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

(B) The root mean square speed 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.
Since $T$ is the same for all three vessels,$v_{rms} \propto \frac{1}{\sqrt{M}}$.
The molar masses are:
Neon (monoatomic): $M_1 \approx 20 \text{ g/mol}$.
Chlorine (diatomic): $M_2 \approx 71 \text{ g/mol}$.
Uranium hexafluoride (polyatomic): $M_3 \approx 352 \text{ g/mol}$.
Since $M_1 < M_2 < M_3$,it follows that $\frac{1}{\sqrt{M_1}} > \frac{1}{\sqrt{M_2}} > \frac{1}{\sqrt{M_3}}$.
Therefore,$v_{rms}(\text{mono}) > v_{rms}(\text{dia}) > v_{rms}(\text{poly})$.

(A) The root mean square speed $(V_{rms})$ of gas molecules is given by the formula: $V_{rms} = \sqrt{\frac{3 k_{B} T}{m}}$.
Given:
Temperature $T = 27^{\circ} C = 27 + 273 = 300 \, K$.
Mass of a nitrogen molecule $m = 4.6 \times 10^{-26} \, kg$.
Boltzmann constant $k_{B} = 1.4 \times 10^{-23} \, J K^{-1}$.
Substituting the values into the formula:
$V_{rms} = \sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{4.6 \times 10^{-26}}}$
$V_{rms} = \sqrt{\frac{1260 \times 10^{-23}}{4.6 \times 10^{-26}}}$
$V_{rms} = \sqrt{273.91 \times 10^{3}} \approx \sqrt{273910} \approx 523.36 \, m/s$.
Thus,the approximate speed is $523 \, m/s$.

(C) The root mean square (r.m.s.) speed 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.
Since the temperature $T$ is the same for both gases,we have $v_{rms} \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the r.m.s. speeds is $\frac{v_{Ar}}{v_{Cl}} = \sqrt{\frac{M_{Cl}}{M_{Ar}}}$.
Given $v_{Cl} = 490\,m/s$,$M_{Cl} = 70.9\,u$,and $M_{Ar} = 39.9\,u$.
Substituting the values: $v_{Ar} = 490 \times \sqrt{\frac{70.9}{39.9}}$.
$v_{Ar} = 490 \times \sqrt{1.7769} \approx 490 \times 1.333 = 653.17\,m/s$.
Rounding to the nearest provided option,we get $651.7\,m/s$.

(A) The root mean square (rms) speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
The average speed of a gas molecule is given by $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.
According to the problem,$v_{rms} = \left(1+\frac{5}{x}\right)^{\frac{1}{2}} v_{avg}$.
Substituting the formulas: $\sqrt{\frac{3RT}{M}} = \left(1+\frac{5}{x}\right)^{\frac{1}{2}} \sqrt{\frac{8RT}{\pi M}}$.
Squaring both sides: $\frac{3RT}{M} = \left(1+\frac{5}{x}\right) \frac{8RT}{\pi M}$.
Canceling $\frac{RT}{M}$ from both sides: $3 = \left(1+\frac{5}{x}\right) \frac{8}{\pi}$.
Given $\pi = \frac{22}{7}$,we have $3 = \left(1+\frac{5}{x}\right) \frac{8}{22/7} = \left(1+\frac{5}{x}\right) \frac{8 \times 7}{22} = \left(1+\frac{5}{x}\right) \frac{56}{22} = \left(1+\frac{5}{x}\right) \frac{28}{11}$.
$\Rightarrow 1+\frac{5}{x} = 3 \times \frac{11}{28} = \frac{33}{28}$.
$\Rightarrow \frac{5}{x} = \frac{33}{28} - 1 = \frac{5}{28}$.
$\Rightarrow x = 28$.

(C) The $rms$ speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.
Let the initial temperature be $T_1 = -50 + 273 = 223\,K$ and the initial speed be $v_1 = v$.
The speed is increased by $3$ times,so the final speed $v_2 = v + 3v = 4v$.
Using the relation $\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$,we have $\frac{v}{4v} = \sqrt{\frac{223}{T_2}}$.
Squaring both sides,$\frac{1}{16} = \frac{223}{T_2}$.
Thus,$T_2 = 223 \times 16 = 3568\,K$.
Converting to Celsius,$T_2 = 3568 - 273 = 3295^{\circ}C$.

(D) The root mean square (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 absolute temperature in Kelvin, and $M$ is the molar mass of the gas.
Given, the r.m.s. velocity of hydrogen $(H_2)$ at temperature $T_H$ is equal to the r.m.s. velocity of oxygen $(O_2)$ at temperature $T_O = 47^{\circ} C = 47 + 273 = 320 \,K$.
The molar mass of hydrogen $M_H = 2 \,g/mol$ and the molar mass of oxygen $M_O = 32 \,g/mol$.
Equating the velocities: $\sqrt{\frac{3RT_H}{M_H}} = \sqrt{\frac{3RT_O}{M_O}}$.
Squaring both sides: $\frac{T_H}{M_H} = \frac{T_O}{M_O}$.
Substituting the values: $\frac{T_H}{2} = \frac{320}{32}$.
$T_H = 2 \times 10 = 20 \,K$.

(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 $(T)$ and gas constant $(R)$ are the same for both gases, the relationship between velocity and molar mass $(M)$ is $V_{rms} \propto \frac{1}{\sqrt{M}}$.
Let $V_H$ be the velocity of hydrogen $(M_H = 2 \,g/mol)$ and $V_O$ be the velocity of oxygen $(M_O = 32 \,g/mol)$.
Then, $\frac{V_H}{V_O} = \sqrt{\frac{M_O}{M_H}}$.
Substituting the given values: $\frac{2}{V_O} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
Therefore, $V_O = \frac{2}{4} = 0.5 \,km/s$.

(B) The root mean square speed $(V_{rms})$ of a gas is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M_w}}$.
Since the gases are in the same sample,they are at the same temperature $(T)$.
Therefore,$V_{rms} \propto \frac{1}{\sqrt{M_w}}$.
The ratio of the root mean square speed of helium $(He)$ to oxygen $(O_2)$ is:
$\frac{V_{He}}{V_{O_2}} = \sqrt{\frac{M_{w, O_2}}{M_{w, He}}}$.
Given the molar mass of helium $(M_{w, He} = 4 \ g/mol)$ and oxygen $(M_{w, O_2} = 32 \ g/mol)$:
$\frac{V_{He}}{V_{O_2}} = \sqrt{\frac{32}{4}} = \sqrt{8} = 2\sqrt{2}$.
Thus,the ratio is $\frac{2\sqrt{2}}{1}$.

(D) The root mean square (rms) speed of a gas 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 both gases are in the same container at the same temperature $T = 300 \ K$,the ratio of their rms speeds is:
$\frac{v_{\text{rms}} \text{ (helium)}}{v_{\text{rms}} \text{ (argon)}} = \frac{\sqrt{\frac{3RT}{M_{\text{He}}}}}{\sqrt{\frac{3RT}{M_{\text{Ar}}}}} = \sqrt{\frac{M_{\text{Ar}}}{M_{\text{He}}}}$
Given $M_{\text{He}} = 4 \ amu$ and $M_{\text{Ar}} = 40 \ amu$,we have:
$\frac{v_{\text{rms}} \text{ (helium)}}{v_{\text{rms}} \text{ (argon)}} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$.

(A) The root mean square velocity $(V_{rms})$ of gas molecules is given by the formula:
$V_{rms} = \sqrt{\frac{3RT}{M}}$
Squaring both sides,we get the mean squared velocity $(V_{rms}^2)$:
$V_{rms}^2 = \frac{3RT}{M}$
Since $R$ (universal gas constant) and $M$ (molar mass of the gas) are constants for a particular ideal gas,we can write:
$V_{rms}^2 \propto T$
This equation is of the form $y = mx$,where $y = V_{rms}^2$,$x = T$,and $m = \frac{3R}{M}$ is the slope.
Therefore,the graph of mean squared velocity versus temperature is a straight line passing through the origin.

(C) The $r.m.s.$ velocity of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since vapour density $d$ is proportional to molar mass $M$ $(d = \frac{M}{2})$,we have $v_{rms} \propto \frac{1}{\sqrt{d}}$.
Given the ratio of vapour densities $\frac{d_1}{d_2} = \frac{4}{25}$.
The ratio of $r.m.s.$ velocities is $\frac{v_1}{v_2} = \sqrt{\frac{d_2}{d_1}}$.
Substituting the values: $\frac{v_1}{v_2} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.

(A) The root mean square velocity is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that $(V_{rms})_{O_2} = (V_{rms})_{H_2}$,we have $\sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}}$.
Squaring both sides and canceling common terms,we get $\frac{T_{O_2}}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}}$.
Here,$M_{O_2} = 32 \ g/mol$,$M_{H_2} = 2 \ g/mol$,and $T_{H_2} = 20 K$.
Substituting the values: $\frac{T_{O_2}}{32} = \frac{20}{2}$.
$T_{O_2} = 10 \times 32 = 320 K$.

(C) The pressure exerted by an ideal gas is given by the formula $P = \frac{1}{3} \rho v_{rms}^2$,where $\rho$ is the density and $v_{rms}$ is the root mean square velocity.
Since the pressures $P_1$ and $P_2$ are equal,we have $P_1 = P_2$.
Thus,$\frac{1}{3} \rho_1 v_{rms,1}^2 = \frac{1}{3} \rho_2 v_{rms,2}^2$.
This simplifies to $\rho_1 v_{rms,1}^2 = \rho_2 v_{rms,2}^2$.
Rearranging for the ratio of velocities,we get $\frac{v_{rms,1}^2}{v_{rms,2}^2} = \frac{\rho_2}{\rho_1}$.
Given the ratio of densities $\rho_1 : \rho_2 = 1 : 16$,we have $\frac{\rho_2}{\rho_1} = \frac{16}{1} = 16$.
Taking the square root on both sides,$\frac{v_{rms,1}}{v_{rms,2}} = \sqrt{16} = 4$.
Therefore,the ratio of their r.m.s. velocities is $4:1$.

(B) The formula for r.m.s. velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given,$v_{H_2} = 4 \times v_{O_2}$.
Temperature of Oxygen $T_{O_2} = 47 + 273 = 320 \ K$.
Molar mass of Hydrogen $M_{H_2} = 2 \ g/mol$ and Oxygen $M_{O_2} = 32 \ g/mol$.
Using the relation $\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = 4 \times \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$.
Squaring both sides: $\frac{T_{H_2}}{M_{H_2}} = 16 \times \frac{T_{O_2}}{M_{O_2}}$.
Substituting the values: $\frac{T_{H_2}}{2} = 16 \times \frac{320}{32}$.
$\frac{T_{H_2}}{2} = 16 \times 10 = 160$.
$T_{H_2} = 320 \ K$.
Converting back to Celsius: $T(^{\circ}C) = 320 - 273 = 47^{\circ} C$.

(C) The root mean square (r.m.s.) velocity of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
In an isothermal process,the temperature $T$ of the gas remains constant.
Since $v_{rms}$ depends only on the temperature $T$ (assuming the gas composition $M$ remains constant),if $T$ is constant,then $v_{rms}$ must also remain constant.
Therefore,when a gas is compressed isothermally,the r.m.s. velocity of its molecules remains the same.

(B) The root mean square ($R$.$M$.$S$.) velocity 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 $v_1$ be the $R$.$M$.$S$. velocity at $T_1 = 100 \ K$ and $v_2$ be the $R$.$M$.$S$. velocity at $T_2 = 400 \ K$.
Given $v_1 = x$.
Using the proportionality: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{v_2}{x} = \sqrt{\frac{400}{100}} = \sqrt{4} = 2$.
Therefore,$v_2 = 2x$.

(B) The $r.m.s.$ speed $(v_{rms})$ of gas molecules is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
From this formula,it is clear that $v_{rms} \propto \sqrt{T}$.
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,$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the given values: $\frac{v_2}{v_1} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2$.
Therefore,$v_2 = 2v_1$.
This means the $r.m.s.$ speed at $800 \ K$ is twice the value at $200 \ K$.

(C) 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: $v_{rms} = \sqrt{\frac{v_1^2 + v_2^2 + v_3^2 + v_4^2 + v_5^2 + v_6^2}{N}}$
Given speeds: $2, 5, 3, 6, 3, 5 \,m/s$.
Number of molecules $N = 6$.
Sum of squares: $2^2 + 5^2 + 3^2 + 6^2 + 3^2 + 5^2 = 4 + 25 + 9 + 36 + 9 + 25 = 108$.
Mean of squares: $\frac{108}{6} = 18$.
$v_{rms} = \sqrt{18} \approx 4.24 \,m/s$.

(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 $v_{rms} \propto \sqrt{T}$.
Let $v_1$ be the velocity at $T_1 = 100 \ K$ and $v_2$ be the velocity at $T_2 = 400 \ K$.
Given $v_1 = x$.
Using the proportionality,$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{v_2}{x} = \sqrt{\frac{400}{100}} = \sqrt{4} = 2$.
Therefore,$v_2 = 2x$.

(C) The root mean square (r.m.s.) speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that the r.m.s. speeds of oxygen $(O_2)$ and helium $(He)$ are equal,we have:
$\sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{He}}{M_{He}}}$
Squaring both sides and simplifying,we get:
$\frac{T_{O_2}}{M_{O_2}} = \frac{T_{He}}{M_{He}}$
Given $T_{He} = 57^{\circ} C = 57 + 273 = 330 \ K$,$M_{O_2} = 32$,and $M_{He} = 4$.
Substituting the values:
$\frac{T_{O_2}}{32} = \frac{330}{4}$
$T_{O_2} = \frac{330 \times 32}{4} = 330 \times 8 = 2640 \ K$.

(C) The root mean square (r.m.s.) velocity 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 $v_1$ be the r.m.s. velocity at $T_1 = 100 \ K$ and $v_2$ be the r.m.s. velocity at $T_2 = 400 \ K$.
Given $v_1 = x$.
Using the proportionality $v_{rms} \propto \sqrt{T}$,we have: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{v_2}{x} = \sqrt{\frac{400}{100}} = \sqrt{4} = 2$.
Therefore,$v_2 = 2x$.

(B) The $r.m.s.$ velocity of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$.
Given,$T_{O_2} = 47^{\circ} C = 47 + 273 = 320 \ K$.
Let $V_H$ be the $r.m.s.$ velocity of hydrogen and $V_O$ be that of oxygen.
According to the problem,$V_H = 4.5 \times V_O$.
Substituting the formula: $\sqrt{\frac{3RT_H}{M_H}} = 4.5 \times \sqrt{\frac{3RT_O}{M_O}}$.
Squaring both sides: $\frac{3RT_H}{M_H} = (4.5)^2 \times \frac{3RT_O}{M_O}$.
Canceling $3R$ from both sides: $\frac{T_H}{M_H} = 20.25 \times \frac{T_O}{M_O}$.
Given $M_H = 2$ and $M_O = 32$: $\frac{T_H}{2} = 20.25 \times \frac{320}{32}$.
$\frac{T_H}{2} = 20.25 \times 10 = 202.5$.
$T_H = 202.5 \times 2 = 405 \ K$.
Converting to Celsius: $T_H = 405 - 273 = 132^{\circ} C$.

(A) Given:
Temperature $T = 27^{\circ} C = 27 + 273 = 300 \ K$.
Root mean square velocity $v_{rms} = 61 \ m/s$.
Gas constant $R = 8.31 \ J \ mol^{-1} \ K^{-1}$.
The formula for $v_{rms}$ is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $M$ is the molar mass in $kg/mol$.
Squaring both sides,we get $v_{rms}^2 = \frac{3RT}{M}$.
Rearranging for $M$,we get $M = \frac{3RT}{v_{rms}^2}$.
Substituting the values: $M = \frac{3 \times 8.31 \times 300}{61 \times 61}$.
$M = \frac{7479}{3721} \approx 2.01 \ kg/mol$.
Since the molar mass is approximately $2 \ kg/mol$,the molecular weight is $2 \ g/mol$.

(C) Initial temperature $T_1 = -80^{\circ} C = -80 + 273 = 193 \ K$.
The root mean square (r.m.s.) speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.
Let the initial speed be $v_1$ and the final speed be $v_2$. We are given that the speed is increased by $2$ times,meaning $v_2 = v_1 + 2v_1 = 3v_1$.
Using the relation $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$,we have $\frac{3v_1}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
$3 = \sqrt{\frac{T_2}{193}} \implies 9 = \frac{T_2}{193}$.
$T_2 = 9 \times 193 = 1737 \ K$.
Converting to Celsius: $T_2 = 1737 - 273 = 1464^{\circ} C$.

(B) The mean square velocity of a gas molecule is given by $\langle v^2 \rangle = \frac{3kT}{m}$.
For gas $A$,the mean square of the $x$-component is $\omega^2 = \langle v_x^2 \rangle$. Since $\langle v^2 \rangle = \langle v_x^2 \rangle + \langle v_y^2 \rangle + \langle v_z^2 \rangle$ and $\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle$,we have $\langle v^2 \rangle = 3\omega^2$.
Thus,$3\omega^2 = \frac{3kT}{m} \implies \omega^2 = \frac{kT}{m}$...$(i)$
For gas $B$,the mean square velocity is $V^2 = \frac{3kT}{2m}$...(ii)
Dividing $(i)$ by (ii):
$\frac{\omega^2}{V^2} = \frac{kT/m}{3kT/2m} = \frac{kT}{m} \times \frac{2m}{3kT} = \frac{2}{3}$.

(D) According to the kinetic theory of gases,the absolute temperature $T$ of an ideal gas is related to the average kinetic energy of its molecules.
The average kinetic energy per molecule is given by $\frac{1}{2} m v_{rms}^2 = \frac{3}{2} k_B T$,where $m$ is the mass of a molecule,$v_{rms}$ is the root mean square velocity,and $k_B$ is the Boltzmann constant.
From this relation,we can see that $T \propto v_{rms}^2$.
Therefore,the absolute temperature is directly proportional to the mean square velocity of the molecules.

(A) The root mean square (r.m.s.) speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
From this relation,we can see that $v_{rms} \propto \sqrt{T}$.
Given initial temperature $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$.
Given final temperature $T_2 = 927^{\circ} C = 927 + 273 = 1200 \ K$.
Taking the ratio of the speeds:
$\frac{v_{rms_2}}{v_{rms_1}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2$.
Therefore,$v_{rms_2} = 2 \cdot v_{rms_1}$.
The r.m.s. speed becomes twice the initial speed.

(C) The root mean square (r.m.s.) speed of gas molecules is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M}}$.
Since the gas is the same,$V_{rms} \propto \sqrt{T}$.
First,convert the temperatures from Celsius to Kelvin:
$T_1 = 127^{\circ} C = 127 + 273 = 400 \ K$
$T_2 = 527^{\circ} C = 527 + 273 = 800 \ K$
The ratio of the r.m.s. speeds is:
$\frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{400}{800}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Thus,the ratio is $1: \sqrt{2}$.

(C) The initial temperature $T_1 = -68^{\circ} C = -68 + 273 = 205 \ K$.
The root mean square (r.m.s.) velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.
Let the initial velocity be $(v_{rms})_1$ and the final velocity be $(v_{rms})_2 = 2(v_{rms})_1$.
Using the ratio: $\frac{(v_{rms})_2}{(v_{rms})_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $2 = \sqrt{\frac{T_2}{205}}$.
Squaring both sides: $4 = \frac{T_2}{205}$.
Therefore,$T_2 = 4 \times 205 = 820 \ K$.
Converting back to Celsius: $T_2 = 820 - 273 = 547^{\circ} C$.

(A) The $r.m.s.$ speed of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \frac{1}{\sqrt{M}}$.
Let the first gas have molecular mass $M_1 = 32$ and $r.m.s.$ speed $(v_{rms})_1$.
Let the second gas have molecular mass $M_2$ and $r.m.s.$ speed $(v_{rms})_2 = 4(v_{rms})_1$.
Using the relation $\frac{(v_{rms})_2}{(v_{rms})_1} = \sqrt{\frac{M_1}{M_2}}$,we substitute the given values:
$4 = \sqrt{\frac{32}{M_2}}$.
Squaring both sides,we get $16 = \frac{32}{M_2}$.
Therefore,$M_2 = \frac{32}{16} = 2$.

(D) The root mean square (r.m.s.) velocity $v_{rms}$ of gas molecules is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
From this relation,it is clear that $v_{rms} \propto \sqrt{T}$.
Let the initial temperature be $T_1 = T$ and the final temperature be $T_2 = 5T$.
Let the initial r.m.s. velocity be $v_1$ and the final r.m.s. velocity be $v_2$.
Then,$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{5T}{T}} = \sqrt{5}$.
Therefore,$v_2 = \sqrt{5} v_1$.
Thus,the r.m.s. velocity becomes $\sqrt{5}$ times the initial velocity.

(A) The root mean square (r.m.s.) speed of a gas molecule is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
This implies that $v_{rms} \propto \sqrt{T}$.
Given that the r.m.s. speed at temperature $T$ is $2$ times the r.m.s. speed at $320 \,K$,we can write:
$\frac{v_T}{v_{320}} = 2$
Since $v \propto \sqrt{T}$,we have:
$\frac{\sqrt{T}}{\sqrt{320}} = 2$
Squaring both sides:
$\frac{T}{320} = 2^2$
$\frac{T}{320} = 4$
$T = 4 \times 320 \,K = 1280 \,K$.

(A) The root mean square (rms) velocity $V$ of a gas molecule is given by the formula:
$V = \sqrt{\frac{3 k_B T}{m}}$
where $k_B$ is the Boltzmann constant,$T$ is the absolute temperature,and $m$ is the mass of one molecule.
Squaring both sides,we get:
$V^2 = \frac{3 k_B T}{m}$
Since $k_B$ and $m$ are constants for a given gas,we can write:
$V^2 \propto T$
Or,$\frac{V^2}{T} = \frac{3 k_B}{m} = \text{constant}$
Therefore,the correct relation is $\frac{V^2}{T} = \text{constant}$.

(D) The root mean square (r.m.s.) speed of gas molecules is given by the formula $v = \sqrt{\frac{3RT}{M}}$.
From this relation, it is clear that $v \propto \sqrt{T}$.
Given initial temperature $T_1 = 140 \,K$ and initial r.m.s. speed $v_1 = v$.
Final temperature $T_2 = 560 \,K$.
We can write the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{v_2}{v} = \sqrt{\frac{560}{140}} = \sqrt{4} = 2$.
Therefore, the new r.m.s. speed $v_2 = 2 v$.

(B) The process is isothermal,which means the temperature $T$ of the gas remains constant throughout the process.
The $rms$ speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $R$ (universal gas constant),$M$ (molar mass),and $T$ (temperature) are constant,the $rms$ speed remains unchanged.

(B) The root mean square (r.m.s.) velocity of an ideal gas is given by the formula:
$V = \sqrt{\frac{3RT}{M_0}}$
Squaring both sides, we get:
$V^2 = \frac{3RT}{M_0}$
Rearranging the terms to isolate the variables $V$ and $T$:
$\frac{V^2}{T} = \frac{3R}{M_0}$
Since $R$ (universal gas constant) and $M_0$ (molar mass) are constants for a given gas, the ratio $\frac{3R}{M_0}$ is also a constant.
Therefore, $\frac{V^2}{T} = \text{constant}$.

(A) The root mean square speed $(v_{rms})$ 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.
Since $M = m \times N_A$ (where $m$ is the mass of one molecule and $N_A$ is Avogadro's number),we can substitute this into the equation:
$v_{rms} = \sqrt{\frac{3RT}{mN_A}}$
Since $3$,$R$,and $N_A$ are constants,we have:
$v_{rms} \propto \sqrt{\frac{T}{m}}$
$v_{rms} \propto m^{-\frac{1}{2}} T^{\frac{1}{2}}$

(B) The pressure exerted by an ideal gas is given by the formula $P = \frac{1}{3} \rho c^2$,where $\rho$ is the density and $c$ is the root mean square (rms) speed.
Since the pressures are equal,we have $P_1 = P_2$.
Therefore,$\frac{1}{3} \rho_1 c_1^2 = \frac{1}{3} \rho_2 c_2^2$.
This simplifies to $\frac{c_1^2}{c_2^2} = \frac{\rho_2}{\rho_1}$.
Given the ratio of densities $\rho_1 : \rho_2 = 1 : 16$,we have $\frac{\rho_2}{\rho_1} = \frac{16}{1} = 16$.
Substituting this into the equation,we get $\frac{c_1^2}{c_2^2} = 16$.
Taking the square root on both sides,we find $\frac{c_1}{c_2} = \sqrt{16} = 4$.
Thus,the ratio of their rms speeds $(c_1: c_2)$ is $4: 1$.

(A) The root mean square (rms) speed 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 rms speed depends only on the temperature $T$ and the molar mass $M$ of the gas,it is independent of the pressure $P$ of the gas.
Given that the temperature remains constant,the rms speed will remain unchanged even if the pressure is increased.
Therefore,the new rms speed will be $V$.

(D) The root mean square (r.m.s) speed of a gas molecule is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Here,$R$ is the universal gas constant,$T$ is the absolute temperature in Kelvin,and $M$ is the molar mass.
For helium $(He)$: $M_{He} = 4$,$T_{He} = 327^{\circ} C = 327 + 273 = 600 \ K$.
For oxygen $(O_2)$: $M_{O} = 32$,$T_{O} = 27^{\circ} C = 27 + 273 = 300 \ K$.
The ratio of the r.m.s speeds is:
$\frac{v_{He}}{v_{O}} = \sqrt{\frac{T_{He}}{T_{O}} \times \frac{M_{O}}{M_{He}}} = \sqrt{\frac{600}{300} \times \frac{32}{4}} = \sqrt{2 \times 8} = \sqrt{16} = 4$.
Thus,the ratio is $4:1$.

(B) The root mean square ($R$.$M$.$S$.) velocity of a gas molecule is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that the $R$.$M$.$S$. velocities of hydrogen $(H_2)$ and oxygen $(O_2)$ are equal,we have $v_{H} = v_{O}$.
Substituting the formula: $\sqrt{\frac{3RT_H}{M_H}} = \sqrt{\frac{3RT_O}{M_O}}$.
Squaring both sides and simplifying: $\frac{T_H}{M_H} = \frac{T_O}{M_O}$.
Given $T_O = 47^{\circ} C = 47 + 273 = 320 \ K$,$M_H = 2$,and $M_O = 32$.
Rearranging for $T_H$: $T_H = T_O \times \frac{M_H}{M_O}$.
Substituting the values: $T_H = 320 \times \frac{2}{32} = 320 \times \frac{1}{16} = 20 \ K$.

(A) The r.m.s. velocity of a gas molecule is given by the formula $C = \sqrt{\frac{3RT}{M}}$,which implies $C \propto \sqrt{T}$.
Given initial temperature $T_{1} = 27^{\circ} C = 27 + 273 = 300 \ K$.
Given final temperature $T_{2} = 327^{\circ} C = 327 + 273 = 600 \ K$.
The ratio of the velocities is $\frac{C_{2}}{C_{1}} = \sqrt{\frac{T_{2}}{T_{1}}}$.
Substituting the values,we get $\frac{C_{2}}{C_{1}} = \sqrt{\frac{600}{300}} = \sqrt{2}$.

(B) The formula for the root mean square ($R.M.S.$) velocity of gas molecules is given by $C = \sqrt{\frac{3RT}{M}}$.
From this relation, we can see that $C \propto \sqrt{\frac{T}{M}}$.
Let the initial velocity be $C = 200 \,m/s$ at temperature $T$ and molecular weight $M$.
Given the new conditions: $T' = \frac{T}{2}$ and $M' = 2M$.
The new $R.M.S.$ velocity $C'$ is given by the ratio:
$\frac{C'}{C} = \sqrt{\frac{T'}{T} \times \frac{M}{M'}} = \sqrt{\frac{T/2}{T} \times \frac{M}{2M}} = \sqrt{\frac{1}{2} \times \frac{1}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore, $C' = \frac{C}{2} = \frac{200}{2} = 100 \,m/s$.

(C) The root mean square (rms) speed of an $O_2$ molecule is given by $u = \sqrt{\frac{3RT}{M}}$,where $T$ is the temperature and $M$ is the molar mass of $O_2$.
When the temperature is doubled $(T' = 2T)$ and the molecule dissociates into two atoms,the molar mass of the oxygen atom becomes $M' = \frac{M}{2}$.
The new rms speed $u'$ is given by $u' = \sqrt{\frac{3RT'}{M'}} = \sqrt{\frac{3R(2T)}{M/2}}$.
Simplifying this,we get $u' = \sqrt{4 \cdot \frac{3RT}{M}} = 2 \sqrt{\frac{3RT}{M}}$.
Since $u = \sqrt{\frac{3RT}{M}}$,we have $u' = 2u$.

(C) The average velocity $(v_{av})$ is calculated as the arithmetic mean of the velocities:
$v_{av} = \frac{v_{1} + v_{2} + v_{3} + v_{4}}{N} = \frac{1 + 3 + 5 + 7}{4} = \frac{16}{4} = 4 \ km/s$
The root mean square $(RMS)$ velocity $(v_{rms})$ is calculated as the square root of the mean of the squares of the velocities:
$v_{rms} = \sqrt{\frac{v_{1}^{2} + v_{2}^{2} + v_{3}^{2} + v_{4}^{2}}{N}} = \sqrt{\frac{1^{2} + 3^{2} + 5^{2} + 7^{2}}{4}} = \sqrt{\frac{1 + 9 + 25 + 49}{4}} = \sqrt{\frac{84}{4}} = \sqrt{21} \approx 4.583 \ km/s$
The difference between the $RMS$ velocity and the average velocity is:
$v_{rms} - v_{av} = 4.583 \ km/s - 4 \ km/s = 0.583 \ km/s$

(A) The root mean square velocity $(v_{rms})$ of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $R$ and $M$ are constants for a given gas,we have $v_{rms} \propto \sqrt{T}$.
This implies $T \propto v_{rms}^2$.
Given that the final velocity $v_2$ is half of the initial velocity $v_1$,i.e.,$v_2 = \frac{v_1}{2}$.
Therefore,$\frac{T_2}{T_1} = \left(\frac{v_2}{v_1}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
The initial temperature $T_1 = 327^{\circ} C = 327 + 273 = 600 \ K$.
Thus,$T_2 = \frac{T_1}{4} = \frac{600 \ K}{4} = 150 \ K$.
Converting back to Celsius: $T_2 = 150 - 273 = -123^{\circ} C$.

(A) The $rms$ speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.
Let $v_1$ be the $rms$ speed at $NTP$ $(T_1 = 273 \ K)$ and $v_2$ be the $rms$ speed at temperature $T_2$.
Given that $v_2 = 2v_1$,we have:
$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$
$2 = \sqrt{\frac{T_2}{273}}$
Squaring both sides:
$4 = \frac{T_2}{273}$
$T_2 = 4 \times 273 = 1092 \ K$
Converting to Celsius:
$T_2(^{\circ}C) = 1092 - 273 = 819^{\circ} C$.

(A) The average speed $(v_{av})$ of gas molecules is given by the formula:
$v_{av} = \sqrt{\frac{8RT}{\pi M}}$
where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
Assuming the temperature $T$ is constant for both gases,the average speed is inversely proportional to the square root of the molar mass:
$v_{av} \propto \frac{1}{\sqrt{M}}$
Let $v_{1}$ be the speed of $SO_{2}$ and $v_{2}$ be the speed of the unknown gas. Given $v_{2} = 4v_{1}$ and $M_{1} = 64$:
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{M_{1}}{M_{2}}}$
$4 = \sqrt{\frac{64}{M_{2}}}$
Squaring both sides:
$16 = \frac{64}{M_{2}}$
$M_{2} = \frac{64}{16} = 4$
The gas with a molar mass of $4$ is Helium $(He)$.
Therefore,the correct option is $A$.

(B) The root mean square (rms) speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
From this relation,we can see that $v_{rms} \propto \sqrt{T}$.
Let the initial rms speed be $v_1$ and the initial temperature be $T_1$. Let the final rms speed be $v_2$ and the final temperature be $T_2$.
Given that the rms speed increases by $25 \%$,we have $v_2 = v_1 + 0.25v_1 = 1.25v_1$.
Since $v \propto \sqrt{T}$,we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values,we get $1.25 = \sqrt{\frac{T_2}{T_1}}$.
Squaring both sides,we get $(1.25)^2 = \frac{T_2}{T_1}$,which gives $1.5625 = \frac{T_2}{T_1}$.
This implies $T_2 = 1.5625 T_1$.
The percentage increase in temperature is given by $\frac{T_2 - T_1}{T_1} \times 100 \%$.
Percentage increase $= (1.5625 - 1) \times 100 \% = 0.5625 \times 100 \% = 56.25 \%$.