).
$1/2 = 4 \times n_{O} \implies n_{O} = 1/8 \ mole$.
Since we started with $1 \ mole$ of oxygen,the fraction of oxygen that escapes is $1/8$.

(D) For a monoatomic gas,the adiabatic index $\gamma = C_P/C_V = 5/3 \approx 1.67$.
Since both gases $A$ and $B$ are monoatomic and are mixed in equal volumes at the same temperature and pressure,the resulting mixture will also behave as a monoatomic gas.
Therefore,the ratio of specific heats $(C_P/C_V)$ for the mixture remains $1.67$.

(B) $1$. The average kinetic energy per mole of an ideal gas is given by the formula $KE_{avg} = \frac{3}{2} RT$.
$2$. Since the temperature $T$ is constant at $400 \ K$ for all gases in the mixture,the average kinetic energy per mole for all gases $(X, Y, Z)$ must be identical $(KE_{avg} = \frac{3}{2} R(400))$.
$3$. Therefore,the statement that the average kinetic energy per mole of gas $X$ is the highest is physically impossible,as it must be equal to the others.
$4$. Option $A$ is possible depending on the molar masses.
$5$. Option $C$ is a property of ideal gases (they cannot be liquefied).
$6$. Option $D$ is correct based on the ideal gas law $P_X = \frac{n_X RT}{V}$.

(B) Given: $n = \frac{1}{24.63} \ mol$,$PV^3 = C$,$V_1 = 1 \ L$,$T_1 = 300 \ K$,$V_2 = 2 \ L$.
Since $PV = nRT$,$P = \frac{nRT}{V}$. Substituting into $PV^3 = C$,we get $\frac{nRT}{V} \cdot V^3 = C$,so $TV^2 = K'$ (constant).
$T_1 V_1^2 = T_2 V_2^2$ $\Rightarrow 300 \times (1)^2 = T_2 \times (2)^2$ $\Rightarrow T_2 = \frac{300}{4} = 75 \ K$.
Work done $w = -\int_{V_1}^{V_2} P \ dV = -\int_{1}^{2} \frac{C}{V^3} \ dV$. Since $P_1 V_1^3 = C$,$P_1 = \frac{nRT_1}{V_1} = \frac{(1/24.63) \times 0.0821 \times 300}{1} = 1 \ atm$. Thus $C = 1 \times 1^3 = 1 \ L^2 \cdot atm$.
$w = -\int_{1}^{2} V^{-3} \ dV = -[\frac{V^{-2}}{-2}]_1^2 = \frac{1}{2} [\frac{1}{4} - 1] = -0.375 \ L \cdot atm$.
Change in internal energy $\Delta U = nC_v \Delta T = \frac{1}{24.63} \times \frac{3}{2} \times 0.0821 \times (75 - 300) = -1.125 \ L \cdot atm$.
Using $\Delta U = q + w$,we get $-1.125 = q - 0.375 \Rightarrow q = -0.75 \ L \cdot atm$.
Since $1 \ L \cdot atm = 101.3 \ J$,$q = -0.75 \times 101.3 = -75.975 \ J$.
The heat lost is $75.9 \ J$.

(C) The Clausius-Clapeyron equation is given by: $\ln \frac{P_{2}}{P_{1}} = \frac{\Delta H_{vap}}{R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right)$
Given: $P_{1} = 1\ atm$,$P_{2} = 2\ atm$,$T_{1} = -173 + 273 = 100\ K$,$\Delta H_{vap} = 200\ cal/mol$,$R = 2\ cal/mol\cdot K$,$\ln 2 = 0.7$.
Substituting the values:
$0.7 = \frac{200}{2} \left( \frac{1}{100} - \frac{1}{T_{2}} \right)$
$0.7 = 100 \left( 0.01 - \frac{1}{T_{2}} \right)$
$0.007 = 0.01 - \frac{1}{T_{2}}$
$\frac{1}{T_{2}} = 0.01 - 0.007 = 0.003$
$T_{2} = \frac{1}{0.003} = 333.33\ K$
Converting to Celsius: $T_{2} = 333.33 - 273 = 60.33\ ^{\circ}C$.

(C) $1$. In the given phase diagram,the line $OB$ represents the solid-liquid equilibrium (melting/freezing curve). At pressure $P_2$,the intersection with the $OB$ line is at temperature $T_4$,which is the melting point. Thus,statement $A$ is correct.
$2$. At pressure $P_2$,the region between the solid-liquid curve $(OB)$ and the liquid-vapour curve $(OA)$ represents the liquid phase. For $T_4 < T < T_5$ at $P_2$,the substance is in the liquid state. Thus,statement $B$ is correct.
$3$. At temperature $T_3$,the pressure $P_1$ lies below the liquid-vapour curve $(OA)$,meaning the substance is in the vapour state,not the liquid state. The vapour pressure of the liquid at $T_3$ would be higher than $P_1$. Thus,statement $C$ is incorrect.
$4$. Sublimation occurs at pressures below the triple point $(O)$. Therefore,the sublimation temperature is always less than the triple point temperature. Thus,statement $D$ is correct.
$5$. The incorrect statement is $C$.

(A) According to Graham's Law of Effusion,the rate of effusion of a gas is directly proportional to its partial pressure and inversely proportional to the square root of its molar mass.
$\frac{r_{He}}{r_{CH_4}} = \frac{p_{He}}{p_{CH_4}} \times \sqrt{\frac{M_{CH_4}}{M_{He}}}$
Given the molar ratio $n_{He} : n_{CH_4} = 4 : 1$,the partial pressures are proportional to the mole fractions,so $\frac{p_{He}}{p_{CH_4}} = \frac{4}{1}$.
The molar masses are $M_{He} = 4 \ g/mol$ and $M_{CH_4} = 16 \ g/mol$.
Substituting these values:
$\frac{r_{He}}{r_{CH_4}} = \frac{4}{1} \times \sqrt{\frac{16}{4}} = 4 \times \sqrt{4} = 4 \times 2 = 8$.
Thus,the ratio of the rates of effusion is $8 : 1$.