Speed (velocity) of Gas (rms, mean and Most probable speed) Questions in English

(B) The root mean square velocity $(v_{rms})$ of a gas is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since the temperature $T$ is constant,$v_{rms} \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the root mean square velocities $v_1$ and $v_2$ for two gases with molecular weights $M_1$ and $M_2$ is:
$\frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}}$.

(C) The formula for $r.m.s.$ velocity is given by ${v_{rms}} = \sqrt{\frac{3RT}{M}}$.
Since the temperature $T$ is constant,we have ${v_{rms}} \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the $r.m.s.$ velocities of ${O_2}$ and ${H_2}$ is given by:
$\frac{{{{({v_{rms}})}_{{O_2}}}}}{{{{({v_{rms}})}_{{H_2}}}}} = \sqrt{\frac{{{M_{{H_2}}}}}{{{M_{{O_2}}}}}}$
Given that ${M_{{O_2}}} = 32$ and ${M_{{H_2}}} = 2$,we substitute the values:
$\frac{{400}}{{{{({v_{rms}})}_{{H_2}}}}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$
Solving for ${({v_{rms}})_{{H_2}}}$,we get:
${({v_{rms}})_{{H_2}}} = 400 \times 4 = 1600\, m/s$.

(C) The formula for $r.m.s.$ velocity is $v_{rms} = \sqrt{\frac{3P}{\rho}}$,where $\rho = \frac{m}{V}$.
Substituting $\rho$,we get $v_{rms} = \sqrt{\frac{3PV}{m}}$.
Given: $P = 24 \times 10^5 \, \text{dyne/cm}^2 = 24 \times 10^4 \, \text{N/m}^2$,$V = 10 \, \text{litre} = 10 \times 10^{-3} \, \text{m}^3$,$m = 20 \, \text{gm} = 20 \times 10^{-3} \, \text{kg}$.
$v_{rms} = \sqrt{\frac{3 \times (24 \times 10^4) \times (10 \times 10^{-3})}{20 \times 10^{-3}}} = \sqrt{\frac{3 \times 24 \times 10^4 \times 10^{-2}}{20 \times 10^{-3}}} = \sqrt{\frac{72 \times 10^2}{20 \times 10^{-3}}} = \sqrt{3.6 \times 10^6} = 600 \, \text{m/s}$.

(A) The formula for $r.m.s.$ velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
From this relation,we can see that $v_{rms} \propto \frac{1}{\sqrt{M}}$,assuming temperature $T$ and gas constant $R$ are constant.
Comparing Hydrogen $(H_2)$ and Oxygen $(O_2)$,the molar mass of Hydrogen $(M_{H_2} \approx 2 \ g/mol)$ is significantly lower than the molar mass of Oxygen $(M_{O_2} \approx 32 \ g/mol)$.
Since the $r.m.s.$ velocity is inversely proportional to the square root of the molar mass,the gas with the lower molecular mass will have a higher $r.m.s.$ velocity.
Therefore,the $r.m.s.$ velocity will be greater for Hydrogen.

(A) The average velocity $(v_{avg})$ of gas molecules is given by the formula: $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.
Here,$R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
Since $R$,$\pi$,and $M$ are constants for a given gas,the relationship simplifies to $v_{avg} \propto \sqrt{T}$.
Therefore,the average velocity of gas molecules is proportional to the square root of the absolute temperature.

(D) The root mean square speed $(v_{rms})$ of gas molecules is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms}$ is the same for both gases and $R$ is a constant,we have $v_{rms} \propto \sqrt{\frac{T}{M}}$.
Therefore,$\frac{T_{O_2}}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}}$.
Given $T_{H_2} = 200 \, K$,$M_{H_2} = 2 \, g/mol$,and $M_{O_2} = 32 \, g/mol$.
Substituting the values: $\frac{T_{O_2}}{32} = \frac{200}{2}$.
$T_{O_2} = 100 \times 32 = 3200 \, K$.

(A) The root mean square velocity is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
This implies that $v_{rms} \propto \sqrt{\frac{T}{M}}$.
Given: $\frac{v_{He}}{v_{H_2}} = \frac{5}{7}$,$T_{H_2} = 0^{\circ}C = 273 \, K$,$M_{He} = 4 \, g/mol$,and $M_{H_2} = 2 \, g/mol$.
Using the ratio: $\frac{v_{He}}{v_{H_2}} = \sqrt{\frac{T_{He}}{M_{He}} \times \frac{M_{H_2}}{T_{H_2}}} = \frac{5}{7}$.
Squaring both sides: $\frac{T_{He}}{4} \times \frac{2}{273} = \frac{25}{49}$.
$T_{He} = \frac{25}{49} \times \frac{4 \times 273}{2} = \frac{25 \times 2 \times 273}{49} = \frac{50 \times 273}{49} \approx 278.57 \, K$.
Converting to Celsius: $278.57 - 273 = 5.57^{\circ}C$,which is approximately $0^{\circ}C$ given the options provided.

(A) 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 in Kelvin,and $M$ is the molar mass of the gas.
Let $T_H$ be the temperature of hydrogen and $T_O$ be the temperature of oxygen. Given $T_O = 47^{\circ}C = 47 + 273 = 320 \; K$.
The molar mass of hydrogen $(H_2)$ is $M_H = 2 \; g/mol$ and the molar mass of oxygen $(O_2)$ is $M_O = 32 \; g/mol$.
According to the problem,the $RMS$ velocities are equal:
$\sqrt{\frac{3RT_H}{M_H}} = \sqrt{\frac{3RT_O}{M_O}}$
Squaring both sides and simplifying:
$\frac{T_H}{M_H} = \frac{T_O}{M_O}$
Substituting the values:
$\frac{T_H}{2} = \frac{320}{32}$
$\frac{T_H}{2} = 10$
$T_H = 20 \; K$.

(C) The root mean square $(r.m.s.)$ speed of a gas molecule is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $R$ and $M$ are constants for a given gas,the $r.m.s.$ speed is directly proportional to the square root of the absolute temperature: $v_{rms} \propto \sqrt{T}$.
Let $v_1$ be the $r.m.s.$ speed at $T_1 = 200 \, K$ and $v_2$ be the $r.m.s.$ speed at $T_2 = 800 \, K$.
Then,$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2$.
Therefore,$v_2 = 2v_1$,which means the $r.m.s.$ speed at $800 \, K$ is twice the value at $200 \, K$.

(D) The root mean square velocity $(v_{rms})$ of a gas molecule is given by the formula:
$v_{rms} = \sqrt{\frac{3kT}{m}}$
where $k$ is the Boltzmann constant,$T$ is the absolute temperature,and $m$ is the mass of the molecule.
From this expression,it is clear that $v_{rms}$ is inversely proportional to the square root of the mass of the molecule.
Therefore,$v_{rms} \propto \frac{1}{\sqrt{m}}$.

(B) The $r.m.s.$ speed of gas molecules is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$.
This implies that $V_{rms} \propto \sqrt{T}$.
Given initial temperature $T_{1} = 0^{\circ}C = 273 \; K$.
Let the final temperature be $T_{2}$.
We want the final $r.m.s.$ speed to be double the initial speed,so $V_{rms_{2}} = 2V_{rms_{1}}$.
Using the relation $\frac{V_{rms_{2}}}{V_{rms_{1}}} = \sqrt{\frac{T_{2}}{T_{1}}}$,we get $2 = \sqrt{\frac{T_{2}}{273}}$.
Squaring both sides,$4 = \frac{T_{2}}{273}$.
Therefore,$T_{2} = 4 \times 273 = 1092 \; K$.
To convert this back to Celsius,$T(^{\circ}C) = T(K) - 273 = 1092 - 273 = 819^{\circ}C$.

(C) The root mean square $(R.M.S)$ velocity of gas molecules is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
In an isothermal process,the temperature $T$ of the gas remains constant.
Since $R$ (universal gas constant) and $M$ (molar mass of the gas) are also constants,the $v_{rms}$ depends only on the temperature $T$.
Because the temperature does not change during isothermal expansion,the root mean square velocity of the molecules will remain unchanged.

(A) The root mean square $(R.M.S.)$ velocity of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $R$,$T$,and $M$ are constants for a given gas,$v_{rms} \propto \sqrt{T}$.
Let $v_1$ be the velocity at $T_1 = 100^{\circ}C = 373\; K$ and $v_2$ be the velocity at $T_2$.
Given $v_2 = 2v_1$.
Using the proportionality,$\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$.
Substituting the values: $\frac{v_1}{2v_1} = \sqrt{\frac{373}{T_2}}$.
$\frac{1}{2} = \sqrt{\frac{373}{T_2}}$.
Squaring both sides: $\frac{1}{4} = \frac{373}{T_2}$.
$T_2 = 373 \times 4 = 1492\; K$.
To convert the temperature to Celsius: $T(^{\circ}C) = T(K) - 273$.
$T_2 = 1492 - 273 = 1219^{\circ}C$.

(A) The formula for $r.m.s.$ speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Rearranging for molar mass $M$,we get $M = \frac{3RT}{v_{rms}^2}$.
Given: $R = 8.3\, J/(mol \cdot K)$,$T = 300\, K$ (room temperature),and $v_{rms} = 1920\, m/s$.
Substituting the values: $M = \frac{3 \times 8.3 \times 300}{(1920)^2}$.
$M = \frac{7470}{3686400} \approx 0.002026\, kg/mol \approx 2 \times 10^{-3}\, kg/mol = 2\, g/mol$.
The molar mass of $H_2$ is $2\, g/mol$. Therefore,the gas is $H_2$.

(C) The $r.m.s.$ velocity of gas molecules is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
This shows that $v_{rms} \propto \sqrt{T}$,where $T$ is the absolute temperature in Kelvin.
Note that the $r.m.s.$ velocity is independent of the pressure of the gas.
Given:
Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \, K$.
Initial $r.m.s.$ velocity $v_1 = 200 \, m/s$.
Final temperature $T_2 = 127^{\circ}C = 127 + 273 = 400 \, K$.
Using the proportionality $v_2 / v_1 = \sqrt{T_2 / T_1}$:
$v_2 = v_1 \times \sqrt{\frac{T_2}{T_1}}$
$v_2 = 200 \times \sqrt{\frac{400}{300}}$
$v_2 = 200 \times \sqrt{\frac{4}{3}}$
$v_2 = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \, m/s$.

(A) The root mean square speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
The average speed is given by $\bar{v} = \sqrt{\frac{8RT}{\pi M}}$.
Taking the ratio of the two speeds:
$\frac{\bar{v}}{v_{rms}} = \frac{\sqrt{\frac{8RT}{\pi M}}}{\sqrt{\frac{3RT}{M}}} = \sqrt{\frac{8}{3\pi}}$.
Calculating the value:
$\sqrt{\frac{8}{3 \times 3.14}} = \sqrt{\frac{8}{9.42}} \approx \sqrt{0.849} \approx 0.921$.
Therefore,$\bar{v} = 0.92\,v_{rms}$.

(D) The most probable speed is defined as the speed possessed by the maximum number of molecules in a sample.
Given the speeds of the five molecules: $2, 1.5, 1.6, 1.6, 1.2 \, km/s$.
By observing the data,the speed $1.6 \, km/s$ appears twice,while all other speeds appear only once.
Since the speed $1.6 \, km/s$ occurs most frequently,it is the most probable speed.
Therefore,the correct option is $D$.

(C) The root mean square velocity is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since the velocities are equal, we have $\sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{N_2}}{M_{N_2}}}$.
Squaring both sides and simplifying, we get $\frac{T_{O_2}}{M_{O_2}} = \frac{T_{N_2}}{M_{N_2}}$, which implies $T_{O_2} = T_{N_2} \times \frac{M_{O_2}}{M_{N_2}}$.
Given $T_{N_2} = 0 + 273 = 273\,K$, $M_{O_2} = 32\,g/mol$, and $M_{N_2} = 28\,g/mol$.
Substituting the values: $T_{O_2} = 273 \times \frac{32}{28} = 273 \times \frac{8}{7} = 39 \times 8 = 312\,K$.
Converting back to Celsius: $312 - 273 = 39\,^\circ C$.

(A) The root mean square velocity $(v_{rms})$ is given by the formula: $v_{rms} = \sqrt{\frac{v_1^2 + v_2^2 + v_3^2 + v_4^2 + v_5^2}{n}}$.
Substituting the given values: $v_{rms} = \sqrt{\frac{1^2 + 2^2 + 3^2 + 4^2 + 5^2}{5}} = \sqrt{\frac{1 + 4 + 9 + 16 + 25}{5}} = \sqrt{\frac{55}{5}} = \sqrt{11} \ km/s$.
The average velocity $(v_{av})$ is given by the formula: $v_{av} = \frac{v_1 + v_2 + v_3 + v_4 + v_5}{n}$.
Substituting the given values: $v_{av} = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3 \ km/s$.
The ratio of $v_{rms}$ to $v_{av}$ is $\frac{\sqrt{11}}{3}$,which is $\sqrt{11} : 3$.

(D) The mean velocity (average speed) of gas molecules is given by the formula: $v_{av} = \sqrt{\frac{8RT}{\pi M}}$.
Since the temperature $T$ is the same for both gases,the mean velocity is inversely proportional to the square root of the molar mass $M$: $v_{av} \propto \frac{1}{\sqrt{M}}$.
For hydrogen $(H_2)$,the molar mass $M_{H_2} = 2 \text{ g/mol}$.
For helium $(He)$,the molar mass $M_{He} = 4 \text{ g/mol}$.
Taking the ratio of the mean velocities:
$\frac{v_{H_2}}{v_{He}} = \sqrt{\frac{M_{He}}{M_{H_2}}} = \sqrt{\frac{4}{2}} = \sqrt{2}$.
Therefore,$v_{H_2} = \sqrt{2} \, v_{He}$.

(A) The root mean square speed $(v_{rms})$ of an ideal gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
This implies that $v_{rms} \propto \sqrt{T}$,where $T$ is the absolute temperature in Kelvin.
Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$.
Final temperature $T_2 = 927^{\circ}C = 927 + 273 = 1200 \ K$.
Using the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{v_2}{v_1} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2$.
Therefore,$v_2 = 2v_1$. The root mean square speed becomes twice the initial speed.

(B) According to the kinetic theory of gases,the root mean square $(r.m.s.)$ velocity of gas molecules 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.
From this expression,it is clear that ${v_{rms}} \propto \sqrt{T}$.
Therefore,the $r.m.s.$ velocity is directly proportional to the square root of the absolute temperature.

(D) The root mean square velocity $(v_{rms})$ of a gas molecule 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.
Since $R$ and $T$ are constant for both gases,we have $v_{rms} \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the root mean square velocities of oxygen $(O_2)$ and hydrogen $(H_2)$ is given by $\frac{(v_{rms})_{O_2}}{(v_{rms})_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}}$.
The molar mass of $H_2$ is $M_{H_2} = 2 \ g/mol$ and the molar mass of $O_2$ is $M_{O_2} = 32 \ g/mol$.
Substituting these values,we get $\frac{(v_{rms})_{O_2}}{(v_{rms})_{H_2}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
Thus,the ratio is $1:4$.

(A) The formula for $r.m.s.$ velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms}$ and $R$ are constant for both gases,we have $T \propto M$.
Therefore,$\frac{T_{N_2}}{T_{O_2}} = \frac{M_{N_2}}{M_{O_2}}$.
Given $T_{O_2} = 127^{\circ}C = 127 + 273 = 400 \ K$.
The molar mass of $N_2$ is $M_{N_2} = 28 \ g/mol$ and for $O_2$ is $M_{O_2} = 32 \ g/mol$.
Substituting the values: $\frac{T_{N_2}}{400} = \frac{28}{32}$.
$T_{N_2} = 400 \times \frac{28}{32} = 400 \times 0.875 = 350 \ K$.
Converting back to Celsius: $T_{N_2} = 350 - 273 = 77^{\circ}C$.

(B) The formula for $r.m.s.$ velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
At $NTP$,the temperature $T = 273.15 \ K$ (often approximated as $273 \ K$ or $300 \ K$ depending on the context; using $273 \ K$ is standard for $NTP$).
For Nitrogen $(N_2)$,the molar mass $M = 28 \times 10^{-3} \ kg/mol$.
The gas constant $R = 8.314 \ J/(mol \cdot K)$.
Substituting the values: $v_{rms} = \sqrt{\frac{3 \times 8.314 \times 273}{28 \times 10^{-3}}} \approx \sqrt{\frac{6811.266}{0.028}} \approx \sqrt{243259.5} \approx 493 \ m/s$.
Given the options provided,$517 \ m/s$ is the standard textbook value often cited for $N_2$ at $300 \ K$ $(v_{rms} = \sqrt{\frac{3 \times 8.314 \times 300}{0.028}} \approx 517 \ m/s)$. Thus,the correct option is $B$.

(C) The formula for the average speed of gas molecules is given by ${v_{av}} = \sqrt{\frac{8RT}{\pi M}}$.
Since the average speeds are equal,we have $\sqrt{\frac{8R{T_{H_2}}}{\pi {M_{H_2}}}} = \sqrt{\frac{8R{T_{O_2}}}{\pi {M_{O_2}}}}$.
Squaring both sides and simplifying,we get $\frac{{T_{H_2}}}{{M_{H_2}}} = \frac{{T_{O_2}}}{{M_{O_2}}}$,which implies ${T_{H_2}} = {T_{O_2}} \times \frac{{M_{H_2}}}{{M_{O_2}}}$.
Given ${T_{O_2}} = 31 + 273 = 304\,K$,${M_{H_2}} = 2\,g/mol$,and ${M_{O_2}} = 32\,g/mol$.
Substituting the values: ${T_{H_2}} = 304 \times \frac{2}{32} = 304 \times \frac{1}{16} = 19\,K$.
Converting back to Celsius: $19 - 273 = -254\,^oC$.

(B) The root mean square velocity $(v_{rms})$ of an ideal gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}.$
Since $R$ (universal gas constant) and $M$ (molar mass) are constants,we have $v_{rms} \propto \sqrt{T}.$
Given initial temperature $T_1 = 120 \ K$ and initial velocity $v_1 = v.$
Final temperature $T_2 = 480 \ K.$
Using the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}.$
Substituting the values: $\frac{v_2}{v} = \sqrt{\frac{480}{120}} = \sqrt{4} = 2.$
Therefore,$v_2 = 2v.$

(D) The root mean square velocity $(v_{rms})$ of an ideal gas is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$ or $v_{rms} = \sqrt{\frac{3P}{\rho}}$.
From the relation $v_{rms} = \sqrt{\frac{3RT}{M}}$,it is clear that $v_{rms}$ depends only on the temperature $(T)$ and the molar mass $(M)$ of the gas.
Since the temperature is kept constant,the $r.m.s.$ velocity remains independent of the pressure $(P)$.
Therefore,even if the pressure is increased to $2P$,the $r.m.s.$ velocity remains $v$.

(C) The root mean square speed $(v_{rms})$ of gas molecules is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature in Kelvin,and $M$ is the molar mass of the gas.
From this expression,it is clear that $v_{rms}$ depends only on the temperature $(T)$ and the molar mass $(M)$ of the gas.
It is independent of the pressure $(P)$ of the gas.
Therefore,$v_{rms} \propto \sqrt{T}$.

(A) The $r.m.s.$ speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since the $r.m.s.$ speeds are equal, we have $\sqrt{\frac{3RT_{He}}{M_{He}}} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}}$.
This implies $\frac{T_{He}}{M_{He}} = \frac{T_{H_2}}{M_{H_2}}$, or $T_{He} = T_{H_2} \times \frac{M_{He}}{M_{H_2}}$.
Given that $T_{H_2} = 273 \, K$ (at $NTP$, $0^\circ C = 273 \, K$), $M_{He} = 4 \, g/mol$, and $M_{H_2} = 2 \, g/mol$.
Substituting the values: $T_{He} = 273 \times \frac{4}{2} = 273 \times 2 = 546 \, K$.
To convert to Celsius: $T(^\circ C) = 546 - 273 = 273^\circ C$.

(A) The presence of an atmosphere on any planet implies that $C_{rms} << V_e$.
If the root mean square velocity $(C_{rms})$ of the gas molecules were equal to or greater than the escape velocity $(V_e)$,the molecules would possess enough kinetic energy to overcome the planet's gravitational pull.
Consequently,the gas molecules would escape into space,and the planet would not be able to retain an atmosphere.
Therefore,for an atmosphere to exist,the average speed of the molecules must be significantly lower than the escape velocity.

(C) The average speed of a gas molecule is given by the formula: $\langle C \rangle = \sqrt{\frac{8RT}{\pi M}}$.
Since the temperature $T$ is the same for both gases,the average speed is inversely proportional to the square root of the molar mass $M$,i.e.,$\langle C \rangle \propto \frac{1}{\sqrt{M}}$.
The molar mass of atomic hydrogen $(H)$ is $M_H = 1 \, g/mol$ and the molar mass of helium $(He)$ is $M_{He} = 4 \, g/mol$.
Taking the ratio of the average speeds:
$\frac{\langle C_H \rangle}{\langle C_{He} \rangle} = \sqrt{\frac{M_{He}}{M_H}} = \sqrt{\frac{4}{1}} = 2$.
Therefore,$\langle C_H \rangle = 2 \langle C_{He} \rangle$.

(D) The root mean square $(r.m.s.)$ velocity of gas molecules 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 (molecular weight) of the gas.
From this expression,it is clear that for a given temperature $T$,the $r.m.s.$ velocity is inversely proportional to the square root of the molecular weight $(M)$.
Therefore,${v_{rms}} \propto \frac{1}{\sqrt{M}}$.

(A) The $r.m.s.$ velocity of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
From this,we can see that $v_{rms} \propto \sqrt{T}$.
Given initial temperature $T_1 = 27^\circ C = 27 + 273 = 300 \ K$.
Given final temperature $T_2 = 327^\circ C = 327 + 273 = 600 \ K$.
Using the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{v_2}{v_1} = \sqrt{\frac{600}{300}} = \sqrt{2}$.
Therefore,the $r.m.s.$ velocity increases by a factor of $\sqrt{2}$.

(D) The root mean square speed $(v_{rms})$ is defined as the square root of the mean of the squares of the individual speeds.
Formula: $v_{rms} = \sqrt{\frac{v_1^2 + v_2^2 + v_3^2 + v_4^2 + v_5^2}{N}}$
Given speeds: $v_1 = 2, v_2 = 3, v_3 = 4, v_4 = 5, v_5 = 6$ and $N = 5$.
Calculate the squares: $2^2 = 4, 3^2 = 9, 4^2 = 16, 5^2 = 25, 6^2 = 36$.
Sum of squares: $4 + 9 + 16 + 25 + 36 = 90$.
Mean of squares: $\frac{90}{5} = 18$.
$v_{rms} = \sqrt{18} \approx 4.24$.

(A) The $r.m.s.$ velocity of a gas molecule is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the temperature,and $M$ is the molar mass.
Since $T$ is constant,$v_{rms} \propto \frac{1}{\sqrt{M}}$.
For hydrogen molecule $(H_2)$,$M_{H_2} = 2 \ g/mol$.
For helium atom $(He)$,$M_{He} = 4 \ g/mol$.
The ratio is $\frac{(v_{rms})_{H_2}}{(v_{rms})_{He}} = \sqrt{\frac{M_{He}}{M_{H_2}}} = \sqrt{\frac{4}{2}} = \sqrt{2} = \frac{\sqrt{2}}{1}$.
Thus,the ratio is $\sqrt{2} : 1$.

(C) The root mean square velocity $(v_{rms})$ 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.
From this,we see that $v_{rms} \propto \frac{1}{\sqrt{M}}$.
For oxygen $(O_2)$,$M_{O_2} = 32 \ g/mol$. For hydrogen $(H_2)$,$M_{H_2} = 2 \ g/mol$.
Given $v_{O_2} = C$,we have the ratio: $\frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}}$.
Substituting the values: $\frac{v_{H_2}}{C} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
Therefore,$v_{H_2} = 4C \ m/s$.

(B) The $R.M.S.$ velocity of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $R$ and $M$ are constants,we have $v_{rms} \propto \sqrt{T}$.
Let the initial temperature be $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$.
Let the initial $R.M.S.$ velocity be $v_1$ and the final $R.M.S.$ velocity be $v_2 = 2v_1$.
Using the proportionality relation: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $2 = \sqrt{\frac{T_2}{300}}$.
Squaring both sides: $4 = \frac{T_2}{300}$.
Therefore,$T_2 = 1200 \ K$.
Converting back to Celsius: $T_2(^{\circ}C) = 1200 - 273 = 927^{\circ}C$.

(B) The $r.m.s.$ velocity of an ideal gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
This implies that $v_{rms} \propto \sqrt{T}$.
Initial temperature $T_1 = 927 + 273 = 1200 \ K$.
Final temperature $T_2 = 27 + 273 = 300 \ K$.
The ratio of the velocities is $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{300}{1200}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,the $r.m.s.$ velocity becomes half of the initial value.

(C) The root mean square $(r.m.s.)$ speed of gas molecules is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M}}$.
Here,$R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
From the formula,it is evident that $V_{rms}$ depends only on the temperature $T$ and the nature of the gas (molar mass $M$).
It does not depend on the number of molecules or the density of the gas.
Since the temperature remains constant and the gas type does not change,the $r.m.s.$ speed remains unchanged regardless of the amount of gas leaking out.
Therefore,the $r.m.s.$ speed of the remaining molecules will be $400 \; m/s$.

(B) The root mean square velocity of gas molecules is given by $V_{rms} = \sqrt{\frac{3P}{\rho}}$.
The velocity of sound in an ideal gas is given by $V_s = \sqrt{\frac{\gamma P}{\rho}}$.
Dividing the two expressions,we get:
$\frac{V_{rms}}{V_s} = \frac{\sqrt{\frac{3P}{\rho}}}{\sqrt{\frac{\gamma P}{\rho}}} = \sqrt{\frac{3}{\gamma}}$.
Therefore,$V_{rms} = \sqrt{\frac{3}{\gamma}} \times V_s$.

(B) The root mean square speed $(v_{rms})$ is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
For two different gases,the ratio of their $v_{rms}$ speeds is: $\frac{(v_{rms})_{O_2}}{(v_{rms})_{H_2}} = \sqrt{\frac{T_{O_2}}{T_{H_2}} \times \frac{M_{H_2}}{M_{O_2}}}$.
Given: $T_{H_2} = 300 \ K$,$T_{O_2} = 900 \ K$,$M_{H_2} = 2 \ g/mol$,$M_{O_2} = 32 \ g/mol$,and $(v_{rms})_{H_2} = 1930 \ m/s$.
Substituting the values: $\frac{(v_{rms})_{O_2}}{1930} = \sqrt{\frac{900}{300} \times \frac{2}{32}} = \sqrt{3 \times \frac{1}{16}} = \frac{\sqrt{3}}{4}$.
Therefore,$(v_{rms})_{O_2} = 1930 \times \frac{\sqrt{3}}{4} \approx 1930 \times \frac{1.732}{4} \approx 836 \ m/s$.

(A) The $r.m.s.$ speed of a gas is given by the formula $C = \sqrt{\frac{3RT}{M}}$.
For gas $A$,$C_A = \sqrt{\frac{3RT_A}{M_A}}$.
For gas $B$,$C_B = \sqrt{\frac{3RT_B}{M_B}}$.
The ratio of the $r.m.s.$ speeds is $\frac{C_A}{C_B} = \frac{\sqrt{\frac{3RT_A}{M_A}}}{\sqrt{\frac{3RT_B}{M_B}}} = \sqrt{\frac{T_A}{M_A} \cdot \frac{M_B}{T_B}}$.
Given that $\frac{T_A}{M_A} = 4 \cdot \frac{T_B}{M_B}$,we have $\frac{T_A/M_A}{T_B/M_B} = 4$.
Substituting this into the ratio formula: $\frac{C_A}{C_B} = \sqrt{4} = 2$.

(B) The expressions for the three speeds of a gas at temperature $T$ are given by:
${v_{rms}} = \sqrt{\frac{3RT}{M}}$
${v_{av}} = \sqrt{\frac{8RT}{\pi M}}$
${v_{mp}} = \sqrt{\frac{2RT}{M}}$
Comparing the coefficients:
${v_{rms}} = \sqrt{3} \cdot \sqrt{\frac{RT}{M}} \approx 1.732 \cdot \sqrt{\frac{RT}{M}}$
${v_{av}} = \sqrt{\frac{8}{3.14}} \cdot \sqrt{\frac{RT}{M}} \approx \sqrt{2.546} \cdot \sqrt{\frac{RT}{M}} \approx 1.596 \cdot \sqrt{\frac{RT}{M}}$
${v_{mp}} = \sqrt{2} \cdot \sqrt{\frac{RT}{M}} \approx 1.414 \cdot \sqrt{\frac{RT}{M}}$
Thus,the relationship is ${v_{rms}} > {v_{av}} > {v_{mp}}$.

(B) The root-mean-square velocity $(v_{rms})$ of gas molecules is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
At a given temperature $(T)$,$v_{rms} \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass of the gas.
The molar masses are: $M_H = 2 \text{ g/mol}$,$M_N = 28 \text{ g/mol}$,and $M_O = 32 \text{ g/mol}$.
Since $M_H < M_N < M_O$,it follows that $\frac{1}{\sqrt{M_H}} > \frac{1}{\sqrt{M_N}} > \frac{1}{\sqrt{M_O}}$.
Therefore,the relationship between the velocities is ${V_H} > {V_N} > {V_O}$.

(B) The mean velocity of gas molecules is given by the formula $v_{av} = \sqrt{\frac{8RT}{\pi M}}$.
From this expression,we can see that the mean velocity is inversely proportional to the square root of the molar mass: $v_{av} \propto \frac{1}{\sqrt{M}}$.
Given that the mass of a $He$ atom $(M_{He})$ is $4$ times the mass of a hydrogen atom $(M_H)$,we have $M_{He} = 4M_H$.
Taking the ratio of the mean velocities:
$\frac{v_{He}}{v_H} = \sqrt{\frac{M_H}{M_{He}}} = \sqrt{\frac{M_H}{4M_H}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,$v_{He} = \frac{1}{2} v_H$.

(D) The root mean square (r.m.s.) speed is defined as the square root of the mean of the squares of the individual speeds.
Formula: $v_{rms} = \sqrt{\frac{\sum v_i^2}{N}}$
Given speeds are $6, 4, 2, 0, -2, -4, -6 \, m/s$ and $N = 7$.
Calculating the sum of squares: $v_{rms} = \sqrt{\frac{6^2 + 4^2 + 2^2 + 0^2 + (-2)^2 + (-4)^2 + (-6)^2}{7}}$
$v_{rms} = \sqrt{\frac{36 + 16 + 4 + 0 + 4 + 16 + 36}{7}}$
$v_{rms} = \sqrt{\frac{112}{7}}$
$v_{rms} = \sqrt{16} = 4 \, m/s$.

(A) The root mean square velocity $(v_{rms})$ of a gas is given by the formula $v_{rms} = \sqrt{\frac{3P}{\rho}}$,where $P$ is the pressure and $\rho$ is the density of the gas.
Since the pressure $P$ is constant,we have $v_{rms} \propto \frac{1}{\sqrt{\rho}}$.
Therefore,the ratio of the rms velocities of hydrogen $(v_H)$ and oxygen $(v_O)$ is given by $\frac{v_H}{v_O} = \sqrt{\frac{\rho_O}{\rho_H}}$.
Given that the ratio of vapour density $\frac{\rho_H}{\rho_O} = \frac{1}{16}$,it follows that $\frac{\rho_O}{\rho_H} = 16$.
Substituting this value,we get $\frac{v_H}{v_O} = \sqrt{16} = 4$.

(B) The root mean square (r.m.s.) speed of gas molecules is defined by the formula:
${v_{rms}} = \sqrt {\frac{{3RT}}{M}} $
Since $\sqrt 3 \approx 1.732$,we can rewrite the expression as:
${v_{rms}} = \sqrt 3 \sqrt {\frac{{RT}}{M}} = 1.73\sqrt {\frac{{RT}}{M}} $
Therefore,the correct option is $B$.

(A) The $r.m.s.$ speed of an ideal gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $R$ and $M$ are constants,$v_{rms} \propto \sqrt{T}$.
Therefore,the ratio of speeds is $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Given $T_1 = 27^\circ C = 27 + 273 = 300 \ K$ and $T_2 = 227^\circ C = 227 + 273 = 500 \ K$.
Given $v_1 = 400 \ m/s$.
Substituting the values: $\frac{V_s}{400} = \sqrt{\frac{500}{300}} = \sqrt{\frac{5}{3}}$.
$V_s = 400 \times \sqrt{1.666} \approx 400 \times 1.291 = 516.4 \ m/s$.
Rounding to the nearest integer,$V_s = 516 \ m/s$.