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

(B) The formula for $rms$ velocity is $V_{rms} = \sqrt{\frac{3RT}{M_w}}$.
Given for $H_2$: $T_1 = 27 + 273 = 300\, K$,$M_{w1} = 2\, g/mol$,$V_{rms1} = 100\, m/s$.
Given for $O_2$: $T_2 = 327 + 273 = 600\, K$,$M_{w2} = 32\, g/mol$.
Taking the ratio: $\frac{V_{rms2}}{V_{rms1}} = \sqrt{\frac{T_2}{T_1} \times \frac{M_{w1}}{M_{w2}}}$.
Substituting the values: $\frac{V_{rms2}}{100} = \sqrt{\frac{600}{300} \times \frac{2}{32}} = \sqrt{2 \times \frac{1}{16}} = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}}$.
$V_{rms2} = 100 \times \frac{1}{2\sqrt{2}} = \frac{50}{\sqrt{2}}\, m/s$.

(D) The expressions for the speeds are:
$v_{rms} = \sqrt{\frac{3kT}{m}}$
$\bar{v} = \sqrt{\frac{8kT}{\pi m}} \approx \sqrt{\frac{2.55kT}{m}}$
$v_p = \sqrt{\frac{2kT}{m}}$
Comparing these,we get $v_p < \bar{v} < v_{rms}$.
For the average kinetic energy of a molecule:
$E_k = \frac{1}{2} m v_{rms}^2 = \frac{1}{2} m \left( \sqrt{\frac{3kT}{m}} \right)^2 = \frac{3}{2} kT$.
Since $v_p^2 = \frac{2kT}{m}$,we have $kT = \frac{1}{2} m v_p^2$.
Substituting this into the expression for $E_k$:
$E_k = \frac{3}{2} \left( \frac{1}{2} m v_p^2 \right) = \frac{3}{4} m v_p^2$.

(D) The $rms$ velocity of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
For the $rms$ velocities to be equal, we have $\sqrt{\frac{3RT_{N_2}}{M_{N_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$.
This simplifies to $\frac{T_{N_2}}{M_{N_2}} = \frac{T_{O_2}}{M_{O_2}}$.
Given $T_{O_2} = 127 + 273 = 400 \, K$, $M_{O_2} = 32 \, g/mol$, and $M_{N_2} = 28 \, g/mol$.
Substituting the values: $\frac{T_{N_2}}{28} = \frac{400}{32}$.
$T_{N_2} = \frac{400 \times 28}{32} = 12.5 \times 28 = 350 \, K$.
Converting back to Celsius: $T(^\circ C) = 350 - 273 = 77 ^\circ C$.

(D) The root mean square velocity $(c_{rms})$ of gas molecules is given by the formula: $c_{rms} = \sqrt{\frac{3RT}{M}}$.
Since the temperature $(T)$ is the same for both gases at $NTP$,we have $c_{rms} \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the velocities is given by: $\frac{(c_{rms})_{H_2}}{(c_{rms})_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}}$.
Given $(c_{rms})_{H_2} = 1.84 \, km/s$,$M_{H_2} = 2$,and $M_{O_2} = 32$.
Substituting the values: $\frac{1.84}{(c_{rms})_{O_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
Thus,$(c_{rms})_{O_2} = \frac{1.84}{4} = 0.46 \, km/s$.

(D) The root mean square speed $(v_{rms})$ is given by $\sqrt{3RT/M}$,and the most probable speed $(v_{mp})$ is given by $\sqrt{2RT/M}$. Since $\sqrt{3} \neq \sqrt{2}$,these speeds are not the same. Thus,the Assertion is incorrect.
The Maxwell-Boltzmann distribution curve for molecular speeds is skewed to the right (asymmetric),meaning it has a long tail towards higher speeds. Therefore,the distribution is not symmetrical. Thus,the Reason is also incorrect.

(A) At a fixed temperature,the average kinetic energy $\frac{1}{2} m \langle v^2 \rangle$ is constant. Thus,the average speed $v_{avg} \propto \frac{1}{\sqrt{m}}$.
The molar mass of $UF_6$ with $U^{235}$ is $M_1 = 235 + 6 \times 19 = 235 + 114 = 349 \text{ units}$.
The molar mass of $UF_6$ with $U^{238}$ is $M_2 = 238 + 6 \times 19 = 238 + 114 = 352 \text{ units}$.
Since $M_1 < M_2$,the $UF_6$ gas containing $U^{235}$ will have the larger average speed.
The ratio of speeds is given by $\frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{352}{349}} \approx \sqrt{1.008596} \approx 1.004288$.
The percentage difference is $\frac{\Delta v}{v} \times 100 = (1.004288 - 1) \times 100 \approx 0.43 \%$.

(B) The root mean square (rms) speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given temperature of helium,$T_{He} = -20^{\circ}C = 253 \; K$.
Atomic mass of argon,$M_{Ar} = 39.9 \; u$.
Atomic mass of helium,$M_{He} = 4.0 \; u$.
We are given that $(v_{rms})_{Ar} = (v_{rms})_{He}$.
Therefore,$\sqrt{\frac{3RT_{Ar}}{M_{Ar}}} = \sqrt{\frac{3RT_{He}}{M_{He}}}$.
Squaring both sides and simplifying,we get $\frac{T_{Ar}}{M_{Ar}} = \frac{T_{He}}{M_{He}}$.
Rearranging for $T_{Ar}$,we get $T_{Ar} = T_{He} \times \frac{M_{Ar}}{M_{He}}$.
Substituting the values: $T_{Ar} = 253 \times \frac{39.9}{4.0}$.
$T_{Ar} = 253 \times 9.975 = 2523.675 \; K$.
Rounding to significant figures,$T_{Ar} \approx 2.52 \times 10^{3} \; K$.

(N/A) The average kinetic energy of one molecule of an ideal gas is given by the relation: $\langle \frac{1}{2} m v^{2} \rangle = \frac{3}{2} k_{B} T$.
Here,$k_{B}$ is the Boltzmann constant,$T$ is the absolute temperature,$m$ is the mass of the molecule,and $v$ is the speed of the gas molecule.
If $T = 0$,then $\langle \frac{1}{2} m v^{2} \rangle = 0$,which implies $\langle v^{2} \rangle = 0$,and consequently,the root-mean-square speed $v_{rms} = \sqrt{\langle v^{2} \rangle} = 0$.
Therefore,absolute temperature is defined as the temperature at which the root-mean-square speed $(v_{rms})$ of gas molecules becomes zero.
Using the relation $v_{rms} = \sqrt{\frac{3P}{\rho}}$,where $P$ is pressure and $\rho$ is density,we can see that at $T = 0$,the pressure $P$ must also be zero.

(N/A) The $rms$ (root mean square) value is defined as the square root of the mean of the squares of the values.
$v_{rms}$ is the square root of the mean of the squares of the speeds of gas molecules.
According to the kinetic theory of gases,the pressure $P$ of an ideal gas is given by:
$P = \frac{1}{3} \rho \langle v^2 \rangle$
where $\rho$ is the density of the gas and $\langle v^2 \rangle$ is the mean square speed.
Rearranging the equation for $\langle v^2 \rangle$:
$\langle v^2 \rangle = \frac{3P}{\rho}$
Since $v_{rms} = \sqrt{\langle v^2 \rangle}$,we substitute the expression for $\langle v^2 \rangle$:
$v_{rms} = \sqrt{\frac{3P}{\rho}}$

(N/A) The average kinetic energy of one molecule of an ideal gas is given by:
$<\frac{1}{2} m v^{2}> = \frac{3}{2} k_{B} T$
Where $m$ is the mass of the molecule,$v$ is the velocity,$k_{B}$ is the Boltzmann constant,and $T$ is the absolute temperature.
Rearranging the equation for the mean square velocity $\langle v^{2} \rangle$:
$\langle v^{2} \rangle = \frac{3 k_{B} T}{m}$
The root mean square velocity $v_{rms}$ is defined as the square root of the mean square velocity:
$v_{rms} = \sqrt{\langle v^{2} \rangle} = \sqrt{\frac{3 k_{B} T}{m}}$
From this expression,it is evident that for a given temperature $T$:
$v_{rms} \propto \frac{1}{\sqrt{m}}$
Thus,at a constant temperature,lighter molecules possess higher root mean square speeds compared to heavier molecules.

(N/A) The average kinetic energy of one molecule of an ideal gas is given by:
$<\frac{1}{2} m v^{2}> = \frac{3}{2} k_{B} T$
Here,$m$ is the mass of one molecule,$v$ is the velocity,$k_{B}$ is the Boltzmann constant,and $T$ is the absolute temperature.
We know that the Boltzmann constant $k_{B} = \frac{R}{N_{A}}$,where $R$ is the universal gas constant and $N_{A}$ is the Avogadro number.
Substituting $k_{B}$ into the equation:
$<\frac{1}{2} m v^{2}> = \frac{3}{2} (\frac{R}{N_{A}}) T$
Multiplying both sides by $2$:
$m = \frac{3 R T}{N_{A}}$
Since the molar mass $M_{0} = m \times N_{A}$,we can write $m = \frac{M_{0}}{N_{A}}$. Substituting this:
$(\frac{M_{0}}{N_{A}}) = \frac{3 R T}{N_{A}}$
Canceling $N_{A}$ from both sides:
$M_{0} = 3 R T$
$ = \frac{3 R T}{M_{0}}$
By definition,the root mean square velocity $v_{rms} = \sqrt{}$.
Therefore,$v_{rms} = \sqrt{\frac{3 R T}{M_{0}}}$.

(A) The root mean square speed $(v_{rms})$ of an ideal gas is given by the formula:
$v_{rms} = \sqrt{\frac{3RT}{M}}$
Where $R = 8.314 \ J \ mol^{-1} \ K^{-1}$ is the universal gas constant,$T = 300 \ K$ is the temperature,and $M$ is the molar mass of nitrogen $(N_2)$.
For nitrogen gas $(N_2)$,the molar mass $M = 28 \ g/mol = 28 \times 10^{-3} \ kg/mol$.
Substituting the values into the formula:
$v_{rms} = \sqrt{\frac{3 \times 8.314 \times 300}{28 \times 10^{-3}}}$
$v_{rms} = \sqrt{\frac{7482.6}{0.028}}$
$v_{rms} = \sqrt{267235.7}$
$v_{rms} \approx 516.95 \ m/s$
Rounding to the nearest whole number,$v_{rms} \approx 517 \ m/s$.

(N/A) $v_{rms}$ stands for the root mean square speed of gas molecules. It is defined as the square root of the mean of the squares of the speeds of individual molecules in a gas.
$(1)$ In terms of density $(\rho)$ and pressure $(P)$: $v_{rms} = \sqrt{\frac{3P}{\rho}}$
$(2)$ In terms of mass of a molecule $(m)$: $v_{rms} = \sqrt{\frac{3k_BT}{m}}$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
$(3)$ In terms of molar mass $(M)$: $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant and $T$ is the absolute temperature.

(B) The root mean square speed $(v_{rms})$ of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
For air at room temperature $(T \approx 300 \ K)$,the molar mass $(M)$ is approximately $29 \times 10^{-3} \ kg/mol$.
Substituting these values: $v_{rms} = \sqrt{\frac{3 \times 8.314 \times 300}{29 \times 10^{-3}}} \approx 460 \ m/s$.
This value is of the order of $10^2 \ m/s$ to $10^3 \ m/s$,but typically,for common gases like nitrogen or oxygen at standard conditions,it is approximately $500 \ m/s$,which falls under the order of $10^2 \ m/s$ (specifically closer to $10^2$ than $10^3$ in magnitude).
Therefore,the correct order is $10^2 \ m/s$.

(C) The root mean square velocity $(v_{rms})$ of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \frac{1}{\sqrt{M}}$ and vapour density $\rho \propto M$,we have $v_{rms} \propto \frac{1}{\sqrt{\rho}}$.
Given the ratio of vapour densities $\rho_1 : \rho_2 = 6 : 9$.
Therefore,the ratio of their $v_{rms}$ is $\frac{(v_{rms})_1}{(v_{rms})_2} = \sqrt{\frac{\rho_2}{\rho_1}}$.
Substituting the values: $\frac{(v_{rms})_1}{(v_{rms})_2} = \sqrt{\frac{9}{6}} = \sqrt{\frac{3}{2}}$.
Thus,the ratio is $\sqrt{3} : \sqrt{2}$.

(NO CHANGE) The root mean square speed $(v_{rms})$ of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$ or $v_{rms} = \sqrt{\frac{3kT}{m}}$.
Here,$R$ is the universal gas constant,$T$ is the absolute temperature,$M$ is the molar mass,$k$ is the Boltzmann constant,and $m$ is the mass of one molecule.
Since $v_{rms}$ depends only on the temperature of the gas and the mass of the molecules,it is independent of the number of molecules or the pressure/volume of the gas.
Therefore,there will be no change in $v_{rms}$.

(N/A) The $rms$ speed of molecules of a gas is given by:
$c = \sqrt{\frac{3P}{\rho}} = \sqrt{\frac{3RT}{M_0}}$
(Where $M_0$ is the molar mass of the gas and $R$ is the universal gas constant).
The speed of sound in a gas is given by:
$v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma RT}{M_0}} \quad ... (1)$
(Where $\gamma$ is the adiabatic index or ratio of specific heats).
Taking the ratio of $c$ and $v$:
$\frac{c}{v} = \frac{\sqrt{\frac{3RT}{M_0}}}{\sqrt{\frac{\gamma RT}{M_0}}} = \sqrt{\frac{3}{\gamma}}$
For all diatomic gases,the adiabatic index $\gamma = \frac{C_P}{C_V} = \frac{7}{5} = 1.4$.
Substituting this value:
$\frac{c}{v} = \sqrt{\frac{3}{7/5}} = \sqrt{\frac{15}{7}}$
Since $\sqrt{\frac{15}{7}}$ is a constant value and does not contain the temperature $T$,the ratio $\frac{c}{v}$ is constant and independent of temperature for all diatomic gases.

(A) The root mean square speed of a gas molecule is given by the formula $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
Since $v_{\text{rms}} \propto \frac{1}{\sqrt{M}}$,the gas with the lower molar mass will have a higher $v_{\text{rms}}$ at the same temperature.
The molar mass of hydrogen $(H_2)$ is approximately $2 \text{ g/mol}$,while the average molar mass of air is approximately $28.97 \text{ g/mol}$.
Since the molar mass of hydrogen is significantly lower than that of air,hydrogen will have a higher $v_{\text{rms}}$.

(C) The root mean square speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$,we have the ratio $\frac{(v_{rms})_{2}}{(v_{rms})_{1}} = \sqrt{\frac{T_{2}}{T_{1}}}$.
Given that the temperature is increased to $3$ times,$T_{2} = 3T_{1}$.
Substituting this into the ratio,we get $\frac{(v_{rms})_{2}}{(v_{rms})_{1}} = \sqrt{\frac{3T_{1}}{T_{1}}} = \sqrt{3}$.
Therefore,$(v_{rms})_{2} = \sqrt{3}(v_{rms})_{1} \approx 1.732(v_{rms})_{1}$.
The change in $v_{rms}$ is $(v_{rms})_{2} - (v_{rms})_{1} = 1.732(v_{rms})_{1} - 1(v_{rms})_{1} = 0.732(v_{rms})_{1}$.
Thus,the $v_{rms}$ increases by $0.732$ times the initial value.

(A) According to Graham's law of effusion,the rate of effusion of a gas is inversely proportional to the square root of its molar mass $(Rate \propto 1/\sqrt{M})$.
Since the molar mass of hydrogen $(M_{H_2} = 2 \ g/mol)$ is much smaller than the molar mass of carbon dioxide $(M_{CO_2} = 44 \ g/mol)$,the hydrogen gas will effuse out of the container faster.
Alternatively,the root mean square speed $(v_{\text{rms}})$ is given by $v_{\text{rms}} = \sqrt{3RT/M}$. Since $v_{\text{rms}} \propto 1/\sqrt{M}$,hydrogen has a higher $v_{\text{rms}}$ than carbon dioxide at the same temperature,leading to a higher rate of effusion.

(N/A) The average kinetic energy of gas molecules is given by the relation $K.E. = \frac{1}{2} m v_{rms}^{2} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
At absolute zero temperature,$T = 0 \ K$.
Substituting this into the equation,we get $\frac{1}{2} m v_{rms}^{2} = \frac{3}{2} k_B (0) = 0$.
Since the mass of the molecule $m \neq 0$,it follows that $v_{rms}^{2} = 0$,which implies $v_{rms} = 0$.
Therefore,at absolute zero temperature,the root mean square velocity of the molecules becomes zero,meaning their motion ceases.

(N/A) According to the kinetic theory of gases,the pressure $P$ is given by $P = \frac{1}{3} \rho v_{rms}^{2}$.
Therefore,$v_{rms} = \sqrt{\frac{3P}{\rho}}$.
Substituting the values $P = 1.01 \times 10^{5} \ Pa$ and $\rho = 0.09 \ kg/m^{3}$:
$v_{rms} = \sqrt{\frac{3 \times 1.01 \times 10^{5}}{0.09}} = \sqrt{\frac{3.03 \times 10^{5}}{0.09}} = \sqrt{33.67 \times 10^{5}} = \sqrt{3.367 \times 10^{6}} \approx 1834.9 \ m/s$.
The average kinetic energy of $1 \ mole$ of an ideal gas is given by $E = \frac{3}{2} RT$.
Using the relation $PV = RT$ for $1 \ mole$,we have $E = \frac{3}{2} PV$.
Since $\rho = \frac{M}{V}$,where $M$ is the molar mass of $H_{2}$ $(2 \times 10^{-3} \ kg/mol)$,the volume of $1 \ mole$ is $V = \frac{M}{\rho} = \frac{2 \times 10^{-3}}{0.09} \approx 0.0222 \ m^{3}$.
Thus,$E = \frac{3}{2} \times (1.01 \times 10^{5}) \times (0.0222) \approx 3363.3 \ J$.

(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,and $M$ is the molar mass of the gas.
Since the temperature $T$ of a gas in equilibrium is greater than $0 \ K$,the ${v_{rms}}$ value is always positive and non-zero.
Therefore,the statement that ${v_{rms}}$ is zero is incorrect; it depends on the temperature and the molar mass of the gas.

(A) 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,and $M$ is the molar mass of the gas.
From this relation,it is clear that $v_{rms} \propto \frac{1}{\sqrt{M}}$.
Therefore,for a given temperature $T$,the speed of gas molecules is inversely proportional to the square root of their molar mass.
As a result,a lighter gas (with a smaller molar mass $M$) will have a higher speed compared to a heavier gas.

(A) The root mean square speed $({v_{rms}})$ of gas molecules is derived from the kinetic theory of gases as ${v_{rms}} = \sqrt{\frac{3P}{\rho}}$, where $P$ is pressure and $\rho$ is density. Thus, $(a)$ matches with $(iii)$.
For $1 \text{ mole}$ of an ideal gas, the formula is ${v_{rms}} = \sqrt{\frac{3RT}{M_0}}$, where $R$ is the universal gas constant, $T$ is temperature, and $M_0$ is molar mass. Thus, $(b)$ matches with $(i)$.
For a single molecule of gas, the formula is ${v_{rms}} = \sqrt{\frac{3k_BT}{m}}$, where $k_B$ is the Boltzmann constant and $m$ is the mass of one molecule. Thus, $(c)$ matches with $(ii)$.
Therefore, the correct matching is $(a-iii, b-i, c-ii)$.

(B) The root mean square speed $(v_{rms})$ of gas molecules is given by the formula:
$v_{rms} = \sqrt{\frac{3RT}{M}}$
From this relation,it is clear that $v_{rms} \propto \sqrt{T}$.
Note that $v_{rms}$ is independent of the pressure of the gas.
Given:
$T_1 = 27^{\circ} \ C = 27 + 273 = 300 \ K$
$T_2 = 127^{\circ} \ C = 127 + 273 = 400 \ K$
$(v_{rms})_1 = 100 \ m/s$
Using the ratio formula:
$\frac{(v_{rms})_1}{(v_{rms})_2} = \sqrt{\frac{T_1}{T_2}}$
$\frac{100}{(v_{rms})_2} = \sqrt{\frac{300}{400}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$
$(v_{rms})_2 = \frac{100 \times 2}{\sqrt{3}} = \frac{200}{\sqrt{3}} \ m/s$.

(C) The root mean square $(v_{rms})$ speed for $n$ molecules is given by the formula:
$v_{rms} = \sqrt{\frac{v_{1}^{2} + v_{2}^{2} + \dots + v_{n}^{2}}{n}}$
For two molecules,the formula becomes:
$v_{rms} = \sqrt{\frac{v_{1}^{2} + v_{2}^{2}}{2}}$
Given:
$v_{1} = 9 \times 10^{6} \ m/s$
$v_{2} = 1 \times 10^{6} \ m/s$
Substituting the values:
$v_{rms} = \sqrt{\frac{(9 \times 10^{6})^{2} + (1 \times 10^{6})^{2}}{2}}$
$v_{rms} = \sqrt{\frac{81 \times 10^{12} + 1 \times 10^{12}}{2}}$
$v_{rms} = \sqrt{\frac{82 \times 10^{12}}{2}}$
$v_{rms} = \sqrt{41 \times 10^{12}}$
$v_{rms} = \sqrt{41} \times 10^{6} \ m/s$
Since $\sqrt{41} \approx 6.403$,
$v_{rms} \approx 6.40 \times 10^{6} \ m/s$

(N/A) The root mean square speed is given by $v_{rms} = \sqrt{\frac{\sum n_i v_i^2}{\sum n_i}}$.
Using the given data:
$v_{rms} = \sqrt{\frac{10(200)^2 + 20(400)^2 + 40(600)^2 + 20(800)^2 + 10(1000)^2}{10+20+40+20+10}}$
$v_{rms} = \sqrt{\frac{10(4 \times 10^4) + 20(16 \times 10^4) + 40(36 \times 10^4) + 20(64 \times 10^4) + 10(100 \times 10^4)}{100}}$
$v_{rms} = \sqrt{\frac{10^4(40 + 320 + 1440 + 1280 + 1000)}{100}} = \sqrt{408000} \approx 638.75 \ m/s$.
Using $v_{rms} = \sqrt{\frac{3k_B T}{m}}$, we get $T = \frac{m v_{rms}^2}{3k_B} = \frac{3.0 \times 10^{-26} \times (638.75)^2}{3 \times 1.38 \times 10^{-23}} \approx 0.098 \ K$.
$(b)$ Removing molecules with $v = 1000 \ m/s$, the new distribution has $90$ molecules total.
$v_{rms}' = \sqrt{\frac{10(200)^2 + 20(400)^2 + 40(600)^2 + 20(800)^2}{90}}$
$v_{rms}' = \sqrt{\frac{10^4(40 + 320 + 1440 + 1280)}{90}} = \sqrt{\frac{3080000}{90}} \approx 585.18 \ m/s$.
$T' = \frac{m (v_{rms}')^2}{3k_B} = \frac{3.0 \times 10^{-26} \times (585.18)^2}{3 \times 1.38 \times 10^{-23}} \approx 0.082 \ K$.

(B) The average speed of molecules of light gases like $H_2$ and $He$ at atmospheric temperature is greater than the escape velocity of Earth $(v_e \approx 11.2 \ km/s)$.
Because their thermal velocity exceeds the escape velocity,these gas molecules easily overcome Earth's gravitational pull and escape into space.
Therefore,they are found in negligible amounts in the Earth's atmosphere.

(A) The root mean square $(rms)$ speed of a gas molecule is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that the $rms$ speed of $H_{2}$ is equal to the $rms$ speed of $N_{2}$:
$V_{rms(H_{2})} = V_{rms(N_{2})}$
$\sqrt{\frac{3RT_{H_{2}}}{M_{H_{2}}}} = \sqrt{\frac{3RT_{N_{2}}}{M_{N_{2}}}}$
Squaring both sides and canceling common terms $(3R)$:
$\frac{T_{H_{2}}}{M_{H_{2}}} = \frac{T_{N_{2}}}{M_{N_{2}}}$
Given $T_{N_{2}} = 300^{\circ}C = 300 + 273 = 573 \ K$,$M_{N_{2}} = 28 \ g/mol$,and $M_{H_{2}} = 2 \ g/mol$:
$\frac{T_{H_{2}}}{2} = \frac{573}{28}$
$T_{H_{2}} = \frac{573 \times 2}{28} = \frac{573}{14} \approx 40.928 \ K$.
Rounding to the nearest integer,we get $41 \ K$.

(B) The relative velocity between two molecules with velocity $V$ is given by $\left|V_{\text{rel}}\right| = \sqrt{V^2 + V^2 - 2V^2 \cos \theta} = 2V \sin(\theta/2)$.
The average relative velocity is $\langle V_{\text{rel}} \rangle = \frac{\int_{0}^{\pi} 2V \sin(\theta/2) d\theta}{\int_{0}^{\pi} d\theta} = \frac{4V}{\pi}$.
Since the average velocity of a molecule is $\langle V \rangle = \sqrt{\frac{8RT}{\pi M}}$,the average relative velocity is $\langle V_{\text{rel}} \rangle = \frac{4}{\pi} \sqrt{\frac{8RT}{\pi M}}$.
Given $R = 8.314 \, J/(mol \cdot K)$,$T = 300 \, K$,and $M = 28 \times 10^{-3} \, kg/mol$ for $N_2$:
$\langle V_{\text{rel}} \rangle = \frac{4}{\pi} \sqrt{\frac{8 \times 8.314 \times 300}{\pi \times 28 \times 10^{-3}}} \approx \frac{4}{3.14} \times 476.5 \approx 606 \, m/s$.

(C) The root mean square (rms) velocity of an ideal gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
The average velocity of an ideal gas is given by $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.
To find the ratio of rms velocity to average velocity,we divide the two expressions:
$\frac{v_{rms}}{v_{avg}} = \frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{8RT}{\pi M}}}$.
Simplifying the expression,the terms $R$,$T$,and $M$ cancel out:
$\frac{v_{rms}}{v_{avg}} = \sqrt{\frac{3RT}{M} \times \frac{\pi M}{8RT}} = \sqrt{\frac{3\pi}{8}}$.
Thus,the ratio is $\sqrt{\frac{3\pi}{8}}$.

(C) The root mean square speed $(v_{rms})$ of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $R$ and $M$ are constants for a given gas,$v_{rms} \propto \sqrt{T}$.
Note that $v_{rms}$ is independent of pressure.
Given $T_1 = 27^{\circ} C = 300\, K$ and $(v_{rms})_1 = 200\, ms^{-1}$.
Given $T_2 = 127^{\circ} C = 400\, K$.
Using the ratio: $\frac{(v_{rms})_2}{(v_{rms})_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.
Therefore,$(v_{rms})_2 = \frac{2}{\sqrt{3}} \times (v_{rms})_1 = \frac{2}{\sqrt{3}} \times 200 = \frac{400}{\sqrt{3}}\, ms^{-1}$.
Comparing this with $\frac{x}{\sqrt{3}}$,we get $x = 400$.

(A) The formula for the $rms$ speed of gas molecules is $V_{RMS} = \sqrt{\frac{3RT}{M_{W}}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M_{W}$ is the molar mass of the gas.
Since the temperature $T$ is the same for all gases,we have $V_{RMS} \propto \frac{1}{\sqrt{M_{W}}}$.
The molar masses are: $M_{H} = 2 \ g/mol$,$M_{O} = 32 \ g/mol$,and $M_{C} = 44 \ g/mol$.
Since $M_{H} < M_{O} < M_{C}$,it follows that $\frac{1}{\sqrt{M_{H}}} > \frac{1}{\sqrt{M_{O}}} > \frac{1}{\sqrt{M_{C}}}$.
Therefore,$V_{H} > V_{O} > V_{C}$.

(A) The $rms$ speed of gas molecules is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the temperature in Kelvin,and $M$ is the molar mass of the gas.
Since the temperature $T$ is the same for both gases,we have $V_{rms} \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the $rms$ speeds is: $\frac{(V_{rms})_{O_2}}{(V_{rms})_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}}$.
The molar mass of oxygen $(O_2)$ is $32 \; {g/mol}$ and the molar mass of hydrogen $(H_2)$ is $2 \; {g/mol}$.
Substituting the values: $\frac{160}{(V_{rms})_{H_2}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
Thus,$(V_{rms})_{H_2} = 160 \times 4 = 640 \; {m/s}$.

(D) The root mean square speed $(v_{\text{rms}})$ of a gas molecule is given by the formula: $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
Since the temperature $T$ is the same for all gases in the mixture,$v_{\text{rms}} \propto \frac{1}{\sqrt{M}}$.
Given the masses $m_{A} < m_{B} < m_{C}$,it follows that $v_{A} > v_{B} > v_{C}$.
Taking the reciprocal of these speeds,we get: $\frac{1}{v_{A}} < \frac{1}{v_{B}} < \frac{1}{v_{C}}$.

(B) The root mean square speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
The most probable speed is given by $v_{p} = \sqrt{\frac{2RT}{M}}$.
Dividing the two expressions,we get:
$\frac{v_{rms}}{v_{p}} = \frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{2RT}{M}}} = \sqrt{\frac{3}{2}}$.
Therefore,$v_{rms} = \sqrt{\frac{3}{2}} v_{p}$.

(B) The root mean square velocity $(V_{rms})$ of a gas is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that the temperature is doubled,$T' = 2T$.
When oxygen molecules $(O_2)$ dissociate into atomic oxygen $(O)$,the molar mass is halved,so $M' = M/2$.
Substituting these into the proportionality $V_{rms} \propto \sqrt{\frac{T}{M}}$,we get:
$\frac{(V_{rms})_{atomic}}{(V_{rms})_{molecular}} = \sqrt{\frac{T'}{M'} \cdot \frac{M}{T}} = \sqrt{\frac{2T}{M/2} \cdot \frac{M}{T}} = \sqrt{4} = 2$.
Therefore,the velocity of atomic oxygen becomes $2$ times the initial velocity of the oxygen molecules.

(D) Statement $I$: In an ideal gas,molecules move in random directions. For every molecule with momentum $\vec{p}$,there is a statistically equal probability of finding a molecule with momentum $-\vec{p}$. Thus,the average momentum $\vec{p}_{avg} = 0$,which is independent of temperature.
Statement $II$: The rms speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Initially,$v = \sqrt{\frac{3RT}{M_{O_2}}}$.
When temperature is doubled $(T' = 2T)$ and $O_2$ dissociates into $O$ atoms,the molar mass becomes $M' = \frac{M_{O_2}}{2}$.
The new rms speed $v' = \sqrt{\frac{3R(2T)}{M_{O_2}/2}} = \sqrt{4 \cdot \frac{3RT}{M_{O_2}}} = 2v$.
Therefore,Statement $I$ is false and Statement $II$ is true.

(C) The root mean square speed $(V_{\text{rms}})$ of a particle is given by the formula: $V_{\text{rms}} = \sqrt{\frac{3kT}{m}}$.
Given:
Mass of particle $(m)$ = $5 \times 10^{-17} \, kg$
Boltzmann constant $(k)$ = $1.38 \times 10^{-23} \, J \, K^{-1}$
Temperature at $NTP$ $(T)$ = $293 \, K$
Substituting the values:
$V_{\text{rms}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 293}{5 \times 10^{-17}}}$
$V_{\text{rms}} = \sqrt{\frac{1213.02 \times 10^{-23}}{5 \times 10^{-17}}}$
$V_{\text{rms}} = \sqrt{242.604 \times 10^{-6}}$
$V_{\text{rms}} \approx 15.57 \times 10^{-3} \, m/s$
Converting to $mm/s$ $(1 \, m/s = 1000 \, mm/s)$:
$V_{\text{rms}} \approx 15.57 \, mm/s \approx 15 \, mm/s$.

(D) The atmosphere on a planet is possible only if the root mean square speed of the gas molecules $(v_{rms})$ is significantly less than the escape speed $(v_e)$ of the planet.
If $v_{rms} \geq v_e$,the gas molecules possess enough kinetic energy to overcome the gravitational pull of the planet and will escape into space.
Therefore,for a planet to retain an atmosphere,the condition must be $v_{rms} < v_e$.

(A) The Maxwell-Boltzmann speed distribution curve describes the distribution of speeds of molecules in a gas at a given temperature.
The most probable speed $V_p$ is given by the formula $V_p = \sqrt{\frac{2RT}{M}}$,which implies $V_p \propto \sqrt{T}$.
As the temperature $T$ increases,the most probable speed $V_p$ increases,meaning the peak of the curve shifts towards the right (higher speeds) along the $V$-axis.
Additionally,since the total area under the curve (representing the total number of molecules) must remain constant,an increase in temperature causes the curve to flatten and broaden.
Comparing the given options,the graph in option $A$ correctly shows that for $T_1 > T_2$,the curve for $T_1$ has its peak at a higher speed and is broader than the curve for $T_2$.

(D) The correct answer is $D$.
Maxwell's speed distribution curve is asymmetric (skewed to the right) because the speed of gas molecules ranges from $0$ to $\infty$.
Since the lowest possible speed is $0$ and the highest is $\infty$,the curve cannot be symmetric about the most probable speed $(V_{mp})$.
Option $A$ is correct as the distribution depends on temperature $T$.
Option $B$ is correct as the relationship between root mean square speed,average speed,and most probable speed is $V_{rms} > V_{av} > V_{mp}$.
Option $C$ is correct as the integral of the distribution function over all speeds gives the total number of molecules $N$.

(B) The $r.m.s.$ speed of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $R$ and $M$ are constant,$V_{rms} \propto \sqrt{T}$.
Let $V_1$ be the $r.m.s.$ speed at $STP$ $(T_1 = 273 \, K)$ and $V_2$ be the $r.m.s.$ speed at temperature $T_2$.
Given that $V_2 = V_1 + 10\% \text{ of } V_1 = 1.1 V_1$.
Using the relation $\frac{V_2}{V_1} = \sqrt{\frac{T_2}{T_1}}$,we get $1.1 = \sqrt{\frac{T_2}{273}}$.
Squaring both sides,$(1.1)^2 = \frac{T_2}{273}$,which gives $1.21 = \frac{T_2}{273}$.
$T_2 = 1.21 \times 273 = 330.33 \, K$.
To convert to Celsius: $T(^{\circ}C) = T(K) - 273.15 = 330.33 - 273.15 \approx 57.18^{\circ} C$.
Rounding to the nearest provided option,the correct answer is $57.3^{\circ} C$.

(C) The velocity of sound in a gas is given by $v_s = \sqrt{\frac{\gamma RT}{M}}$.
The $r.m.s.$ velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Taking the ratio of the two equations:
$\frac{v_s}{v_{rms}} = \sqrt{\frac{\gamma RT / M}{3RT / M}} = \sqrt{\frac{\gamma}{3}}$.
Given $\gamma = 1.5$ and $v_s = 600 \, m/s$,we substitute these values:
$\frac{600}{v_{rms}} = \sqrt{\frac{1.5}{3}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Solving for $v_{rms}$:
$v_{rms} = 600\sqrt{2} \, m/s$.

(B) The average speed $v_{av}$ is calculated as:
$v_{av} = \frac{500 + 600 + 700 + 800 + 900}{5} = \frac{3500}{5} = 700 \ m/s$
The root mean square speed $v_{rms}$ is calculated as:
$v_{rms} = \sqrt{\frac{500^2 + 600^2 + 700^2 + 800^2 + 900^2}{5}}$
$v_{rms} = \sqrt{\frac{250000 + 360000 + 490000 + 640000 + 810000}{5}}$
$v_{rms} = \sqrt{\frac{2550000}{5}} = \sqrt{510000} \approx 714.14 \ m/s$
Rounding to the nearest integer,$v_{rms} \approx 714 \ m/s$.
Thus,the root mean square speed is $714 - 700 = 14 \ m/s$ higher than the average speed.

(C) The root mean square (rms) 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 the formula,it is clear that $V_{RMS}$ is directly proportional to the square root of the absolute temperature:
$V_{RMS} \propto \sqrt{T}$
Therefore,the correct option is $C$.

(B) The root mean square (rms) speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
The average speed of gas molecules is given by $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.
According to the problem,$v_{rms} = \sqrt{\frac{\alpha+5}{\alpha}} \times v_{avg}$.
Substituting the formulas: $\sqrt{\frac{3RT}{M}} = \sqrt{\frac{\alpha+5}{\alpha}} \times \sqrt{\frac{8RT}{\pi M}}$.
Squaring both sides: $\frac{3RT}{M} = \left(\frac{\alpha+5}{\alpha}\right) \times \frac{8RT}{\pi M}$.
Canceling $\frac{RT}{M}$ from both sides: $3 = \left(\frac{\alpha+5}{\alpha}\right) \times \frac{8}{\pi}$.
Given $\pi = \frac{22}{7}$,so $\frac{8}{\pi} = \frac{8 \times 7}{22} = \frac{56}{22} = \frac{28}{11}$.
Thus,$3 = \left(\frac{\alpha+5}{\alpha}\right) \times \frac{28}{11}$.
$33\alpha = 28(\alpha + 5)$.
$33\alpha = 28\alpha + 140$.
$5\alpha = 140$.
$\alpha = 28$.

(C) According to the ideal gas law $PV = nRT$,since $P$,$V$,and $T$ are the same for all vessels,the number of moles $n$ is the same. By Avogadro's law,the number of molecules is the same in all vessels.
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,and $M$ is the molar mass of the gas.
Since $R$ and $T$ are constant,$V_{rms} \propto \frac{1}{\sqrt{M}}$.
The molar masses are:
$M_{He} = 4 \text{ g/mol}$
$M_{F_2} = 38 \text{ g/mol}$
$M_{SF_6} = 146 \text{ g/mol}$
Since helium has the smallest molar mass,it will have the largest root mean square speed.

(D) The root mean square (r.m.s.) speed of an ideal gas is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M}}$.
From this relation,we can see that $V_{rms} \propto \sqrt{T}$.
Let $T_1 = 200 \ K$ and $T_2 = 800 \ K$.
Given $V_{rms,1} = v_0$.
We have the ratio: $\frac{V_{rms,2}}{V_{rms,1}} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{V_{rms,2}}{v_0} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2$.
Therefore,$V_{rms,2} = 2 v_0$.