Mix Examples- States of Matter Questions in English

(B) Let the weight of both gases be $x \, g$.
The number of moles of $CH_4$ is $n_{CH_4} = \frac{x}{16}$.
The number of moles of $H_2$ is $n_{H_2} = \frac{x}{2}$.
The mole fraction of $H_2$ is given by $X_{H_2} = \frac{n_{H_2}}{n_{H_2} + n_{CH_4}} = \frac{x/2}{x/2 + x/16} = \frac{1/2}{9/16} = \frac{1}{2} \times \frac{16}{9} = \frac{8}{9}$.
According to Dalton's Law of Partial Pressures,the partial pressure of a gas is equal to its mole fraction multiplied by the total pressure,so $P_{H_2} = X_{H_2} \times P_T = \frac{8}{9} P_T$.

(A) For an ideal gas undergoing isothermal expansion,the temperature $T$ is constant.
$(a)$ According to Boyle's Law,$P = nRT / V_m$. Thus,$P$ vs $1/V_m$ is a straight line passing through the origin. This represents isothermal expansion.
$(b)$ $P$ vs $V_m$ should be a rectangular hyperbola $(P \propto 1/V_m)$,not a straight line. This plot is incorrect.
$(c)$ For an ideal gas,$PV_m = RT$. Since $T$ is constant,$PV_m$ is constant regardless of $P$. This plot is correct.
$(d)$ For an ideal gas,internal energy $U$ depends only on temperature $(U = f(T))$. Since the process is isothermal,$U$ must be constant. The plot shows $U$ increasing with $V_m$,which is incorrect.
Therefore,plots $(b)$ and $(d)$ do not represent isothermal expansion.

(B) Dalton's law of partial pressure is applicable only to non-reacting gases.
$NH_3$ and $HCl$ react with each other to form solid ammonium chloride $(NH_4Cl)$:
$NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)$
Since they react,this mixture does not obey Dalton's law.

(C) Step $1$: Calculate the number of moles of each gas.
$n_{CH_4} = \frac{4 \ g}{16 \ g/mol} = 0.25 \ mol$
$n_{He} = \frac{2 \ g}{4 \ g/mol} = 0.5 \ mol$
Step $2$: Calculate the total number of moles.
$n_{total} = n_{CH_4} + n_{He} = 0.25 + 0.5 = 0.75 \ mol = \frac{3}{4} \ mol$
Step $3$: Calculate the mole fraction of $He$.
$X_{He} = \frac{n_{He}}{n_{total}} = \frac{0.5}{0.75} = \frac{2}{3}$
Step $4$: Calculate the partial pressure of $He$ at $NTP$.
At $NTP$,the total pressure $P_{total} = 1 \ atm$.
$P_{He} = X_{He} \times P_{total} = \frac{2}{3} \times 1 \ atm = \frac{2}{3} \ atm$.

(A) clathrate is defined as a cage compound. In these compounds,gaseous atoms are trapped within the cavities of the crystal lattice of another substance.

(C) Helium $II$ behaves as a superfluid. It is a form of liquid helium that has zero viscosity and flows without friction.

(B) Liquid $He$ (specifically $He-II$) behaves as a superfluid at very low temperatures. Due to its zero viscosity and high thermal conductivity,it exhibits the 'creeping' effect,allowing it to climb up the walls of a glass vessel.

(B) According to Graham's Law of Effusion,the rate of effusion $(r)$ of a gas is directly proportional to its partial pressure $(P)$ and inversely proportional to the square root of its molar mass $(M)$: $r \propto \frac{P}{\sqrt{M}}$.
Given the molar ratio of $He : CH_4$ is $x : 1$,the partial pressures are $P_{He} = \frac{x}{x+1} \times 20 \ bar$ and $P_{CH_4} = \frac{1}{x+1} \times 20 \ bar$.
The ratio of the rates of effusion is given by: $\frac{r_{He}}{r_{CH_4}} = \frac{P_{He}}{P_{CH_4}} \times \sqrt{\frac{M_{CH_4}}{M_{He}}}$.
Substituting the values: $\frac{8}{1} = \left( \frac{x}{1} \right) \times \sqrt{\frac{16}{4}}$.
$8 = x \times \sqrt{4} = 2x$.
Therefore,$x = 4$.

(C) The ideal gas equation is $PV = nRT$,where $n = \frac{w}{M_w}$.
Substituting $n$,we get $PV = \frac{w}{M_w} RT$,which rearranges to $P = \frac{w}{V} \frac{RT}{M_w} = \frac{dRT}{M_w}$,where $d$ is the density.
Therefore,$d = \frac{PM_w}{RT}$.
Given $\frac{d_A}{d_B} = 2$ and $\frac{M_{wA}}{M_{wB}} = \frac{1}{2}$.
From the formula,$\frac{P_A}{P_B} = \frac{d_A}{d_B} \times \frac{M_{wB}}{M_{wA}}$.
Substituting the values: $\frac{P_A}{P_B} = 2 \times \frac{1}{1/2} = 2 \times 2 = 4$.
Thus,the ratio of partial pressures $\frac{P_A}{P_B}$ is $4 : 1$.

(B) Initial state: $P_{total, 1} = 830 \, mm \, Hg$,$P_{H_2O, 1} = 30 \, mm \, Hg$.
Pressure of dry gas $P_{gas, 1} = P_{total, 1} - P_{H_2O, 1} = 830 - 30 = 800 \, mm \, Hg$.
Temperature $T_1 = T$.
Final state: Temperature $T_2 = T - 0.01T = 0.99T$.
Using Gay-Lussac's Law for the dry gas: $\frac{P_{gas, 1}}{T_1} = \frac{P_{gas, 2}}{T_2}$.
$\frac{800}{T} = \frac{P_{gas, 2}}{0.99T} \implies P_{gas, 2} = 800 \times 0.99 = 792 \, mm \, Hg$.
New total pressure $P_{total, 2} = P_{gas, 2} + P_{H_2O, 2} = 792 + 25 = 817 \, mm \, Hg$.

(C) The resultant pressure $P$ is given by the formula $P = \frac{P_1 V_1 + P_2 V_2}{V_1 + V_2}$.
Substituting the given values:
$P = \frac{(1000 \, torr \times 1.2 \, L) + (500 \, torr \times 3.8 \, L)}{1.2 \, L + 3.8 \, L}$.
$P = \frac{1200 + 1900}{5.0} \, torr$.
$P = \frac{3100}{5} \, torr = 620 \, torr$.

(D) The average molar mass of the mixture is given by $M = \frac{dRT}{P}$.
Given $d = 0.543\, g\, L^{-1}$,$R = 0.0821\, L\, atm\, K^{-1}\, mol^{-1}$,$T = 300\, K$,and $P = 1\, atm$.
$M = \frac{0.543 \times 0.0821 \times 300}{1} \approx 13.38\, g\, mol^{-1}$.
Let $x$ be the mass fraction of $He$. Then the mass fraction of $O_2$ is $(1-x)$.
The average molar mass $M$ is related to the mole fractions $y_1$ and $y_2$ by $M = y_1 M_1 + y_2 M_2$.
Using the relation between mass fraction $(w)$ and mole fraction $(y)$: $y_1 = \frac{w_1/M_1}{w_1/M_1 + w_2/M_2}$.
$\frac{1}{M} = \frac{w_1}{M_1} + \frac{w_2}{M_2} = \frac{x}{4} + \frac{1-x}{32}$.
$\frac{1}{13.38} = \frac{8x + 1 - x}{32} = \frac{7x + 1}{32}$.
$7x + 1 = \frac{32}{13.38} \approx 2.391$.
$7x = 1.391 \implies x \approx 0.1987$.
Percentage by mass of $He = 0.1987 \times 100 \approx 19.9\% \approx 20.2\%$ (considering rounding differences in constants).

(C) Let the mass of both ethyne $(C_2H_2)$ and benzene $(C_6H_6)$ be $w \ g$.
Molar mass of ethyne $(C_2H_2) = 2 \times 12 + 2 \times 1 = 26 \ g/mol$.
Molar mass of benzene $(C_6H_6) = 6 \times 12 + 6 \times 1 = 78 \ g/mol$.
Number of moles of ethyne $(n_1) = w/26$.
Number of moles of benzene $(n_2) = w/78$.
Partial pressure is proportional to the number of moles $(P \propto n)$ at constant volume and temperature.
$P_{ethyne} / P_{benzene} = n_1 / n_2 = (w/26) / (w/78) = 78 / 26 = 3$.
So,$P_{ethyne} = 3 \times P_{benzene} = 3 \times 600 \ torr = 1800 \ torr$.
Total pressure $(P_{total}) = P_{ethyne} + P_{benzene} = 1800 \ torr + 600 \ torr = 2400 \ torr$.

(C) Given: Volume of $N_2$ $(V)$ = $48 \, L$,Temperature $(T)$ = $27 \, ^oC = 300 \, K$,Mass of water lost = $1.20 \, g$,Molar mass of $H_2O$ = $18 \, g/mol$.
Moles of water vapor $(n_{H_2O})$ = $\frac{1.20}{18} = 0.0667 \, mol$.
Moles of $N_2$ $(n_{N_2})$ = $\frac{PV}{RT} = \frac{1 \times 48}{0.0821 \times 300} = 1.9488 \, mol$.
Total moles $(n_{total})$ = $n_{N_2} + n_{H_2O} = 1.9488 + 0.0667 = 2.0155 \, mol$.
Using Dalton's Law of Partial Pressures: $P_{H_2O} = \frac{n_{H_2O}}{n_{total}} \times P_{total}$.
Assuming $P_{total} = 1 \, atm$: $P_{H_2O} = \frac{0.0667}{2.0155} \times 1 \approx 0.0331 \, atm$.
Rounding to the nearest option,the value is $0.034 \, atm$.

(B) Given: Total pressure $P_{total} = 760\, mm\, Hg$ at $V_1 = 100\, mL$.
Water vapor pressure $P_{H_2O} = 355.5\, mm\, Hg$.
Partial pressure of $O_2$ in the first vessel: $P_{O_2(1)} = P_{total} - P_{H_2O} = 760 - 355.5 = 404.5\, mm\, Hg$.
According to Boyle's Law at constant temperature: $P_{O_2(1)} \times V_1 = P_{O_2(2)} \times V_2$.
$404.5 \times 100 = P_{O_2(2)} \times 50$.
$P_{O_2(2)} = \frac{404.5 \times 100}{50} = 404.5 \times 2 = 809\, mm\, Hg$.

(A) According to Graham's Law of Diffusion,the rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass $(Rate \propto \frac{1}{\sqrt{M}})$.
Since the gases are in a container,if there is any leakage or diffusion,the gas with the lowest molar mass will diffuse the fastest.
The molar masses are: $M(H_2) = 2 \ g/mol$,$M(CH_4) = 16 \ g/mol$,and $M(SO_2) = 64 \ g/mol$.
Since $M(H_2) < M(CH_4) < M(SO_2)$,the rate of diffusion follows the order: $Rate(H_2) > Rate(CH_4) > Rate(SO_2)$.
After $3$ hours,the gas that diffuses the fastest will have the lowest amount remaining in the container.
Therefore,the amount remaining follows the order: $n(SO_2) > n(CH_4) > n(H_2)$.
Since partial pressure is directly proportional to the number of moles $(P \propto n)$,the order of partial pressures will be $P_{SO_2} > P_{CH_4} > P_{H_2}$.

(A) According to Graham's Law of diffusion,the rate of diffusion $r \propto \frac{1}{\sqrt{M}}$.
Given $\frac{r_X}{r_Y} = \frac{1}{2}$.
Therefore,$\frac{r_X}{r_Y} = \sqrt{\frac{M_Y}{M_X}} = \frac{1}{2}$.
Squaring both sides,$\frac{M_Y}{M_X} = \frac{1}{4}$,which implies $M_X = 4M_Y$.
Let $M_Y = M$,then $M_X = 4M$.
Moles of $X$ $(n_X)$ = $\frac{2}{4M} = \frac{0.5}{M}$.
Moles of $Y$ $(n_Y)$ = $\frac{3}{M}$.
Mole fraction of $X$ $(x_X)$ = $\frac{n_X}{n_X + n_Y} = \frac{0.5/M}{0.5/M + 3/M} = \frac{0.5}{3.5} = \frac{5}{35} = \frac{1}{7}$.

(C) According to the Ideal Gas Equation,$PV = nRT = (\frac{w}{M})RT$.
Rearranging for density $(d = \frac{w}{V})$,we get $d = \frac{PM}{RT}$.
Since $P$,$T$,and $R$ are the same for both gases,the density $d$ depends only on the molar mass $M$.
$M(O_2) = 32\, g/mol$ and $M(CH_3OH) = 32\, g/mol$.
Since the molar masses are identical,the density $d$ will be the same for both gases.
Rate of diffusion depends on molar mass $(r \propto \frac{1}{\sqrt{M}})$,which is also the same here.
However,density is the most direct property derived from the given conditions.

(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: $r \propto \frac{1}{\sqrt{M}}$.
Since the time $t$ is the same for both gases,the amount diffused $n$ (in moles) is proportional to the rate of diffusion: $n \propto r$.
Therefore,$\frac{n_{H_2}}{n_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}}$.
The molar mass of $O_2$ is $32 \ g/mol$ and $H_2$ is $2 \ g/mol$.
Let $w$ be the mass of $H_2$ in grams. The number of moles $n = \frac{w}{M}$.
$\frac{w_{H_2} / 2}{4 / 32} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
$\frac{w_{H_2}}{2} = 4 \times \frac{4}{32} = 4 \times 0.125 = 0.5$.
$w_{H_2} = 0.5 \times 2 = 1 \ g$.

(D) According to Graham's Law of Effusion,the rate of effusion $r$ is proportional to $\frac{P}{\sqrt{M}}$.
Since distance $d = r \times t$,and time $t$ is constant $(5 \ s)$,the distance $d$ is proportional to the rate $r$.
Therefore,$d \propto \frac{P}{\sqrt{M}}$.
For hydrogen $(H_2)$: $d_1 = 10.5 \ cm$,$P_1 = 1 \ atm$,$M_1 = 2 \ g/mol$.
For oxygen $(O_2)$: $d_2 = ?$,$P_2 = 2 \ atm$,$M_2 = 32 \ g/mol$.
Using the ratio: $\frac{d_2}{d_1} = \frac{P_2}{P_1} \times \sqrt{\frac{M_1}{M_2}}$.
$\frac{d_2}{10.5} = \frac{2}{1} \times \sqrt{\frac{2}{32}} = 2 \times \sqrt{\frac{1}{16}} = 2 \times \frac{1}{4} = 0.5$.
$d_2 = 10.5 \times 0.5 = 5.25 \ cm$.

(A) According to Graham's Law of Effusion,the rate of effusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
Since the time $(t)$ is the same for both gases,the rate of effusion is directly proportional to the mass $(w)$ effused: $r = \frac{w}{t}$.
Therefore,$\frac{w_1}{w_2} = \sqrt{\frac{M_2}{M_1}}$.
Given: $w_1 (CH_4) = 1 \ g$,$M_1 (CH_4) = 16 \ g/mol$,$M_2 (SO_2) = 64 \ g/mol$.
Substituting the values: $\frac{1}{w_2} = \sqrt{\frac{64}{16}} = \sqrt{4} = 2$.
Thus,$w_2 = \frac{1}{2} = 0.5 \ g$.

(A) According to Graham's Law of diffusion,the rate of diffusion $r \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass.
Let $M_{N_2} = M$. Then the molar mass of the other gas is $M_x = 2M$.
Given: $r_x = 56 \ mL \ s^{-1}$.
Using the formula: $\frac{r_{N_2}}{r_x} = \sqrt{\frac{M_x}{M_{N_2}}}$.
$\frac{r_{N_2}}{56} = \sqrt{\frac{2M}{M}} = \sqrt{2}$.
$r_{N_2} = 56 \times \sqrt{2} = 56 \times 1.414 = 79.184 \approx 79.19 \ mL \ s^{-1}$.

(B) The rate of diffusion $(r)$ is defined as the volume $(V)$ diffused per unit time $(t)$,and according to Graham's Law,it is inversely proportional to the square root of the molar mass $(M)$.
$\frac{r_1}{r_2} = \frac{V_1 \cdot t_2}{V_2 \cdot t_1} = \sqrt{\frac{M_2}{M_1}}$
Here,$V_1 = 20 \, L$,$t_1 = 60 \, s$,and $M_1 (SO_2) = 64 \, g/mol$.
For $O_2$,$t_2 = 30 \, s$ and $M_2 (O_2) = 32 \, g/mol$.
Substituting the values into the equation:
$\frac{20 \times 30}{V_2 \times 60} = \sqrt{\frac{32}{64}}$
$\frac{600}{60 \cdot V_2} = \sqrt{0.5}$
$\frac{10}{V_2} = 0.707$
$V_2 = \frac{10}{0.707} \approx 14.14 \, L$.
Thus,the volume of $O_2$ diffused is $14.14 \, L$. The correct option is $B$.

(B) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ of a gas is inversely proportional to the square root of its molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
Therefore,gases with the same molar mass will have the same rate of diffusion.
Let us calculate the molar masses of the gases in option $B$:
$M(CO_2) = 12 + 2 \times 16 = 44 \ g/mol$.
$M(N_2O) = 2 \times 14 + 16 = 44 \ g/mol$.
$M(C_3H_8) = 3 \times 12 + 8 \times 1 = 44 \ g/mol$.
Since all three gases have the same molar mass $(44 \ g/mol)$,they will have the same rate of diffusion.

(A) Let the mass of both $CH_4$ and $H_2$ be $x \ g$.
Number of moles of $CH_4$ $(n_{CH_4})$ = $\frac{x}{16} \ mol$.
Number of moles of $H_2$ $(n_{H_2})$ = $\frac{x}{2} \ mol$.
The mole fraction of $CH_4$ $(X_{CH_4})$ is given by $\frac{n_{CH_4}}{n_{CH_4} + n_{H_2}}$.
$X_{CH_4} = \frac{x/16}{x/16 + x/2} = \frac{1/16}{1/16 + 8/16} = \frac{1/16}{9/16} = 1/9$.
According to Dalton's Law of partial pressures,the partial pressure of a gas is equal to its mole fraction multiplied by the total pressure $(P_{CH_4} = X_{CH_4} \times P_{Total})$.
Therefore,the fraction of the total pressure exerted by methane is $1/9$.

(B) The radius ratio is calculated as: $\frac{r_{Na^{+}}}{r_{Cl^{-}}} = \frac{95 \ pm}{181 \ pm} = 0.524$.
Since the radius ratio $0.524$ lies in the range $0.414 - 0.732$, the coordination number of $Na^{+}$ is $6$.

(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is given by $r = \frac{V}{t} \propto \frac{1}{\sqrt{M}}$.
For two gases,the ratio is $\frac{V_1 / t_1}{V_2 / t_2} = \sqrt{\frac{M_2}{M_1}}$.
Given: $V_1 = 80 \ mL$,$t_1 = 2 \ min$,$M_1 (O_2) = 32 \ g/mol$,$t_2 = 3 \ min$,$M_2 (SO_2) = 64 \ g/mol$.
Substituting the values: $\frac{80 / 2}{V_2 / 3} = \sqrt{\frac{64}{32}}$.
$40 \times \frac{3}{V_2} = \sqrt{2}$.
$V_2 = \frac{120}{\sqrt{2}} \ mL$.

(C) To find the boiling point at a different pressure,we use the Clausius-Clapeyron equation: $\ln(\frac{P_2}{P_1}) = \frac{\Delta H_{vap}}{R} (\frac{1}{T_1} - \frac{1}{T_2})$.
Given: $P_1 = 760 \ mm$,$T_1 = 373 \ K$,$P_2 = 23 \ mm$,$\Delta H_{vap} = 40.656 \ kJ/mol = 40656 \ J/mol$,$R = 8.314 \ J/mol \cdot K$.
Substituting the values: $\ln(\frac{23}{760}) = \frac{40656}{8.314} (\frac{1}{373} - \frac{1}{T_2})$.
$-3.498 = 4890.06 (0.00268 - \frac{1}{T_2})$.
$-0.000715 = 0.00268 - \frac{1}{T_2}$.
$\frac{1}{T_2} = 0.00268 + 0.000715 = 0.003395$.
$T_2 = \frac{1}{0.003395} \approx 294.5 \ K$.
Rounding to the nearest provided option,the answer is $298 \ K$.

(D) The specific heat capacity at constant pressure $(C_p)$ is given by $C_p = \frac{\gamma R}{M(\gamma - 1)}$.
For $He$ (monatomic),$\gamma = 1.66$ and $M = 4 \ g/mol$.
For $H_2$ (diatomic),$\gamma = 1.4$ and $M = 2 \ g/mol$.
Calculating $C_p$ for $He$: $C_p = \frac{1.66 \times R}{4 \times 0.66} \approx 0.63R$.
Calculating $C_p$ for $H_2$: $C_p = \frac{1.4 \times R}{2 \times 0.4} = 1.75R$.
Calculating $C_v$ for $H_2$: $C_v = \frac{R}{M(\gamma - 1)} = \frac{R}{2 \times 0.4} = 1.25R$.
Comparing these values,none of the given statements are correct.

(A) Given,$P_1 = 15 \, atm$,$P_2 = 60 \, atm$.
$V_1 = 76 \, cm^3$,$V_2 = 20.5 \, cm^3$.
If the gas is an ideal gas,then according to Boyle's law,it must follow the equation $P_1V_1 = P_2V_2$.
$P_1 \times V_1 = 15 \times 76 = 1140$.
$P_2 \times V_2 = 60 \times 20.5 = 1230$.
Since $P_1V_1 \neq P_2V_2$,the gas behaves non-ideally.
Additionally,processes like dimerization or adsorption on vessel walls would also lead to a deviation from the expected ideal volume change,making all three statements possible explanations for the observed behavior.

(D) The melting point of ice decreases with an increase in pressure. Since atmospheric pressure decreases at high altitudes,the melting point of ice actually increases,meaning it does not melt faster.
Therefore,the assertion is false.
Additionally,atmospheric pressure is lower,not higher,at high altitudes,making the reason also false.
Thus,both the assertion and the reason are incorrect.

(D) The molar mass of water vapor $(H_2O)$ is $18 \ g/mol$,while the average molar mass of dry air (mostly $N_2$ and $O_2$) is approximately $29 \ g/mol$.
Since water vapor replaces heavier air molecules,wet air is actually lighter than dry air,making the Assertion incorrect.
The density of dry air is much lower than the density of liquid water,making the Reason also incorrect.

(A) The use of a pressure cooker reduces cooking time because the increase in pressure inside the cooker increases the boiling point $(b.p.)$ of water.
Since the water boils at a higher temperature,the food cooks faster.

(B) The Assertion is correct because when we touch ice,heat flows from our body (at a higher temperature) to the ice (at a lower temperature),causing a sensation of coldness.
The Reason is also correct because ice is indeed the solid state of $H_2O$.
However,the Reason does not explain why we feel cold; the sensation of cold is due to the heat transfer process,not merely the state of matter.

(B) According to Boyle's Law,$p_{1}V_{1} = p_{2}V_{2}$ at constant temperature.
For $H_{2}$:
$p_{1} = 0.8 \, bar, V_{1} = 0.5 \, L, V_{2} = 1 \, L$
$p_{H_{2}} = \frac{0.8 \times 0.5}{1} = 0.4 \, bar$
For $O_{2}$:
$p_{1} = 0.7 \, bar, V_{1} = 2.0 \, L, V_{2} = 1 \, L$
$p_{O_{2}} = \frac{0.7 \times 2.0}{1} = 1.4 \, bar$
According to Dalton's Law of Partial Pressures,the total pressure $p_{total} = p_{H_{2}} + p_{O_{2}}$.
$p_{total} = 0.4 \, bar + 1.4 \, bar = 1.8 \, bar$.

(D) London dispersion forces are inversely proportional to the sixth power of the distance $(r)$ between the particles: $F \propto \frac{1}{r^{6}}$.
Let the initial force be $x_{1} = \frac{k}{r^{6}}$.
When the distance becomes half,the new distance $r' = \frac{r}{2}$.
The new force $x_{2} = \frac{k}{(r/2)^{6}} = \frac{k \cdot 2^{6}}{r^{6}}$.
Therefore,$x_{2} = 2^{6} \cdot x_{1} = 64 \cdot x_{1}$.
The London forces become $64$ times the original value.

(N/A) The appearance of water vapour during a bath in winter is a physical process known as $condensation$.
When hot water is used for a bath,it releases water vapour into the air.
In winter,the ambient air temperature is significantly lower than the temperature of the hot water vapour.
When this warm water vapour comes into contact with the cold air,it loses heat energy.
As a result,the water vapour molecules slow down and come closer together,changing from the gaseous state to the liquid state,forming tiny water droplets that appear as mist or vapour.

(A) The magnitude of surface tension is directly proportional to the strength of intermolecular attractive forces.
$1$. Hexane $(CH_3(CH_2)_4CH_3)$ is a non-polar molecule,exhibiting only weak London dispersion forces.
$2$. Alcohol $(C_2H_5OH)$ is a polar molecule capable of $H$-bonding,which is stronger than dispersion forces.
$3$. Water $(H_2O)$ is a highly polar molecule with extensive $H$-bonding,making its intermolecular forces the strongest among the three.
Therefore,the increasing order of surface tension is $Hexane < Alcohol < Water$.

(N/A) $(i)$ According to Boyle's law,the pressure of a gas is inversely proportional to its volume if the temperature is kept constant. Thus,the volume of the gas decreases if the pressure on the gas increases while keeping the temperature constant. For example,at $200 \ K$,when pressure is increased from $p_{1}$ to $p_{2}$,the volume of the gas decreases,i.e.,$V_{2} < V_{1}$.
$(ii)$ According to Charles's law,the volume of a gas is directly proportional to its temperature if the pressure is kept constant. Thus,on increasing the temperature,the volume of a gas will increase if the pressure is kept constant. At constant $p$,when we increase the temperature from $200 \ K$ to $400 \ K$,the volume of the gas increases,i.e.,$V_{4} > V_{3}$.

(N/A) Sharp glass edges are heated to make them smooth. On heating,glass melts and the surface of the liquid tends to take a rounded shape at the edges to minimize its surface area. This process is known as fire polishing of glass. This phenomenon is due to the property of surface tension.

(B) Chemical properties of a substance remain the same regardless of its physical state.
The rate of a chemical reaction often depends on the physical state of the reactants.
In many experimental calculations,it is essential to know whether a substance is in a solid,liquid,or gaseous state.
Therefore,it is necessary for chemists to understand the physical laws that govern the behavior of matter in different states.

(C) Intermolecular forces tend to keep the molecules together,whereas thermal energy tends to keep them apart. The three states of matter are the result of the balance between intermolecular forces and the thermal energy of the molecules.

(A) Viscosity is a measure of a fluid's resistance to flow. Generally,for liquids,as the density increases,the intermolecular forces of attraction also increase,which leads to higher viscosity. Therefore,the relationship is expressed as: $\text{Viscosity} \propto \text{Density}$.

(B) When oil is added to water,it spreads and forms a layer on the surface of the water.
This happens because the surface tension of water is higher than that of oil.

(B) When glass is heated,it softens and becomes viscous. Due to surface tension,the sharp edges of the glass tend to minimize their surface area by becoming rounded. This process of heating glass to obtain smooth,rounded edges is known as fire polishing of glass.

(A) Surface tension is a property of liquids that decreases when surface-active agents (surfactants) like soap are added.
Soap molecules reduce the cohesive forces between water molecules.
Therefore,pure water has a higher surface tension compared to soapy water.

(N/A) $(1)$ The graph of pressure versus temperature at constant volume is called an isochore.
$(2)$ $A$ gas that obeys Boyle's law,Charles's law,and Avogadro's law under all conditions of temperature and pressure is called an ideal gas.
$(3)$ The most probable speed $(U_{mp})$ is $\sqrt{2RT/M}$ and the average speed $(U_{av})$ is $\sqrt{8RT/\pi M}$. The ratio $U_{mp} : U_{av} = \sqrt{2} : \sqrt{8/\pi} = 1 : 1.128$.
$(4)$ The root mean square speed $(U_{rms})$ is $\sqrt{3RT/M}$ and the most probable speed $(U_{mp})$ is $\sqrt{2RT/M}$. The ratio $U_{rms} : U_{mp} = \sqrt{3} : \sqrt{2} = 1.224 : 1$.

(N/A) $(1)$ Negative
$(2)$ Boyle temperature or Boyle point
$(3)$ Critical temperature
$(4)$ Poise

(A) Initial moles in flask $I$: $n_I = \frac{2.8 \, g}{28 \, g \, mol^{-1}} = 0.1 \, mol$. Initial moles in flask $II$: $n_{II} = \frac{0.2 \, g}{28 \, g \, mol^{-1}} = \frac{1}{140} \, mol \approx 0.00714 \, mol$.
Total moles $n_{total} = 0.1 + \frac{1}{140} = \frac{15}{140} = \frac{3}{28} \, mol$.
Since the flasks are connected and the system reaches thermal equilibrium,the final temperature $T$ is determined by the heat balance: $n_I C_v (T - 300) + n_{II} C_v (T - 60) = 0$.
$0.1(T - 300) + \frac{1}{140}(T - 60) = 0$.
$14(T - 300) + (T - 60) = 0$ $\Rightarrow 15T = 4260$ $\Rightarrow T = 284 \, K$.
Using $PV = nRT$ for the total system: $P(V_I + V_{II}) = n_{total} RT$.
$P(1 \, L + 2 \, L) = (\frac{3}{28} \, mol) \times (0.0831 \, L \, bar \, K^{-1} \, mol^{-1}) \times 284 \, K$.
$3P = \frac{3 \times 0.0831 \times 284}{28} \approx 2.528 \, bar \cdot L$.
$P = 0.8427 \, bar = 84.27 \times 10^{-2} \, bar$.
Rounding to the nearest integer,$x = 84$.

(C) Initial mass of gas $= 29.0 - 14.8 = 14.2 \ kg$
Mass of gas remaining $= 23.0 - 14.8 = 8.2 \ kg$
Using the ideal gas law $PV = nRT$,where $n = \frac{m}{M}$:
$P_1 V = \left( \frac{m_1}{M} \right) RT$
$P_2 V = \left( \frac{m_2}{M} \right) RT$
Since $V$,$R$,and $T$ are constant,$\frac{P_1}{P_2} = \frac{m_1}{m_2}$
$\frac{3.47}{P_2} = \frac{14.2}{8.2}$
$P_2 = \frac{3.47 \times 8.2}{14.2} \approx 2.003 \ atm$
The nearest integer is $2$.