Molecular speeds Questions in English

(A) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
Calculating the molar masses:
$M(NH_3) = 14 + 3(1) = 17 \ g/mol$
$M(N_2) = 2(14) = 28 \ g/mol$
$M(CO_2) = 12 + 2(16) = 44 \ g/mol$
$M(O_2) = 2(16) = 32 \ g/mol$
Since $NH_3$ has the lowest molar mass,it will have the highest rate of diffusion.
Therefore,the correct option is $(A)$.

(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molecular weight $M$: $r \propto \frac{1}{\sqrt{M}}$.
Given: $M_1 = 64$,$r_2 = 2r_1$.
Using the formula $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$,we get:
$\frac{r_1}{2r_1} = \sqrt{\frac{M_2}{64}}$
$\frac{1}{2} = \sqrt{\frac{M_2}{64}}$
Squaring both sides: $\frac{1}{4} = \frac{M_2}{64}$
$M_2 = \frac{64}{4} = 16$.
Thus,the molecular weight is $16$.

(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: $r \propto \frac{1}{\sqrt{M}}$.
Two gases will diffuse at the same rate if they have the same molar mass.
For option $(D)$: $M_{NO} = 14 + 16 = 30 \ g/mol$ and $M_{C_2H_6} = (2 \times 12) + (6 \times 1) = 30 \ g/mol$.
Since both $NO$ and $C_2H_6$ have the same molecular weight of $30 \ g/mol$,they will diffuse at the same rate.

(C) According to Graham's Law of effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$ and directly proportional to the square root of the temperature $T$ for a given pressure.
$\frac{r_{N_2}}{r_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{N_2}} \times \frac{T_{N_2}}{T_{SO_2}}}$
Given: $r_{N_2} = 1.625 \times r_{SO_2}$,$T_{SO_2} = 50 + 273 = 323\,K$,$M_{N_2} = 28\,g/mol$,$M_{SO_2} = 64\,g/mol$.
Substituting the values:
$1.625 = \sqrt{\frac{64}{28} \times \frac{T_{N_2}}{323}}$
Squaring both sides:
$(1.625)^2 = \frac{64}{28} \times \frac{T_{N_2}}{323}$
$2.640625 = 2.2857 \times \frac{T_{N_2}}{323}$
$T_{N_2} = \frac{2.640625 \times 323}{2.2857} \approx 373\,K$.

(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ of the gas: $r \propto \frac{1}{\sqrt{M}}$.
Since the molar mass of $H_2$ is $2 \ g/mol$ and the molar mass of $O_2$ is $32 \ g/mol$,$H_2$ has a lower molar mass.
Therefore,$H_2$ diffuses faster than $O_2$.

(C) The kinetic gas equation is derived as $PV = \frac{1}{3} m N u_{rms}^2$.
In this derivation,the pressure is related to the mean square velocity of the gas molecules.
By definition,the root mean square velocity $(u_{rms})$ is the square root of the mean of the squares of the velocities of individual molecules,i.e.,$u_{rms} = \sqrt{\frac{u_1^2 + u_2^2 + ... + u_n^2}{n}}$.
Therefore,it is the square root of the average square velocity of the molecules.

(B) As the temperature of a gas increases,the Maxwell-Boltzmann distribution curve shifts to the right and flattens.
$1$. The most probable speed $(v_{mp} = \sqrt{2RT/M})$ increases.
$2$. The peak of the curve shifts to the right and its height decreases,meaning the fraction of molecules possessing the most probable speed decreases.
$3$. The distribution curve becomes broader.
$4$. The total area under the curve represents the total number of molecules,which remains constant regardless of temperature.
Therefore,the statement that the fraction of molecules with the most probable speed increases is $NOT$ true.

(A) The root mean square velocity $(V_{rms})$ is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$.
The average velocity $(V_{av})$ is given by $V_{av} = \sqrt{\frac{8RT}{\pi M}}$.
The ratio $\frac{V_{rms}}{V_{av}}$ is calculated as $\sqrt{\frac{3RT}{M} \times \frac{\pi M}{8RT}} = \sqrt{\frac{3\pi}{8}}$.
Substituting the value of $\pi \approx 3.14$,we get $\sqrt{\frac{3 \times 3.14}{8}} = \sqrt{1.1775} \approx 1.086$.
Thus,the ratio is $1.086 : 1$.

(D) The formulas for the three types of molecular velocities are:
Most probable velocity $(V_{mp})$ = $\sqrt{\frac{2RT}{M}}$
Mean velocity $(V_{avg})$ = $\sqrt{\frac{8RT}{\pi M}}$
Root mean square velocity $(V_{rms})$ = $\sqrt{\frac{3RT}{M}}$
Taking the ratio $V_{mp} : V_{avg} : V_{rms}$:
$= \sqrt{\frac{2RT}{M}} : \sqrt{\frac{8RT}{\pi M}} : \sqrt{\frac{3RT}{M}}$
Dividing by $\sqrt{\frac{RT}{M}}$,we get:
$= \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3}$
Thus,the correct option is $(D)$.

(D) . The root mean square velocity $(V_{rms})$ is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$.
At the same temperature $(T)$,$V_{rms} \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass of the gas.
Since $H_2$ has the lowest molar mass $(M = 2 \ g/mol)$ compared to $SO_2$ $(64 \ g/mol)$,$CO_2$ $(44 \ g/mol)$,and $O_2$ $(32 \ g/mol)$,it will have the maximum $V_{rms}$.

(D) The formula for $RMS$ velocity is $U = \sqrt{\frac{3RT}{M}}$.
Given that the $RMS$ velocity of $SO_2$ is half of that of $He$ at $300 \ K$,we have $\frac{U_{SO_2}}{U_{He}} = \frac{1}{2}$.
Substituting the formula: $\frac{1}{2} = \sqrt{\frac{M_{He}}{M_{SO_2}} \times \frac{T_{SO_2}}{T_{He}}}$.
Substituting the molar masses ($M_{He} = 4 \ g/mol$,$M_{SO_2} = 64 \ g/mol$) and temperature $(T_{He} = 300 \ K)$: $\frac{1}{2} = \sqrt{\frac{4}{64} \times \frac{T_{SO_2}}{300}}$.
Squaring both sides: $\frac{1}{4} = \frac{4}{64} \times \frac{T_{SO_2}}{300}$.
$\frac{1}{4} = \frac{1}{16} \times \frac{T_{SO_2}}{300}$.
$T_{SO_2} = \frac{16 \times 300}{4} = 4 \times 300 = 1200 \ K$.

(C) The $rms$ velocity is given by the formula $U_{rms} = \sqrt{\frac{3RT}{M}}$.
Since the temperature $T$ is the same for both gases,the ratio of $rms$ velocities is inversely proportional to the square root of their molar masses:
$\frac{U_{O_3}}{U_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{O_3}}}$
Given molar mass of oxygen $(O_2)$ is $32 \ g/mol$ and ozone $(O_3)$ is $48 \ g/mol$.
$\frac{U_{O_3}}{U_{O_2}} = \sqrt{\frac{32}{48}} = \sqrt{\frac{2}{3}}$.

(B) The formula for $RMS$ velocity is $U_{rms} = \sqrt{\frac{3RT}{M}}$.
For the velocities to be equal: $\sqrt{\frac{3RT_{SO_2}}{M_{SO_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$.
This simplifies to $\frac{T_{SO_2}}{M_{SO_2}} = \frac{T_{O_2}}{M_{O_2}}$.
Given $M_{SO_2} = 64 \ g/mol$,$M_{O_2} = 32 \ g/mol$,and $T_{O_2} = 303 \ K$.
Substituting the values: $\frac{T_{SO_2}}{64} = \frac{303}{32}$.
$T_{SO_2} = \frac{303 \times 64}{32} = 303 \times 2 = 606 \ K$.

(D) The root mean square velocity $(v_{rms})$ is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the gas constant,$T$ is the temperature in Kelvin,and $M$ is the molar mass of the gas.
Since $v_{rms} \propto \frac{1}{\sqrt{M}}$,the gas with the highest molar mass will have the lowest root mean square velocity.
The molar masses are: $SO_2 = 64 \ g/mol$,$N_2 = 28 \ g/mol$,$O_2 = 32 \ g/mol$,and $Cl_2 = 71 \ g/mol$.
Since $Cl_2$ has the highest molar mass $(71 \ g/mol)$,it possesses the lowest root mean square velocity.
Therefore,the correct option is $(D)$.

(D) The root mean square $(rms)$ velocity of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $PV = nRT = \frac{m}{M}RT$,we have $RT = \frac{PV M}{m}$. Substituting this into the $rms$ formula gives $v_{rms} = \sqrt{\frac{3PV}{m}} = \sqrt{\frac{3PV}{M}}$ (where $M$ is molar mass).
Also,since density $d = \frac{m}{V}$,we have $P = \frac{dRT}{M}$,which implies $\frac{P}{d} = \frac{RT}{M}$. Substituting this into the $rms$ formula gives $v_{rms} = \sqrt{\frac{3P}{d}}$.
Therefore,all the given expressions are correct.

(C) The formula for root mean square velocity $(V_{rms})$ is given by:
$V_{rms} = \sqrt{\frac{3RT}{M}}$
where $R$ is the universal gas constant,$T$ is the temperature,and $M$ is the molar mass (or $m$ as the molecular mass).
From the expression,it is clear that $V_{rms} \propto \frac{1}{\sqrt{m}}$,which can be written as $V_{rms} \propto m^{-1/2}$.

(D) The root mean square velocity $(v_{rms})$ of a gas is given by the formula: $v = \sqrt{\frac{3RT}{M}}$,where $R$ is the gas constant,$T$ is the temperature,and $M$ is the molar mass.
Since the temperature $T$ is the same for both gases,$v \propto \frac{1}{\sqrt{M}}$.
Therefore,$\frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}}$.
Squaring both sides,we get $\frac{v_1^2}{v_2^2} = \frac{M_2}{M_1}$.
Rearranging the terms,we obtain $M_1 v_1^2 = M_2 v_2^2$. Since mass $m$ is proportional to molar mass $M$,the expression holds as $m_1 v_1^2 = m_2 v_2^2$.

(D) The root mean square velocity $(U_{rms})$ is given by the formula $U_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $U_{rms} \propto \sqrt{T}$,the ratio of velocities is $\frac{U_2}{U_1} = \sqrt{\frac{T_2}{T_1}}$.
Given $T_1 = 27 + 273 = 300 \, K$ and $T_2 = 927 + 273 = 1200 \, K$.
Substituting the values: $\frac{U_2}{U_1} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2$.
Therefore,$U_2 = 2 \times U_1$,which means the velocity is doubled.

(C) The root mean square velocity $(U_{rms})$ is given by the formula $U_{rms} = \sqrt{\frac{3RT}{M}}$.
For $H_2$ at $50 \ K$: $U_{H_2} = \sqrt{\frac{3R \times 50}{2}}$.
For $O_2$ at $800 \ K$: $U_{O_2} = \sqrt{\frac{3R \times 800}{32}}$.
The ratio is $\frac{U_{H_2}}{U_{O_2}} = \sqrt{\frac{50}{2} \times \frac{32}{800}} = \sqrt{25 \times 0.04} = \sqrt{1} = 1$.

(D) The root mean square velocity $(U)$ is given by the formula $U = \sqrt{\frac{3RT}{M}}$.
Since $PV = nRT = \frac{m}{M}RT$,we have $P = \frac{m}{V} \cdot \frac{RT}{M} = d \cdot \frac{RT}{M}$,where $d$ is the density.
Thus,$\frac{RT}{M} = \frac{P}{d}$.
Substituting this into the velocity formula,we get $U = \sqrt{\frac{3P}{d}}$.
At constant pressure $(P)$,$U \propto \frac{1}{\sqrt{d}}$.

(D) The average speed $(v_{avg})$ of a gas molecule is given by the formula $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$,where $R$ is the universal gas constant,$T$ is the temperature,and $M$ is the molar mass of the gas.
Since the temperature $T$ is the same for both $SO_2$ and $O_2$,the average speed is inversely proportional to the square root of the molar mass $(v_{avg} \propto \frac{1}{\sqrt{M}})$.
The molar mass of $SO_2$ is $64 \ g/mol$ and the molar mass of $O_2$ is $32 \ g/mol$.
Since $M(SO_2) > M(O_2)$,the average speed of $SO_2$ will be smaller than that of $O_2$.

(B) The formula for $rms$ speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Initially,for $N_2$ molecules: $v_1 = v = \sqrt{\frac{3RT}{M_{N_2}}}$,where $M_{N_2} = 28 \ g/mol$.
Finally,for $N$ atoms: $v_2 = \sqrt{\frac{3R(2T)}{M_N}}$,where $M_N = 14 \ g/mol$.
Taking the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{3R(2T)}{14} \times \frac{28}{3RT}} = \sqrt{2 \times 2} = \sqrt{4} = 2$.
Therefore,$v_2 = 2v$.

(B) The root mean square speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
The average speed is given by $\bar{v} = \sqrt{\frac{8RT}{\pi M}}$.
The most probable speed is given by $v_p = \sqrt{\frac{2RT}{M}}$.
Comparing the coefficients: $\sqrt{3} \approx 1.732$,$\sqrt{\frac{8}{3.14}} \approx \sqrt{2.546} \approx 1.596$,and $\sqrt{2} \approx 1.414$.
Therefore,the correct order is $v_{rms} > \bar{v} > v_p$.

(C) The most probable velocity $(V_{mp})$ is given by $\sqrt{2RT / M}$.
The average velocity $(V_{av})$ is given by $\sqrt{8RT / \pi M}$.
The ratio is $\frac{V_{mp}}{V_{av}} = \frac{\sqrt{2RT / M}}{\sqrt{8RT / \pi M}} = \sqrt{\frac{2RT}{M} \times \frac{\pi M}{8RT}} = \sqrt{\frac{2\pi}{8}} = \sqrt{\frac{\pi}{4}} = \frac{\sqrt{\pi}}{2}$.

(A) The formula for $r.m.s.$ velocity is $V_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $R$ and $M$ are constant,$V_{rms} \propto \sqrt{T}$.
Given $V_1 = v$ at $T_1 = 300 \ K$ and $V_2 = 2v$ at $T_2 = ?$.
Using the ratio: $\frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}}$.
Substituting the values: $\frac{v}{2v} = \sqrt{\frac{300}{T_2}}$.
Squaring both sides: $(\frac{1}{2})^2 = \frac{300}{T_2} \implies \frac{1}{4} = \frac{300}{T_2}$.
Therefore,$T_2 = 300 \times 4 = 1200 \ K$.

(C) The root mean square $(rms)$ velocity of a gas is given by the formula:
$V_{rms} = \sqrt{\frac{3RT}{M}}$
where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
Thus,$V_{rms}$ depends on both the temperature $(T)$ and the molar mass $(M)$ of the gas.

(C) The root mean square velocity of one mole of a monoatomic gas is given by $V_{rms} = \sqrt{\frac{3RT}{M}} \dots (1)$.
The average kinetic energy $(E)$ of one mole of a monoatomic gas is given by $E = \frac{3}{2}RT$.
From this,we can express $RT$ as $RT = \frac{2E}{3} \dots (2)$.
Substituting equation $(2)$ into equation $(1)$:
$V_{rms} = \sqrt{\frac{3}{M} \times \frac{2E}{3}}$
$V_{rms} = \sqrt{\frac{2E}{M}}$.

(A) The formula for average velocity $(v_{avg})$ is $\sqrt{\frac{8RT}{\pi M}}$.
The formula for most probable velocity $(v_{mp})$ is $\sqrt{\frac{2RT}{M}}$.
The ratio is $\frac{v_{avg}}{v_{mp}} = \frac{\sqrt{\frac{8RT}{\pi M}}}{\sqrt{\frac{2RT}{M}}} = \sqrt{\frac{8RT}{\pi M} \times \frac{M}{2RT}} = \sqrt{\frac{4}{\pi}} = \sqrt{1.273} \approx 1.128$.
Therefore,the ratio is $1.128 : 1$.

(D) The formula for root mean square velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given $v_{rms} = 30 R^{1/2}$ and $T = 27 ^\circ C = 300 \, K$.
Squaring both sides,we get $v_{rms}^2 = \frac{3RT}{M} = (30 R^{1/2})^2 = 900 R$.
Equating the two expressions: $900 R = \frac{3RT}{M}$.
Substituting $T = 300 \, K$: $900 R = \frac{3R \times 300}{M}$.
$900 R = \frac{900 R}{M}$,which implies $M = 1 \, g/mol$.
Converting to kilograms: $M = 1 \, g = 0.001 \, kg$.

(A) The expressions for the velocities are as follows:
Most probable velocity $(\alpha) = \sqrt{\frac{2RT}{M}}$
Average velocity $(v) = \sqrt{\frac{8RT}{\pi M}}$
Root mean square velocity $(u) = \sqrt{\frac{3RT}{M}}$
Taking the ratio $\alpha : v : u$:
$\sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3}$
Dividing by $\sqrt{2} \approx 1.414$:
$1 : \sqrt{\frac{4}{\pi}} : \sqrt{1.5}$
$1 : 1.128 : 1.224$

(A) The root mean square speed $(u_{rms})$ is defined as the square root of the mean of the squares of the speeds of all molecules.
Mathematically,it is given by the formula:
$u_{rms} = \sqrt{\frac{\sum n_i C_i^2}{\sum n_i}} = \left[ \frac{n_1 C_1^2 + n_2 C_2^2 + n_3 C_3^2 + .....}{n_1 + n_2 + n_3 + .....} \right]^{1/2}$.

(B) The formula for average speed is $V_{av} = \sqrt{\frac{8RT}{\pi M}}$.
For the average speeds of $CH_4$ and $O_2$ to be equal: $\sqrt{\frac{8RT_{CH_4}}{\pi M_{CH_4}}} = \sqrt{\frac{8RT_{O_2}}{\pi M_{O_2}}}$.
This simplifies to $\frac{T_{CH_4}}{M_{CH_4}} = \frac{T_{O_2}}{M_{O_2}}$.
Given $T_{O_2} = 300 \ K$,$M_{O_2} = 32 \ g/mol$,and $M_{CH_4} = 16 \ g/mol$.
Substituting the values: $\frac{T_{CH_4}}{16} = \frac{300}{32}$.
$T_{CH_4} = \frac{300 \times 16}{32} = 150 \ K$.

(D) The average speed of gas molecules is given by the formula $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.
Since the gas is the same,$M$ is constant. Thus,$v_{avg} \propto \sqrt{T}$.
Given $v_X = 2v_Y$,we have $\sqrt{T_X} = 2\sqrt{T_Y}$,which implies $T_X = 4T_Y$.
Using the ideal gas equation $PV = nRT$,we have $P = \frac{nRT}{V}$.
For flask $X$: $P_X = \frac{1 \cdot R \cdot T_X}{V_X} = \frac{RT_X}{1}$.
For flask $Y$: $P_Y = \frac{1 \cdot R \cdot T_Y}{V_Y} = \frac{RT_Y}{2}$.
Taking the ratio: $\frac{P_X}{P_Y} = \frac{RT_X}{1} \cdot \frac{2}{RT_Y} = 2 \cdot \frac{T_X}{T_Y}$.
Substituting $T_X = 4T_Y$: $\frac{P_X}{P_Y} = 2 \cdot 4 = 8$.
Therefore,$P_X = 8P_Y$.

(A) The most probable velocity $(v_{mp})$ of a gas molecule is defined as the velocity possessed by the maximum fraction of molecules at a given temperature.
It is given by the formula: $v_{mp} = \sqrt{\frac{2RT}{M_w}}$
Where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M_w$ is the molar mass of the gas.

(C) The average velocity $(v_{avg})$ of gas molecules is given by the formula:
$v_{avg} = \sqrt{\frac{8RT}{\pi M_w}}$
where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M_w$ is the molar mass of the gas.
Thus,the correct expression is $(\frac{8RT}{\pi M_w})^{1/2}$.

(A) The three types of molecular speeds are defined as follows:
$1$. Root mean square speed: $U_{rms} = \sqrt{\frac{3RT}{M}}$
$2$. Average speed: $\bar{v} = \sqrt{\frac{8RT}{\pi M}}$
$3$. Most probable speed: $\alpha = \sqrt{\frac{2RT}{M}}$
Comparing the coefficients:
$U_{rms} = \sqrt{3} \approx 1.732$
$\bar{v} = \sqrt{\frac{8}{3.14}} \approx \sqrt{2.546} \approx 1.596$
$\alpha = \sqrt{2} \approx 1.414$
Therefore,the correct order is $U_{rms} > \bar{v} > \alpha$.

(D) The root mean square $(R.M.S.)$ velocity of a gas is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
At a constant temperature $(T)$,the $v_{rms}$ is inversely proportional to the square root of the molar mass $(M)$: $v_{rms} \propto \frac{1}{\sqrt{M}}$.
Therefore,the gas with the lowest molar mass will have the maximum $R.M.S.$ velocity.
Molar masses are:
$O_2 = 32 \, g/mol$
$CO_2 = 44 \, g/mol$
$SO_2 = 64 \, g/mol$
$CO = 28 \, g/mol$
Since $CO$ has the lowest molar mass $(28 \, g/mol)$,it will have the maximum $R.M.S.$ velocity.

(C) The root mean square velocity $(u_{rms})$ is given by the formula: $u_{rms} = \sqrt{\frac{3RT}{M_w}}$.
From this expression,it is clear that $u_{rms} \propto \sqrt{T}$.

(C) The average kinetic energy $(E)$ for one mole of an ideal gas is given by $E = \frac{3}{2}RT$.
We know that $U_{rms} = \sqrt{\frac{3RT}{M}}$.
Squaring both sides,we get $U_{rms}^2 = \frac{3RT}{M}$,which implies $RT = \frac{U_{rms}^2 \times M}{3}$.
Substituting the value of $RT$ into the kinetic energy equation: $E = \frac{3}{2} \times \left( \frac{U_{rms}^2 \times M}{3} \right) = \frac{1}{2} M U_{rms}^2$.
Rearranging for $U_{rms}$: $U_{rms}^2 = \frac{2E}{M}$,so $U_{rms} = \sqrt{\frac{2E}{M}}$.

(D) The root mean square speed $(u_{rms})$ of a gas is given by the formula: $u_{rms} = \sqrt{\frac{3RT}{M}}$.
Here,$R$ is the gas constant,$T$ is the temperature in Kelvin,and $M$ is the molar mass of the gas.
Since $3$,$R$,and $T$ are constant for all gases at the same temperature,$u_{rms} \propto \frac{1}{\sqrt{M}}$.
This means the gas with the highest molar mass will have the lowest root mean square speed.
The molar masses are:
$M(SO_2) = 32 + 2 \times 16 = 64 \, g/mol$
$M(N_2) = 2 \times 14 = 28 \, g/mol$
$M(O_2) = 2 \times 16 = 32 \, g/mol$
$M(Cl_2) = 2 \times 35.5 = 71 \, g/mol$
Comparing the molar masses,$Cl_2$ has the highest molar mass $(71 \, g/mol)$.
Therefore,$Cl_2$ will have the lowest root mean square speed.

(C) The formula for the average speed $(v_{avg})$ of gas molecules is given by $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.
Since the average speeds are equal,we have $\sqrt{\frac{8RT_1}{\pi M_1}} = \sqrt{\frac{8RT_2}{\pi M_2}}$.
This simplifies to $\frac{T_1}{M_1} = \frac{T_2}{M_2}$.
Given $T_1 = 300 \ K$,$M_1 (CH_4) = 16 \ g/mol$,and $M_2 (O_2) = 32 \ g/mol$.
Substituting the values: $\frac{300}{16} = \frac{T_2}{32}$.
$T_2 = \frac{300 \times 32}{16} = 300 \times 2 = 600 \ K$.

(B) The root mean square velocity $(v_{rms})$ of an ideal gas is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
From this relation,we can see that $v_{rms} \propto \sqrt{T}$.
Given: $T_1 = 120 \ K$,$v_1 = v$ and $T_2 = 480 \ K$.
Therefore,$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{v_2}{v} = \sqrt{\frac{480}{120}} = \sqrt{4} = 2$.
Thus,$v_2 = 2v$.

(D) The root mean square velocity $(u_{rms})$ of a gas is given by the formula: $u_{rms} = \sqrt{\frac{3RT}{M_w}}$.
From the ideal gas equation,$PV = nRT = \frac{m}{M_w}RT$,we have $\frac{RT}{M_w} = \frac{PV}{m} = \frac{P}{\rho}$,where $\rho$ (or $d$) is the density of the gas $(d = \frac{m}{V})$.
Substituting this into the $u_{rms}$ formula:
$1$. $u_{rms} = \sqrt{\frac{3RT}{M_w}}$ (Option $A$ is correct).
$2$. $u_{rms} = \sqrt{\frac{3P}{d}}$ (Option $B$ is correct).
$3$. $u_{rms} = \sqrt{\frac{3PV}{M}}$ (Option $C$ is correct,as $PV = nRT$ and $M$ is total mass).
Option $D$ is $(\frac{3P}{dM_w})^{1/2}$,which is dimensionally incorrect as it does not represent velocity.

(A) The formulas for the speeds are:
Most probable speed $(\alpha) = \sqrt{\frac{2RT}{M}}$
Average speed $(v) = \sqrt{\frac{8RT}{\pi M}}$
Root mean square speed $(u) = \sqrt{\frac{3RT}{M}}$
Taking the ratio $\alpha : v : u$:
$\sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3}$
$1.414 : 1.596 : 1.732$
Dividing by $1.414$:
$1 : 1.128 : 1.224$

(C) The formula for $rms$ speed is $u_{rms} = \sqrt{\frac{3RT}{M}}$.
For $H_2$ at $T_1 = 50 \ K$ and $M_1 = 2 \ g/mol$,and $O_2$ at $T_2 = 800 \ K$ and $M_2 = 32 \ g/mol$:
$\frac{u_{H_2}}{u_{O_2}} = \sqrt{\frac{T_1}{M_1} \times \frac{M_2}{T_2}}$
$\frac{u_{H_2}}{u_{O_2}} = \sqrt{\frac{50}{2} \times \frac{32}{800}}$
$\frac{u_{H_2}}{u_{O_2}} = \sqrt{25 \times \frac{1}{25}} = \sqrt{1} = 1$.

(C) The root mean square velocity $(u_{rms})$ is given by the formula: $u_{rms} = \sqrt{\frac{3RT}{M}}$
For $H_2$ at $50 \ K$: $u_{rms(H_2)} = \sqrt{\frac{3R \times 50}{2}}$
For $O_2$ at $800 \ K$: $u_{rms(O_2)} = \sqrt{\frac{3R \times 800}{32}}$
The ratio is: $\frac{u_{rms(H_2)}}{u_{rms(O_2)}} = \sqrt{\frac{3R \times 50}{2} \times \frac{32}{3R \times 800}}$
$= \sqrt{\frac{50}{2} \times \frac{32}{800}} = \sqrt{25 \times 0.04} = \sqrt{1} = 1$

(D) The root mean square speed $(u_{rms})$ of a gas is given by the formula: $u_{rms} = \sqrt{\frac{3RT}{M}}$.
Here,$R$ is the gas constant,$T$ is the temperature in Kelvin,and $M$ is the molar mass of the gas.
To maximize $u_{rms}$,we need a high temperature $(T)$ and a low molar mass $(M)$.
Let's calculate the relative values for each:
$A$: $O_2$ $(M=32)$,$T=273\,K \implies \frac{T}{M} = \frac{273}{32} \approx 8.53$
$B$: $N_2$ $(M=28)$,$T=1273\,K \implies \frac{T}{M} = \frac{1273}{28} \approx 45.46$
$C$: $CH_4$ $(M=16)$,$T=298\,K \implies \frac{T}{M} = \frac{298}{16} \approx 18.63$
$D$: $H_2$ $(M=2)$,$T=223\,K \implies \frac{T}{M} = \frac{223}{2} = 111.5$
Comparing the values of $\frac{T}{M}$,the gas with the highest value is $H_2$ at $-50\,^{\circ}C$.

(A) The average velocity $(v_{avg})$ of gas molecules is given by the formula: $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.
This implies that $v_{avg} \propto \sqrt{T}$.
Given $T_1 = 27 + 273 = 300\,K$ and $v_1 = 0.3\,m/s$.
Given $T_2 = 927 + 273 = 1200\,K$.
Using the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
$\frac{v_2}{0.3} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2$.
$v_2 = 0.3 \times 2 = 0.6\,m/s$.

(B) The root mean square $(rms)$ speed is calculated using the formula: $v_{rms} = \sqrt{\frac{\sum v_i^2}{n}}$.
Given speeds are $v_1 = 2, v_2 = 3, v_3 = 4, v_4 = 5 \, cm/s$ and $n = 4$.
Calculating the sum of squares: $\sum v_i^2 = 2^2 + 3^2 + 4^2 + 5^2 = 4 + 9 + 16 + 25 = 54$.
Now,$v_{rms} = \sqrt{\frac{54}{4}} = \sqrt{\frac{27}{2}} \, cm/s$.