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

(C) The root mean square velocity of gas molecules is given by the relation $v_{rms} \propto \sqrt{T}$.
Given that the initial velocity is $v_1 = v$ at temperature $T_1 = 120\,K$,and the final velocity is $v_2 = 2v$ at temperature $T_2$.
Using the proportionality,we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{2v}{v} = \sqrt{\frac{T_2}{120}}$.
$2 = \sqrt{\frac{T_2}{120}}$.
Squaring both sides: $4 = \frac{T_2}{120}$.
Therefore,$T_2 = 4 \times 120 = 480\,K$.

(D) The $r.m.s.$ speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}.$
Since $R$ and $M$ are constant,$v_{rms} \propto \sqrt{T}.$
Given initial temperature $T_1 = 0^\circ C = 273 \ K$ and initial velocity $(v_{rms})_1 = 0.5 \ km \ sec^{-1}.$
Final temperature $T_2 = 819^\circ C = 819 + 273 = 1092 \ K.$
Using the ratio formula: $\frac{(v_{rms})_1}{(v_{rms})_2} = \sqrt{\frac{T_1}{T_2}}.$
Substituting the values: $\frac{0.5}{(v_{rms})_2} = \sqrt{\frac{273}{1092}}.$
Since $\frac{1092}{273} = 4,$ we have $\frac{0.5}{(v_{rms})_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}.$
Therefore,$(v_{rms})_2 = 0.5 \times 2 = 1 \ km \ sec^{-1}.$

(A) The $r.m.s.$ speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$, we have $\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$.
At $N.T.P.$, the temperature $T_1 = 273 \, K$.
We want the speed to double, so $v_2 = 2v_1$.
Substituting the values: $\frac{v_1}{2v_1} = \sqrt{\frac{273}{T_2}}$.
Squaring both sides: $\frac{1}{4} = \frac{273}{T_2}$.
$T_2 = 4 \times 273 = 1092 \, K$.
To convert to Celsius: $T(^{\circ} C) = 1092 - 273 = 819 \, ^{\circ} C$.

(B) The $r.m.s.$ speed of a 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,at $T_1 = 400\,K$,the speed is $v_1 = v$.
We want to find the temperature $T_2$ at which the speed becomes $v_2 = 2v$.
Using the proportionality $v_{rms} \propto \sqrt{T}$,we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the given values: $\frac{2v}{v} = \sqrt{\frac{T_2}{400}}$.
Squaring both sides: $2^2 = \frac{T_2}{400}$.
$4 = \frac{T_2}{400}$.
Therefore,$T_2 = 4 \times 400 = 1600\,K$.

(B) 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 $T$ is constant,we have ${v_{rms}} \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the ${v_{rms}}$ of ${O_2}$ molecules to that of ${H_2}$ molecules is:
$\frac{{v_{rms, O_2}}}{{v_{rms, H_2}}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}}$
Given that the molar mass of ${H_2}$ is $M_{H_2} = 2 \text{ g/mol}$ and the molar mass of ${O_2}$ is $M_{O_2} = 32 \text{ g/mol}$.
Substituting these values:
$\frac{{v_{rms, O_2}}}{{v_{rms, H_2}}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.

(A) The average speed of a gas molecule is given by the formula $v_{av} = \sqrt{\frac{8RT}{\pi M}}$.
Since $R$ and $T$ are constant for both gases,we have $v_{av} \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the average speeds is $\frac{v_{Gas}}{v_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{Gas}}}$.
Given that the average speed of the gas is $4$ times that of $SO_2$,we have $\frac{v_{Gas}}{v_{SO_2}} = 4$.
Substituting the values: $4 = \sqrt{\frac{64}{M_{Gas}}}$.
Squaring both sides: $16 = \frac{64}{M_{Gas}}$.
Solving for $M_{Gas}$: $M_{Gas} = \frac{64}{16} = 4$.
$A$ gas with a molecular mass of $4$ is Helium $(He)$.

(D) The formula for the root mean square speed $(v_{rms})$ of an ideal gas is given by $v_{rms} = \sqrt{\frac{3P}{\rho}}$,where $P$ is the pressure and $\rho$ is the density.
Rearranging the formula to solve for pressure $P$,we get $P = \frac{v_{rms}^2 \rho}{3}$.
Given $v_{rms} = 3180 \ m/s$ and $\rho = 8.99 \times 10^{-2} \ kg/m^3$.
Substituting the values: $P = \frac{(3180)^2 \times 8.99 \times 10^{-2}}{3}$.
$P = \frac{10112400 \times 0.0899}{3} \approx \frac{909104.76}{3} \approx 303034.9 \ N/m^2$.
Since $1 \ atm = 1.01 \times 10^5 \ N/m^2$,then $P \approx \frac{303034.9}{101000} \approx 3 \ atm$.

(D) For an ideal gas,the speeds are defined as:
${v_{rms}} = \sqrt{\frac{3kT}{m}}$,${v_p} = \sqrt{\frac{2kT}{m}}$,and $\bar{v} = \sqrt{\frac{8kT}{\pi m}}$.
Comparing these values:
${v_p} = \sqrt{2} \sqrt{\frac{kT}{m}} \approx 1.414 \sqrt{\frac{kT}{m}}$
$\bar{v} = \sqrt{\frac{8}{3.14}} \sqrt{\frac{kT}{m}} \approx 1.596 \sqrt{\frac{kT}{m}}$
${v_{rms}} = \sqrt{3} \sqrt{\frac{kT}{m}} \approx 1.732 \sqrt{\frac{kT}{m}}$
Thus,${v_p} < \bar{v} < {v_{rms}}$,which confirms option $(c)$ is correct.
For the average kinetic energy:
${E_{av}} = \frac{1}{2} m v_{rms}^2 = \frac{1}{2} m \left( \frac{3kT}{m} \right) = \frac{3}{2} kT$.
Since ${v_p}^2 = \frac{2kT}{m}$,we have $kT = \frac{1}{2} m v_p^2$.
Substituting this into the energy equation:
${E_{av}} = \frac{3}{2} \left( \frac{1}{2} m v_p^2 \right) = \frac{3}{4} m v_p^2$,which confirms option $(b)$ is correct.
Therefore,both $(b)$ and $(c)$ are correct.

(C) The root mean square speed $(V_{rms})$ 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.
Initially,for the diatomic gas,the speed is $v = \sqrt{\frac{3RT}{M}}$.
When the temperature is doubled,the new temperature becomes $T' = 2T$.
When the diatomic molecules dissociate into two atoms,the molar mass of the new species (atoms) becomes half of the original molar mass,so $M' = \frac{M}{2}$.
The new root mean square speed $V'_{rms}$ is given by:
$V'_{rms} = \sqrt{\frac{3RT'}{M'}} = \sqrt{\frac{3R(2T)}{M/2}} = \sqrt{4 \cdot \frac{3RT}{M}} = 2 \sqrt{\frac{3RT}{M}}$.
Substituting the initial speed $v$,we get $V'_{rms} = 2v$.

(D) Let the mass of a molecule in the $L$ part be $m_1$ and in the $R$ part be $m_2$. Since the separator is diathermic,the temperature $T$ of both gases is the same.
The rms speed of molecules in the $L$ part is $v_{rms} = \sqrt{\frac{3 k_B T}{m_1}}$.
The mean speed of molecules in the $R$ part is $v_{mean} = \sqrt{\frac{8 k_B T}{\pi m_2}}$.
Given that $v_{rms} = v_{mean}$,we have:
$\sqrt{\frac{3 k_B T}{m_1}} = \sqrt{\frac{8 k_B T}{\pi m_2}}$
Squaring both sides:
$\frac{3 k_B T}{m_1} = \frac{8 k_B T}{\pi m_2}$
$\frac{3}{m_1} = \frac{8}{\pi m_2}$
Rearranging to find the ratio $\frac{m_1}{m_2}$:
$\frac{m_1}{m_2} = \frac{3\pi}{8}$

(D) The escape velocity from the Earth's surface is $v_{e} = 11.2 \, km/s = 11.2 \times 10^3 \, m/s$.
The $r.m.s.$ speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Equating the two,we have $\sqrt{\frac{3RT}{M}} = v_{e}$.
Squaring both sides,$\frac{3RT}{M} = v_{e}^2$,which implies $T = \frac{M v_{e}^2}{3R}$.
For hydrogen gas $(H_2)$,the molar mass $M = 2 \times 10^{-3} \, kg/mol$ and the universal gas constant $R = 8.31 \, J/(mol \cdot K)$.
Substituting the values: $T = \frac{(2 \times 10^{-3}) \times (11.2 \times 10^3)^2}{3 \times 8.31}$.
$T = \frac{2 \times 10^{-3} \times 125.44 \times 10^6}{24.93} = \frac{250880}{24.93} \approx 10063 \, K$.

(B) The root mean square speed of gas molecules is given by the formula:
$v_{rms} = \sqrt{\frac{3RT}{M}}$
Squaring both sides,we get:
$v_{rms}^2 = \frac{3RT}{M}$
Since $R$ (universal gas constant) and $M$ (molar mass of the gas) are constants,we can write:
$v_{rms}^2 \propto T$
This equation represents a linear relationship of the form $y = mx$,where $y = v_{rms}^2$,$x = T$,and the slope $m = \frac{3R}{M}$.
Therefore,the graph between $v_{rms}^2$ and $T$ is a straight line passing through the origin.
This corresponds to graph $B$.

(D) The root mean square $(rms)$ velocity $(u)$ of an ideal gas is given by the formula:
$u = \sqrt{\frac{3P}{d}}$
where $P$ is the pressure and $d$ is the density of the gas.
Since the pressure $(P)$ is kept constant,the relationship between the $rms$ velocity $(u)$ and the density $(d)$ is:
$u \propto \frac{1}{\sqrt{d}}$
Therefore,the $rms$ velocity varies inversely with the square root of the density.

(B) In an ideal gas,the velocity vectors of molecules are distributed randomly in all directions.
Due to this symmetry,the average velocity vector $\vec{v}$ is zero,i.e.,$\langle \vec{v} \rangle = 0$.
Similarly,for any odd power of velocity,such as $\langle v^3 \rangle$ or $\langle v^5 \rangle$,the average value is zero because the positive and negative components cancel each other out.
However,for even powers of velocity,such as $\langle v^2 \rangle$ or $\langle v^4 \rangle$,the values are always positive because the square or fourth power of any velocity component is non-negative.
Therefore,$\langle v^4 \rangle$ will not be zero.

(D) The $rms$ velocity $(v_{rms})$ of a gas is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $M$ is the molar mass.
Since the number of moles $n = \frac{m}{M}$,where $m$ is the total mass of the gas,we can write $M = \frac{m}{n}$.
Substituting this into the $rms$ velocity formula: $v_{rms} = \sqrt{\frac{3RT}{(m/n)}} = \sqrt{\frac{3RTn}{m}}$.
For a fixed amount of gas at a constant temperature,$T$,$R$,and $n$ are constants.
Therefore,$v_{rms} \propto \frac{1}{\sqrt{m}}$.

(B) The expressions for the different velocities of gas molecules are given by:
$V_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{3} \cdot \sqrt{\frac{RT}{M}} \approx 1.732 \sqrt{\frac{RT}{M}}$
$V_{av} = \sqrt{\frac{8RT}{\pi M}} = \sqrt{\frac{8}{\pi}} \cdot \sqrt{\frac{RT}{M}} \approx 1.596 \sqrt{\frac{RT}{M}}$
$V_{mp} = \sqrt{\frac{2RT}{M}} = \sqrt{2} \cdot \sqrt{\frac{RT}{M}} \approx 1.414 \sqrt{\frac{RT}{M}}$
Comparing the coefficients,we find that $1.732 > 1.596 > 1.414$.
Therefore,the correct relation is $\nu_{rms} > \nu_{av} > \nu_{mp}$.

(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 absolute temperature,and $M$ is the molar mass of the gas.
Since $R$ and $T$ are constant for all gases at a given temperature,we have $v_{rms} \propto \frac{1}{\sqrt{M}}$.
The molar masses are: $M_H = 2 \ g/mol$,$M_N = 28 \ g/mol$,and $M_O = 32 \ g/mol$.
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,$V_H > V_N > V_O$.

(A) The root mean square $(RMS)$ 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 absolute temperature,and $M$ is the molar mass of the gas.
From this formula,it is clear that $v_{rms}$ depends only on the temperature $T$ and the molar mass $M$ of the gas.
Since the temperature $T$ is kept constant and the molar mass $M$ is a property of the gas,the $RMS$ speed remains unchanged even if the pressure is doubled.
Therefore,the correct option is $A$.

(C) The $rms$ speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$,we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Given $v_2 = 2v_1$,so $\frac{2v_1}{v_1} = \sqrt{\frac{T_2}{T_1}}$,which implies $2 = \sqrt{\frac{T_2}{T_1}}$.
Squaring both sides,$4 = \frac{T_2}{T_1}$,so $T_2 = 4T_1$.
The initial temperature $T_1 = 0^{\circ}C = 273 \ K$.
Therefore,$T_2 = 4 \times 273 \ K = 1092 \ K$.
To convert back to Celsius: $t_2 = 1092 - 273 = 819^{\circ}C$.

(D) The root mean square velocity of an ideal gas is given by the formula $v_{rms} = \sqrt{\frac{3P}{d}}$,where $P$ is the pressure and $d$ is the density of the gas.
Given that the pressure $P$ is constant,we can see that $v_{rms} \propto \frac{1}{\sqrt{d}}$.
Therefore,the correct option is $D$.

(A) The root mean square $(rms)$ 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 absolute temperature,and $M$ is the molar mass of the gas.
Since the temperature $T$ is the same for both gases,$v_{rms} \propto \frac{1}{\sqrt{M}}$.
The molar mass of $H_2$ $(M_{H_2} = 2 \ g/mol)$ is much smaller than the molar mass of $O_2$ $(M_{O_2} = 32 \ g/mol)$.
Therefore,$v_{rms}$ of $H_2$ is greater than $v_{rms}$ of $O_2$.
Regarding average kinetic energy,the formula is $K_{avg} = \frac{3}{2}kT$,which depends only on temperature. Since both gases are at the same temperature,their average kinetic energies are equal.

(A) The formula for $r.m.s.$ speed is given by $\upsilon_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $\upsilon_{rms} \propto \sqrt{\frac{T}{M}}$,we can equate the speeds for hydrogen and oxygen:
$\sqrt{\frac{T_{H_2}}{M_{H_2}}} = \sqrt{\frac{T_{O_2}}{M_{O_2}}}$
Given $T_{O_2} = 47^{\circ}C = 273 + 47 = 320 \ K$,$M_{H_2} = 2 \ g/mol$,and $M_{O_2} = 32 \ g/mol$.
Substituting the values:
$\frac{T_{H_2}}{2} = \frac{320}{32}$
$\frac{T_{H_2}}{2} = 10$
$T_{H_2} = 20 \ K$.

(B) The $r.m.s.$ velocity of a gas is given by the formula: $\upsilon_{rms} = \sqrt{\frac{3RT}{M}}$.
Since the temperature $T$ is constant,$\upsilon_{rms} \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass of the gas.
The molar masses are: $M_H = 2 \ g/mol$,$M_N = 28 \ g/mol$,and $M_O = 32 \ g/mol$.
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,$V_H > V_N > V_O$.

(D) The $rms$ speed of a gas is given by the formula $v = \sqrt{\frac{3RT}{M}}$,where $M$ is the molar mass of the gas.
For the diatomic gas at temperature $T$,the speed is $v = \sqrt{\frac{3RT}{M}}$.
When the gas dissociates into atoms,the molar mass becomes $M' = \frac{M}{2}$.
Let the new temperature be $T'$. The new $rms$ speed is $2v = \sqrt{\frac{3RT'}{M/2}}$.
Squaring both sides of the equation for the new speed: $(2v)^2 = \frac{3RT'}{M/2} \Rightarrow 4v^2 = \frac{6RT'}{M}$.
Since $v^2 = \frac{3RT}{M}$,we substitute this into the equation: $4 \left( \frac{3RT}{M} \right) = \frac{6RT'}{M}$.
Canceling common terms $\frac{3R}{M}$ from both sides,we get $4T = 2T'$,which simplifies to $T' = 2T$.

(D) The initial temperature $T_1 = -73^{\circ}C = (-73 + 273) \, K = 200 \, K$.
The $rms$ speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.
Let the initial $rms$ speed be $v$ and the final $rms$ speed be $2v$.
Using the ratio: $\frac{v_{rms,2}}{v_{rms,1}} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{2v}{v} = \sqrt{\frac{T_2}{200}}$.
Squaring both sides: $4 = \frac{T_2}{200}$.
Therefore,$T_2 = 800 \, K$.
Converting back to Celsius: $T_2(^{\circ}C) = 800 - 273 = 527^{\circ}C$.

(B) The $rms$ speed of a gas 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 $rms$ speed at $S.T.P.$ $(T_1 = 273 \ K)$ and $v_2$ be the $rms$ speed at temperature $T_2$.
Given $v_2 = 2v_1$.
Therefore,$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Squaring both sides,we get $\left(\frac{v_2}{v_1}\right)^2 = \frac{T_2}{T_1}$.
Substituting the values: $(2)^2 = \frac{T_2}{273}$.
$4 = \frac{T_2}{273} \implies T_2 = 4 \times 273 = 1092 \ K$.
To convert to Celsius: $T(^oC) = 1092 - 273 = 819 \ ^oC$.

(A) The $r.m.s.$ velocity of an ideal gas is given by the formula $\upsilon_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $\upsilon_{rms} \propto \sqrt{T}$.
Given initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$.
Given final temperature $T_2 = 927^{\circ}C = 927 + 273 = 1200 \ K$.
Using the ratio $\frac{\upsilon_2}{\upsilon_1} = \sqrt{\frac{T_2}{T_1}}$,we get $\frac{\upsilon_2}{\upsilon_1} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2$.
Therefore,$\upsilon_2 = 2\upsilon_1$,meaning the $r.m.s.$ velocity doubles.

(D) The root mean square $(rms)$ speed is defined as $\upsilon_{rms} = \sqrt{\langle \upsilon^2 \rangle}$.
This is calculated as $\upsilon_{rms} = \sqrt{\frac{\upsilon_1^2 + \upsilon_2^2 + \upsilon_3^2 + \upsilon_4^2 + \upsilon_5^2}{n}}$.
Substituting the given values: $\upsilon_{rms} = \sqrt{\frac{2^2 + 3^2 + 4^2 + 5^2 + 6^2}{5}}$.
$\upsilon_{rms} = \sqrt{\frac{4 + 9 + 16 + 25 + 36}{5}}$.
$\upsilon_{rms} = \sqrt{\frac{90}{5}} = \sqrt{18}$.
$\upsilon_{rms} \approx 4.24$ units.

(D) Since the partition is diathermic,the temperature $T$ of both gases is the same.
Given that the $rms$ speed of molecules in the $L$ part equals the average speed of molecules in the $R$ part:
$V_{rms, L} = V_{avg, R}$
$\sqrt{\frac{3kT}{m_L}} = \sqrt{\frac{8kT}{\pi m_R}}$
Squaring both sides:
$\frac{3kT}{m_L} = \frac{8kT}{\pi m_R}$
$\frac{3}{m_L} = \frac{8}{\pi m_R}$
Rearranging to find the ratio $\frac{m_L}{m_R}$:
$\frac{m_L}{m_R} = \frac{3\pi}{8}$

(C) The root mean square velocity $(v_{rms})$ 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.
Since the temperature $(T)$ remains constant and the nature of the gas $(M)$ does not change,the $v_{rms}$ depends only on the temperature.
Even if the amount of gas in the container decreases due to leakage,the temperature remains constant.
Therefore,the $v_{rms}$ of the remaining gas molecules remains unchanged at $400 \ ms^{-1}$.

(A) Given: Temperature $T = 27^{\circ}C = 27 + 273 = 300 \, K$.
Root mean square speed $\nu_{rms} = 1930 \, m/s$.
Gas constant $R = 8.3 \, J/mol \cdot K$.
The formula for $\nu_{rms}$ is $\nu_{rms} = \sqrt{\frac{3RT}{M}}$.
Squaring both sides,we get $\nu_{rms}^2 = \frac{3RT}{M}$.
Rearranging for molar mass $M$,we get $M = \frac{3RT}{\nu_{rms}^2}$.
Substituting the values: $M = \frac{3 \times 8.3 \times 300}{(1930)^2}$.
$M = \frac{7470}{3724900} \approx 0.002 \, kg/mol = 2 \, g/mol$.
The molar mass of $H_2$ is $2 \, g/mol$.
Therefore,the gas is $H_2$.

(D) The $r.m.s.$ velocity of a gas is given by the formula $\upsilon_{rms} = \sqrt{\frac{3RT}{M}}$,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 both gases,we have $\upsilon_{rms} \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the $r.m.s.$ velocities of oxygen $(O_2)$ and hydrogen $(H_2)$ is given by:
$\frac{(\upsilon_{rms})_{O_2}}{(\upsilon_{rms})_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}}$.
The molar mass of hydrogen $(H_2)$ is $M_{H_2} = 2 \ g/mol$ and the molar mass of oxygen $(O_2)$ is $M_{O_2} = 32 \ g/mol$.
Substituting these values,we get:
$\frac{(\upsilon_{rms})_{O_2}}{(\upsilon_{rms})_{H_2}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
Thus,the ratio is $1:4$.

(D) The root mean square $(rms)$ 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: $v_1 = 2, v_2 = 3, v_3 = 4, v_4 = 5, v_5 = 6$ and $N = 5$.
Calculation: $v_{rms} = \sqrt{\frac{2^2 + 3^2 + 4^2 + 5^2 + 6^2}{5}}$
$v_{rms} = \sqrt{\frac{4 + 9 + 16 + 25 + 36}{5}}$
$v_{rms} = \sqrt{\frac{90}{5}}$
$v_{rms} = \sqrt{18} \approx 4.24$ units.

(A) The formula for $rms$ speed is given by $v_{rms} = \sqrt{\frac{3RT}{M_w}}$,where $R = 8.31 \ J/(mol \cdot K)$,$T = 27 + 273 = 300 \ K$,and $v_{rms} = 412 \ m/s$.
Squaring both sides,we get $v_{rms}^2 = \frac{3RT}{M_w}$.
Rearranging for molar mass $M_w$: $M_w = \frac{3RT}{v_{rms}^2}$.
Substituting the values: $M_w = \frac{3 \times 8.31 \times 300}{412^2} \approx \frac{7479}{169744} \approx 0.044 \ kg/mol = 44 \ g/mol$.
The molar mass of $CO_2$ is $12 + (2 \times 16) = 44 \ g/mol$. Therefore,the gas is $CO_2$.

(D) The formula for $rms$ speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that the $rms$ speed of $H_2$ at temperature $T$ is equal to the $rms$ speed of $N_2$ at $T_2 = 27^{\circ}C = 300 \ K$.
So,$\sqrt{\frac{3RT}{M_{H_2}}} = \sqrt{\frac{3RT_2}{M_{N_2}}}$.
Squaring both sides: $\frac{T}{M_{H_2}} = \frac{T_2}{M_{N_2}}$.
Given $M_{N_2} = 14 \times M_{H_2}$.
Substituting the values: $\frac{T}{M_{H_2}} = \frac{300}{14 \times M_{H_2}}$.
$T = \frac{300}{14} \approx 21.4 \ K$.

(B) The formula for $rms$ speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that the $rms$ speed of hydrogen $(H_2)$ equals the $rms$ speed of oxygen $(O_2)$:
$\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$
Squaring both sides and cancelling common terms $(3R)$:
$\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}$
Here,$M_{H_2} = 2 \ g/mol$,$M_{O_2} = 32 \ g/mol$,and $T_{O_2} = 47 + 273 = 320 \ K$.
Substituting the values:
$\frac{T_{H_2}}{2} = \frac{320}{32}$
$\frac{T_{H_2}}{2} = 10$
$T_{H_2} = 20 \ K$.

(B) The formula for root mean square speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
For hydrogen at $T_1 = 300 \ K$ and $M_1 = 2 \ g/mol$:
$(v_{rms})_H = \sqrt{\frac{3RT_1}{M_1}} = 1930 \ m/s \implies 1930 = \sqrt{\frac{3R(300)}{2}} \quad \dots(1)$
For oxygen at $T_2 = 900 \ K$ and $M_2 = 32 \ g/mol$:
$(v_{rms})_O = \sqrt{\frac{3RT_2}{M_2}} = \sqrt{\frac{3R(900)}{32}}$
$(v_{rms})_O = \sqrt{\frac{3R(300) \times 3}{16 \times 2}} = \sqrt{3} \times \frac{1}{4} \times \sqrt{\frac{3R(300)}{2}}$
Substituting equation $(1)$:
$(v_{rms})_O = \frac{\sqrt{3}}{4} \times 1930 = \frac{1.732}{4} \times 1930 \approx 836 \ m/s$.

(D) The $rms$ velocity of a gas is given by $v_{rms} = \sqrt{\frac{3kT}{M}}$,where $M$ is the mass of a single molecule.
For gas $A$,the mass of a molecule is $m$. Thus,$v_{rms, A}^2 = \frac{3kT}{m}$.
The mean square velocity is $\langle v^2 \rangle = v_{rms}^2$. The $x$-component of the mean square velocity is $\langle v_x^2 \rangle = \frac{1}{3} \langle v^2 \rangle = \frac{1}{3} \frac{3kT}{m} = \frac{kT}{m}$. So,$w^2 = \frac{kT}{m}$.
For gas $B$,the mass of a molecule is $2m$. Thus,$v^2 = v_{rms, B}^2 = \frac{3kT}{2m}$.
Now,the ratio $\frac{w^2}{v^2} = \frac{kT/m}{3kT/2m} = \frac{1}{3/2} = \frac{2}{3} \approx 0.67$.

(A) The $rms$ velocity of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
From this formula,it is clear that $v_{rms} \propto \sqrt{T}$.
The $rms$ velocity is independent of the pressure of the gas.
Given: $T_1 = 27^{\circ}C = 300 \, K$,$v_1 = 200 \, m \, s^{-1}$.
New temperature: $T_2 = 127^{\circ}C = 400 \, K$.
Using the proportionality $v_2 = v_1 \sqrt{\frac{T_2}{T_1}}$:
$v_2 = 200 \sqrt{\frac{400}{300}} = 200 \sqrt{\frac{4}{3}} = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \, m \, s^{-1}$.

(B) According to the definition of $rms$ (root-mean-square) speed:
$\upsilon_{rms} = \sqrt{\langle \upsilon^2 \rangle}$
Here,the mean square speed $\langle \upsilon^2 \rangle$ is given by:
$\langle \upsilon^2 \rangle = \frac{\upsilon_1^2 + \upsilon_2^2 + \upsilon_3^2 + \dots + \upsilon_n^2}{n}$
Therefore,the $rms$ speed is:
$\upsilon_{rms} = [\frac{\upsilon_1^2 + \upsilon_2^2 + \upsilon_3^2 + \dots + \upsilon_n^2}{n}]^{1/2}$

(D) The formula for the $rms$ speed of gas molecules is given by: $v_{rms} = \sqrt{\frac{3RT}{M_w}}$
Given:
Temperature $T = 300 \ K$
Universal gas constant $R = 8.31 \ J/(mol \cdot K)$
Molar mass of hydrogen gas $(H_2)$ $M_w = 2 \ g/mol = 2 \times 10^{-3} \ kg/mol$
Substituting the values into the formula:
$v_{rms} = \sqrt{\frac{3 \times 8.31 \times 300}{2 \times 10^{-3}}}$
$v_{rms} = \sqrt{\frac{7479}{2 \times 10^{-3}}}$
$v_{rms} = \sqrt{3739500}$
$v_{rms} \approx 1933.7 \ m/s$
Rounding to the appropriate significant figures,we get:
$v_{rms} \approx 1.93 \times 10^3 \ m/s$

(D) The root mean square $(rms)$ velocity is defined as the square root of the mean of the squares of the individual velocities.
Formula: $\upsilon_{rms} = \sqrt{\frac{\upsilon_1^2 + \upsilon_2^2 + \upsilon_3^2 + \upsilon_4^2 + \upsilon_5^2}{N}}$
Given velocities: $\upsilon_1 = 2, \upsilon_2 = 3, \upsilon_3 = 4, \upsilon_4 = 5, \upsilon_5 = 6$ and $N = 5$.
Calculation: $\upsilon_{rms} = \sqrt{\frac{2^2 + 3^2 + 4^2 + 5^2 + 6^2}{5}}$
$\upsilon_{rms} = \sqrt{\frac{4 + 9 + 16 + 25 + 36}{5}}$
$\upsilon_{rms} = \sqrt{\frac{90}{5}}$
$\upsilon_{rms} = \sqrt{18}$
$\upsilon_{rms} \approx 4.24$

(D) The escape velocity from the Earth is $v_{e} = 11.2 \, km/s = 11.2 \times 10^{3} \, m/s$.
The $r.m.s.$ speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Equating $v_{rms}$ to the escape velocity $v_{e}$:
$v_{e} = \sqrt{\frac{3RT}{M}}$
Squaring both sides:
$v_{e}^{2} = \frac{3RT}{M}$
Solving for $T$:
$T = \frac{M v_{e}^{2}}{3R}$
Given for hydrogen $(H_{2})$,the molar mass $M = 2 \times 10^{-3} \, kg/mol$ and the universal gas constant $R = 8.31 \, J/(mol \cdot K)$.
Substituting the values:
$T = \frac{(2 \times 10^{-3}) \times (11.2 \times 10^{3})^{2}}{3 \times 8.31}$
$T = \frac{2 \times 10^{-3} \times 125.44 \times 10^{6}}{24.93}$
$T = \frac{250.88 \times 10^{3}}{24.93} \approx 10063 \, K$.

(C) The average speed of gas molecules is given by the formula $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.
For hydrogen $(H_2)$,the molar mass $M_{H_2} = 2 \times 10^{-3} \ kg/mol$.
For oxygen $(O_2)$,the molar mass $M_{O_2} = 32 \times 10^{-3} \ kg/mol$.
Given the temperature of oxygen $T_{O_2} = 31 + 273 = 304 \ K$.
Equating the average speeds: $\sqrt{\frac{8RT_{H_2}}{\pi M_{H_2}}} = \sqrt{\frac{8RT_{O_2}}{\pi M_{O_2}}}$.
Squaring both sides and simplifying: $\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}$.
Substituting the values: $\frac{T_{H_2}}{2} = \frac{304}{32}$.
$T_{H_2} = \frac{304 \times 2}{32} = \frac{304}{16} = 19 \ K$.
Converting to Celsius: $T(^oC) = 19 - 273 = -254^oC$.

(A) The formula for $rms$ speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that the $rms$ speed of hydrogen $(H_2)$ is equal to the $rms$ speed of oxygen $(O_2)$:
$\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$
Squaring both sides and canceling common terms $(3R)$:
$\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}$
Given $T_{O_2} = 47^{\circ}C = 47 + 273 = 320 \ K$.
Molar mass of $H_2$ $(M_{H_2})$ = $2 \ g/mol$.
Molar mass of $O_2$ $(M_{O_2})$ = $32 \ g/mol$.
Substituting the values:
$\frac{T_{H_2}}{2} = \frac{320}{32}$
$\frac{T_{H_2}}{2} = 10$
$T_{H_2} = 20 \ K$.

(D) The $r.m.s.$ velocity of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M_w}}$.
Given,$v_{rms} = v_e = 11.2 \times 10^3 \ m/s$.
The molar mass of hydrogen $(H_2)$ is $M_w = 2 \times 10^{-3} \ kg/mol$.
Equating the two: $\sqrt{\frac{3RT}{M_w}} = 11.2 \times 10^3$.
Squaring both sides: $\frac{3RT}{M_w} = (11.2 \times 10^3)^2$.
$T = \frac{(11.2 \times 10^3)^2 \times M_w}{3R}$.
Substituting $R = 8.314 \ J/(mol \cdot K)$ and $M_w = 2 \times 10^{-3} \ kg/mol$:
$T = \frac{125.44 \times 10^6 \times 2 \times 10^{-3}}{3 \times 8.314} \approx \frac{250.88 \times 10^3}{24.942} \approx 10058 \ K$.
Rounding to the nearest provided option,$T \approx 10075 \ K$.

(D) The $rms$ velocity of the gas on the left side is given by: $\upsilon_{rms} = \sqrt{\frac{3KT}{m_L}}$.
The average velocity of the gas on the right side is given by: $\upsilon_{avg} = \sqrt{\frac{8KT}{\pi m_R}}$.
According to the problem,the $rms$ velocity of the left gas equals the average velocity of the right gas:
$\sqrt{\frac{3KT}{m_L}} = \sqrt{\frac{8KT}{\pi m_R}}$.
Squaring both sides:
$\frac{3KT}{m_L} = \frac{8KT}{\pi m_R}$.
Canceling $KT$ from both sides:
$\frac{3}{m_L} = \frac{8}{\pi m_R}$.
Rearranging for the ratio of masses $\frac{m_L}{m_R}$:
$\frac{m_L}{m_R} = \frac{3\pi}{8}$.

(B) The root mean square velocity of an ideal gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$ or $v_{rms} = \sqrt{\frac{3kT}{m}}$,where $m$ is the mass of the molecule.
The momentum of a molecule is $p = mv$.
For container $A$,$p_A = m_A v_A = m_A \sqrt{\frac{3kT}{m_A}} = \sqrt{3kT m_A}$.
For container $B$,$p_B = m_B v_B = m_B \sqrt{\frac{3kT}{m_B}} = \sqrt{3kT m_B}$.
Taking the ratio of the two momenta:
$\frac{p_A}{p_B} = \frac{\sqrt{3kT m_A}}{\sqrt{3kT m_B}} = \sqrt{\frac{m_A}{m_B}} = \left( \frac{m_A}{m_B} \right)^{1/2}$.
Therefore,$p_A = \left( \frac{m_A}{m_B} \right)^{1/2} p_B$.

(C) Average speed $v_{av} = \frac{0 + 2 + 3 + 4 + 4 + 4 + 5 + 5 + 6 + 9}{10} = \frac{42}{10} = 4.2 \ m/s$.
$rms$ speed $v_{rms} = \sqrt{\frac{0^2 + 2^2 + 3^2 + 4^2 + 4^2 + 4^2 + 5^2 + 5^2 + 6^2 + 9^2}{10}} = \sqrt{\frac{0 + 4 + 9 + 16 + 16 + 16 + 25 + 25 + 36 + 81}{10}} = \sqrt{\frac{228}{10}} = \sqrt{22.8} \approx 4.77 \ m/s$.
Most probable speed $v_{mp}$ is the speed that occurs most frequently. Here,$4$ appears three times,which is more than any other value. Thus,$v_{mp} = 4 \ m/s$.