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

(C) The root mean square speed of gas molecules is given by the formula $\nu_{rms} = \sqrt{\frac{3RT}{M}}$.
Given: Temperature of hydrogen $T_H = 27 \, ^\circ C = 27 + 273 = 300 \, K$.
Molar mass of oxygen $M_{O_2} = 32 \, g \, mol^{-1}$.
Molar mass of hydrogen $M_{H_2} = 2 \, g \, mol^{-1}$.
Since $\nu_{rms(O_2)} = \nu_{rms(H_2)}$,we have $\sqrt{\frac{3RT_O}{M_{O_2}}} = \sqrt{\frac{3RT_H}{M_{H_2}}}$.
Squaring both sides and simplifying,we get $\frac{T_O}{M_{O_2}} = \frac{T_H}{M_{H_2}}$.
Therefore,$T_O = T_H \times \frac{M_{O_2}}{M_{H_2}} = 300 \times \frac{32}{2} = 300 \times 16 = 4800 \, K$.

(D) The average kinetic energy $(E)$ of a gas molecule is directly proportional to the absolute temperature $(T)$: $E \propto T$.
Given $E_1 = 6.21 \times 10^{-21} \ J$ at $T_1 = 300 \ K$. At $T_2 = 600 \ K$,the new kinetic energy $E_2$ is:
$E_2 = E_1 \times (T_2 / T_1) = 6.21 \times 10^{-21} \times (600 / 300) = 12.42 \times 10^{-21} \ J$.
The root mean square speed $(\nu_{rms})$ is directly proportional to the square root of the absolute temperature: $\nu_{rms} \propto \sqrt{T}$.
Given $(\nu_{rms})_1 = 325 \ m/s$ at $T_1 = 300 \ K$. At $T_2 = 600 \ K$,the new speed $(\nu_{rms})_2$ is:
$(\nu_{rms})_2 = (\nu_{rms})_1 \times \sqrt{T_2 / T_1} = 325 \times \sqrt{600 / 300} = 325 \times \sqrt{2} \approx 325 \times 1.414 = 459.6 \ m/s$.

(A) Given temperature,$T = 27^{\circ}C = 273 + 27 = 300 \, K$.
The molar mass of oxygen $(O_2)$ is $M_w = 32 \times 10^{-3} \, kg/mol$ and the universal gas constant is $R = 8.31 \, J \, mol^{-1} K^{-1}$.
The formula for $rms$ speed is $v_{rms} = \sqrt{\frac{3RT}{M_w}}$.
Substituting the values: $v_{rms} = \sqrt{\frac{3 \times 8.31 \times 300}{32 \times 10^{-3}}}$.
$v_{rms} = \sqrt{\frac{7479}{0.032}} = \sqrt{233718.75} \approx 483.44 \, ms^{-1}$.
Thus,the $rms$ speed is approximately $483.5 \, ms^{-1}$.

(D) The root mean square velocity is given by the formula $\nu_{rms} = \sqrt{\frac{3RT}{M}}$,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,$\nu_{rms} \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio is given by $\frac{(\nu_{rms})_{H_2}}{(\nu_{rms})_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}}$.
Given the molar masses $M_{O_2} = 32 \, g/mol$ and $M_{H_2} = 2 \, g/mol$.
Substituting these values,we get $\frac{(\nu_{rms})_{H_2}}{(\nu_{rms})_{O_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.

(A) The formula for the root mean square $(rms)$ speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given: $v_{rms} = 1930 \, m/s$,$T = 27 \, ^\circ C = 300 \, K$,and $R = 8.31 \, J/(mol \cdot K)$.
Rearranging the formula to solve for the molar mass $M$: $M = \frac{3RT}{v_{rms}^2}$.
Substituting the values: $M = \frac{3 \times 8.31 \times 300}{(1930)^2} \approx \frac{7479}{3724900} \approx 0.002 \, kg/mol = 2 \, g/mol$.
The gas with a molar mass of $2 \, g/mol$ is Hydrogen $(H_2)$.

(C) 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.
From this formula,it is clear that $v_{rms}$ depends only on the temperature $T$ and the nature of the gas (molar mass $M$).
It does not depend on the pressure,volume,or the number of molecules (density) of the gas.
Since the temperature remains constant and the gas remains the same,the $rms$ speed will remain unchanged.
Therefore,the new $rms$ speed is $400 \, ms^{-1}$.

(A) The $rms$ 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 in Kelvin,and $M$ is the molar mass.
Given that the $rms$ speed of $H_2$ equals the $rms$ speed of $O_2$:
$v_{rms(H_2)} = v_{rms(O_2)}$
$\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$
Squaring both sides and canceling common terms $(3R)$:
$\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}$
Given values:
$T_{O_2} = 47^{\circ}C = 47 + 273 = 320 \ K$
$M_{O_2} = 32 \ g/mol$
$M_{H_2} = 2 \ g/mol$
Substituting the values:
$\frac{T_{H_2}}{2} = \frac{320}{32}$
$T_{H_2} = \frac{320}{32} \times 2$
$T_{H_2} = 10 \times 2 = 20 \ K$.

(D) The root mean square $(rms)$ speed is given by the formula:
$v_{rms} = \sqrt{\frac{v_1^2 + v_2^2 + v_3^2 + v_4^2 + v_5^2}{N}}$
Given speeds are $v_1 = 2, v_2 = 3, v_3 = 4, v_4 = 5, v_5 = 6 \ km/s$ and $N = 5$.
$v_{rms} = \sqrt{\frac{2^2 + 3^2 + 4^2 + 5^2 + 6^2}{5}}$
$v_{rms} = \sqrt{\frac{4 + 9 + 16 + 25 + 36}{5}}$
$v_{rms} = \sqrt{\frac{90}{5}}$
$v_{rms} = \sqrt{18} \approx 4.24 \ km/s$.

(C) The $rms$ 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.
Initially,for the diatomic molecule: $v = \sqrt{\frac{3RT}{M}}$.
When the temperature is doubled,the new temperature $T' = 2T$.
When the diatomic molecule dissociates into two atoms,the molar mass of the atom becomes $M' = \frac{M}{2}$.
The new $rms$ speed $v'$ is given by: $v' = \sqrt{\frac{3RT'}{M'}} = \sqrt{\frac{3R(2T)}{M/2}} = \sqrt{\frac{12RT}{M}} = 2 \sqrt{\frac{3RT}{M}}$.
Substituting the initial speed $v$: $v' = 2v$.

(C) The $rms$ speed of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
From this,we can see that $v_{rms} \propto \sqrt{T}$.
Note that the $rms$ speed is independent of the pressure of the gas.
Given: $T_1 = 27^{\circ}C = 27 + 273 = 300 \, K$ and $v_1 = 200 \, m/s$.
Given: $T_2 = 127^{\circ}C = 127 + 273 = 400 \, K$.
Using the proportionality $v_1 / v_2 = \sqrt{T_1 / T_2}$:
$\frac{200}{v_2} = \sqrt{\frac{300}{400}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Therefore,$v_2 = \frac{200 \times 2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \, m/s$.

(D) The root mean square speed of molecules in part $L$ is given by: $v_{rms, L} = \sqrt{\frac{3KT}{m_L}}$
The average speed of molecules in part $R$ is given by: $v_{av, R} = \sqrt{\frac{8KT}{\pi m_R}}$
Given that $v_{rms, L} = v_{av, R}$,we equate the two expressions:
$\sqrt{\frac{3KT}{m_L}} = \sqrt{\frac{8KT}{\pi m_R}}$
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 to find the ratio $\frac{m_L}{m_R}$:
$\frac{m_L}{m_R} = \frac{3\pi}{8}$

(A) The root mean square speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$. Since $T$ is constant,$v_{rms} \propto \frac{1}{\sqrt{M}}$.
Given $m_1 > m_2 > m_3$,it follows that $(v_{rms})_1 < (v_{rms})_2 < (v_{rms})_3$.
The average kinetic energy of a gas molecule is given by $\bar{K} = \frac{3}{2}kT$. Since the temperature $T$ is the same for all gases in the mixture,the average kinetic energy depends only on the temperature and is independent of the molecular mass.
Therefore,$(\bar{K})_1 = (\bar{K})_2 = (\bar{K})_3$.

(D) The average kinetic energy of a gas molecule is given by $E = \frac{3}{2} k_B T$,which implies $E \propto T$.
Given $T_1 = 300 \ K$ and $T_2 = 600 \ K$,the new kinetic energy $E_2 = E_1 \times (T_2 / T_1) = 6.21 \times 10^{-21} \times (600 / 300) = 12.42 \times 10^{-21} \ J$.
The $rms$ speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.
Given $v_{rms,1} = 484 \ m/s$,the new $rms$ speed $v_{rms,2} = v_{rms,1} \times \sqrt{T_2 / T_1} = 484 \times \sqrt{600 / 300} = 484 \times \sqrt{2} \approx 484 \times 1.414 = 684.37 \ m/s \approx 684 \ m/s$.
Thus,the correct option is $D$.

(B) The average velocity of a gas molecule is given by the formula $v_{av} = \sqrt{\frac{8kT}{\pi m}}$,where $k$ is the Boltzmann constant,$T$ is the absolute temperature,and $m$ is the mass of one molecule of the gas.
Since the temperature $T$ is the same for all vessels,the average velocity of a specific gas depends only on its molecular mass $m$.
In vessel $A$,the average velocity of $O_2$ is $V_1 = \sqrt{\frac{8kT}{\pi m_{O_2}}}$.
In vessel $C$,the $O_2$ molecules are at the same temperature $T$ and have the same molecular mass $m_{O_2}$ as in vessel $A$.
The presence of $N_2$ in the mixture in vessel $C$ does not affect the individual average velocity of the $O_2$ molecules.
Therefore,the average velocity of $O_2$ in vessel $C$ remains $V_1$.

(A) The $r.m.s.$ velocity of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
This shows that $v_{rms} \propto \sqrt{T}$.
The pressure does not affect the $r.m.s.$ velocity of an ideal gas.
Given $T_1 = 27^o C = 300 \,K$ and $v_1 = 200 \,m s^{-1}$.
Given $T_2 = 127^o C = 400 \,K$.
Using the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
$\frac{v_2}{200} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.
Therefore,$v_2 = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \,m s^{-1}$.

(B) The escape velocity from the Earth's surface is $v_{\text{escape}} = 11200 \, m/s$.
For the oxygen molecules to escape,their $rms$ speed must be equal to the escape velocity:
$v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}} = v_{\text{escape}}$
Substituting the given values:
$11200 = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times T}{2.76 \times 10^{-26}}}$
Squaring both sides:
$(11200)^2 = \frac{3 \times 1.38 \times 10^{-23} \times T}{2.76 \times 10^{-26}}$
$1.2544 \times 10^8 = \frac{4.14 \times 10^{-23} \times T}{2.76 \times 10^{-26}}$
$1.2544 \times 10^8 = 1.5 \times 10^3 \times T$
Solving for $T$:
$T = \frac{1.2544 \times 10^8}{1.5 \times 10^3} \approx 8.360 \times 10^4 \, K$.

(B) The root mean square $(rms)$ speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3k_B T}{m}}$.
For the molecules to escape the Earth's atmosphere,their $v_{rms}$ must be equal to the escape velocity of the Earth,$v_e = 11.2 \, km/s = 1.12 \times 10^4 \, m/s$.
Equating $v_{rms} = v_e$,we get $\sqrt{\frac{3k_B T}{m}} = v_e$.
Squaring both sides: $\frac{3k_B T}{m} = v_e^2$.
Solving for $T$: $T = \frac{m v_e^2}{3k_B}$.
Substituting the given values: $T = \frac{(2.76 \times 10^{-26} \, kg) \times (1.12 \times 10^4 \, m/s)^2}{3 \times 1.38 \times 10^{-23} \, JK^{-1}}$.
$T = \frac{2.76 \times 10^{-26} \times 1.2544 \times 10^8}{4.14 \times 10^{-23}}$.
$T = \frac{3.462144 \times 10^{-18}}{4.14 \times 10^{-23}} \approx 0.83626 \times 10^5 \, K = 8.3626 \times 10^4 \, K$.
Rounding to the nearest option,we get $8.360 \times 10^4 \, K$.

(A) The $r.m.s.$ velocity of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $R$ and $M$ are constant,$v_{rms} \propto \sqrt{T}$.
Initial temperature $T_1 = 27^o C = 27 + 273 = 300 \ K$.
Final temperature $T_2 = 90^o C = 90 + 273 = 363 \ K$.
The ratio of velocities is $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{363}{300}} = \sqrt{1.21} = 1.1$.
The percentage increase is given by $\left( \frac{v_2}{v_1} - 1 \right) \times 100$.
Percentage increase $= (1.1 - 1) \times 100 = 0.1 \times 100 = 10\%$.

(C) The root mean square (r.m.s.) velocity of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $M$ is the molar mass of the gas.
Since $v_{rms} \propto \frac{1}{\sqrt{M}}$,we can write the ratio for two gases as $\frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}}$.
Given that the r.m.s. velocity of the gas $(v_1)$ is $\sqrt{2}$ times the r.m.s. velocity of oxygen $(v_2)$,we have $\frac{v_1}{v_2} = \sqrt{2}$.
The molar mass of oxygen $(O_2)$ is $M_2 = 32 \text{ g/mol}$.
Substituting these values into the ratio: $\sqrt{2} = \sqrt{\frac{32}{M_1}}$.
Squaring both sides: $2 = \frac{32}{M_1}$.
Solving for $M_1$: $M_1 = \frac{32}{2} = 16 \text{ g/mol}$.
The gas with a molar mass of $16 \text{ g/mol}$ is methane $(CH_4)$.
Therefore,the correct option is $C$.

(D) The formula for $r.m.s.$ velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
From this,we can see that $v_{rms} \propto \sqrt{\frac{T}{M}}$.
Let the initial state be $v_1, T_1, M_1$ and the final state be $v_2, T_2, M_2$.
Given: $v_1 = 300 \ m/s$,$M_2 = 2M_1$,and $T_2 = \frac{T_1}{2}$.
Using the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{M_1}{M_2} \times \frac{T_2}{T_1}}$.
Substituting the values: $\frac{v_2}{300} = \sqrt{\frac{M_1}{2M_1} \times \frac{T_1/2}{T_1}} = \sqrt{\frac{1}{2} \times \frac{1}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,$v_2 = \frac{300}{2} = 150 \ m/s$.

(A) The root mean square velocity of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $T$ is the absolute temperature.
Case $(i)$: The absolute temperature $T$ is doubled,so $T' = 2T$. The new velocity is $v'_{rms} = \sqrt{\frac{3R(2T)}{M}} = \sqrt{2} v_{rms} \approx 1.414 v_{rms}$.
Case $(ii)$: The centigrade temperature $t$ is doubled,so $t' = 2t$. The absolute temperature becomes $T' = (2t + 273) = 2(T - 273) + 273 = 2T - 273$. Since $2T - 273 < 2T$,the final absolute temperature in case $(ii)$ is lower than in case $(i)$.
Since $v_{rms} \propto \sqrt{T}$,the increase in velocity is greater when the absolute temperature is doubled compared to when the centigrade temperature is doubled. Therefore,the increase is more in case $(i)$.

(D) The average speed of gas molecules is given by $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.
Since $v_{avg} \propto \sqrt{T}$,if the average speed is doubled,the temperature $T$ must increase by a factor of $2^2 = 4$.
For an ideal gas,the pressure $P$ is related to temperature $T$ by the ideal gas law $PV = nRT$.
In a closed container,the volume $V$ and the number of moles $n$ remain constant.
Therefore,$P \propto T$.
Since the temperature $T$ increases by a factor of $4$,the pressure $P$ will also increase by a factor of $4$.

(A) The root mean square (r.m.s.) speed is given by $V_{rms} = \sqrt{\frac{(2u)^2 + (10u)^2 + (11u)^2}{3}} = \sqrt{\frac{4u^2 + 100u^2 + 121u^2}{3}} = \sqrt{\frac{225u^2}{3}} = \sqrt{75}u = 5\sqrt{3}u \approx 8.66u$.
The mean (average) speed is given by $V_{avg} = \frac{2u + 10u + 11u}{3} = \frac{23u}{3} \approx 7.66u$.
Comparing the two values,$V_{rms} - V_{avg} \approx 8.66u - 7.66u = 1u$.
Therefore,the r.m.s. speed exceeds the mean speed by about $u$.

(D) The average speed $u_{avg}$ is given by the sum of velocities divided by the number of molecules $N$: $u_{avg} = \frac{1+2+3+...+N}{N} = \frac{N(N+1)}{2N} = \frac{N+1}{2}$.
The root mean square speed $u_{rms}$ is given by the square root of the sum of squares of velocities divided by $N$: $u_{rms} = \sqrt{\frac{1^2+2^2+3^2+...+N^2}{N}} = \sqrt{\frac{N(N+1)(2N+1)}{6N}} = \sqrt{\frac{(N+1)(2N+1)}{6}}$.
The ratio of rms speed to average speed is $\frac{u_{rms}}{u_{avg}} = \frac{\sqrt{\frac{(N+1)(2N+1)}{6}}}{\frac{N+1}{2}} = \sqrt{\frac{(N+1)(2N+1)}{6}} \times \frac{2}{N+1} = \sqrt{\frac{2^2(N+1)(2N+1)}{6(N+1)^2}} = \sqrt{\frac{4(2N+1)}{6(N+1)}} = \sqrt{\frac{2(2N+1)}{3(N+1)}}$.

(D) For an ideal gas at temperature $T$,the average kinetic energy per molecule is given by $\frac{1}{2} m v_{rms}^2 = \frac{3}{2} k_B T$.
For gas $A$ with mass $m$ at temperature $T$:
The mean square velocity is $\langle v_A^2 \rangle = \frac{3 k_B T}{m}$.
Since the motion is isotropic,the mean square of the $x$-component is $\langle v_{Ax}^2 \rangle = w^2 = \frac{1}{3} \langle v_A^2 \rangle = \frac{k_B T}{m}$.
For gas $B$ with mass $2m$ at temperature $T$:
The mean square velocity is $v^2 = \langle v_B^2 \rangle = \frac{3 k_B T}{2m}$.
Now,calculating the ratio $w^2 / v^2$:
$\frac{w^2}{v^2} = \frac{k_B T / m}{3 k_B T / 2m} = \frac{1}{m} \times \frac{2m}{3} = \frac{2}{3}$.
Thus,the ratio is $2/3$.

(B) The root mean square (r.m.s.) speed of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $V_{rms} \propto \sqrt{T}$,the ratio is $\frac{V_{rms,2}}{V_{rms,1}} = \sqrt{\frac{T_2}{T_1}}$.
From the ideal gas equation $PV = nRT$,we have $T = \frac{PV}{nR}$.
For the isotherm at $T_2$,choosing a point $(V=2 \text{ m}^3, P=2 \times 10^5 \text{ Pa})$,we get $T_2 \propto P_2V_2 = (2 \times 10^5) \times 2 = 4 \times 10^5$.
For the isotherm at $T_1$,choosing a point $(V=1 \text{ m}^3, P=1 \times 10^5 \text{ Pa})$,we get $T_1 \propto P_1V_1 = (1 \times 10^5) \times 1 = 1 \times 10^5$.
Therefore,$\frac{T_2}{T_1} = \frac{4 \times 10^5}{1 \times 10^5} = 4$.
The ratio of the r.m.s. speeds is $\sqrt{\frac{T_2}{T_1}} = \sqrt{4} = 2$.

(B) The $rms$ velocity of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that the $rms$ velocity of hydrogen $(H_2)$ is equal to that 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 $(3R)$:
$\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}$
Here,$M_{H_2} = 2 \ g/mol$,$M_{O_2} = 32 \ g/mol$,and $T_{O_2} = 47 + 273 = 320 \ K$.
Substituting the values:
$\frac{T_{H_2}}{2} = \frac{320}{32}$
$\frac{T_{H_2}}{2} = 10$
$T_{H_2} = 20 \ K$.

(A) According to the kinetic molecular theory of gases,the root mean square $(rms)$ velocity is given by the formula:
$u_{rms} = \sqrt{\frac{3RT}{M}}$
where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
At a given temperature,$u_{rms} \propto \frac{1}{\sqrt{M}}$.
This implies that the $rms$ velocity is inversely proportional to the square root of the molar mass of the gas.
Comparing the molar masses: $M(H_2) = 2 \ g/mol$,$M(He) = 4 \ g/mol$,$M(N_2) = 28 \ g/mol$,and $M(O_2) = 32 \ g/mol$.
Since $H_2$ has the lowest molar mass,it will have the maximum $rms$ velocity.

(C) 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 nature of the gas (molar mass $M$).
It does not depend on the pressure,volume,or the amount (mass or number of moles) of the gas present in the vessel.
Since the temperature remains constant and the gas composition does not change,the $rms$ speed of the remaining molecules will remain the same as the initial $rms$ speed.
Therefore,the $rms$ speed remains $90 \times 10^2 \ m/s$.

(A) The expressions for the speeds of gas molecules are given by:
$V_{rms} = \sqrt{\frac{3kT}{m}} \approx 1.732 \sqrt{\frac{kT}{m}}$
$\bar{V} = \sqrt{\frac{8kT}{\pi m}} \approx 1.596 \sqrt{\frac{kT}{m}}$
$V_m = \sqrt{\frac{2kT}{m}} \approx 1.414 \sqrt{\frac{kT}{m}}$
Comparing the numerical coefficients,we find that $1.732 > 1.596 > 1.414$.
Therefore,the correct relationship is $V_{rms} > \bar{V} > V_m$.

(C) 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.
From this formula,it is clear that $V_{rms} \propto \sqrt{T}$.
Since the temperature $T$ remains constant (room temperature),the $rms$ speed depends only on the temperature and the nature of the gas.
Changing the pressure of the gas at a constant temperature does not affect the $rms$ speed of the gas molecules.
Therefore,$V_1 = V_2$.

(B) The root mean square velocity $(V_{rms})$ of a gas is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M_W}}$.
From this relation,it is clear that $V_{rms} \propto \frac{1}{\sqrt{M_W}}$,where $M_W$ is the molar mass of the gas.
To have the minimum $V_{rms}$,the gas must have the maximum molar mass $(M_W)$.
The molar masses are: $H_2 = 2 \ g/mol$,$O_2 = 32 \ g/mol$,and $CO_2 = 44 \ g/mol$.
Since $CO_2$ has the highest molar mass among the given options,its root mean square velocity will be the minimum.

(B) 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.
From this formula,it is evident 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.
Since the temperature remains constant at $27\,^{\circ}C$ $(300\,K)$ and the gas remains helium,the $rms$ speed will remain unchanged regardless of the change in pressure from $1\,atm$ to $2\,atm$.
Therefore,the $rms$ speed is $900\,ms^{-1}$.

(B) The root mean square velocity $v_{rms}$ of a gas molecule is given by the formula:
$v_{rms} = \sqrt{\frac{3RT}{M_w}}$
Squaring both sides,we get:
$v_{rms}^2 = \frac{3RT}{M_w}$
Since $R$ (universal gas constant) and $M_w$ (molar mass of the gas) are constants for a given 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 the slope $m = \frac{3R}{M_w}$.
Therefore,the graph between $v_{rms}^2$ and $T$ is a straight line passing through the origin.

(D) The root mean square velocity (thermal velocity) of a gas molecule is given by the formula: $v_{rms} = \sqrt{\frac{3k_B T}{m}}$.
Given values are:
$k_B = 1.4 \times 10^{-23}\,J/K$
$T = 300\,K$
$m_{He} = 7 \times 10^{-27}\,kg$
Substituting these values into the formula:
$v = \sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{7 \times 10^{-27}}}$
$v = \sqrt{\frac{1260 \times 10^{-23}}{7 \times 10^{-27}}}$
$v = \sqrt{180 \times 10^4}$
$v = \sqrt{1.8 \times 10^6}$
$v \approx 1.34 \times 10^3\,m/s$.
Thus,the value closest to the thermal velocity is $1.3 \times 10^3\,m/s$.

(A) The formula for the $r.m.s.$ speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
Given $v_{rms} = 1930 \, m/s$,$T = 300 \, K$ (room temperature),and $R = 8.314 \, J/(mol \cdot K)$.
Squaring both sides: $v_{rms}^2 = \frac{3RT}{M}$.
Rearranging for $M$: $M = \frac{3RT}{v_{rms}^2}$.
Substituting the values: $M = \frac{3 \times 8.314 \times 300}{1930^2} \approx \frac{7482.6}{3724900} \approx 0.002008 \, kg/mol$.
This is approximately $2 \times 10^{-3} \, kg/mol$,which corresponds to $2 \, g/mol$.
The molar mass of $H_2$ is $2 \, g/mol$. Therefore,the gas is $H_2$.

(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 both gases are in the same container at the same temperature $T = 300\, K$, the ratio of their rms speeds is:
$\frac{V_{rms}(\text{helium})}{V_{rms}(\text{argon})} = \frac{\sqrt{\frac{3RT}{M_{He}}}}{\sqrt{\frac{3RT}{M_{Ar}}}} = \sqrt{\frac{M_{Ar}}{M_{He}}}$
Given $M_{He} = 4\, u$ and $M_{Ar} = 40\, u$, we substitute these values:
$\frac{V_{rms}(\text{helium})}{V_{rms}(\text{argon})} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$.

(A) The root mean square velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R = k_B N_A$ is the universal gas constant,$T$ is the temperature,and $M$ is the molar mass of hydrogen $(2 \times 10^{-3} \, kg/mol)$.
The escape velocity from the Earth is given by $v_e = \sqrt{2gR_e}$.
Equating the two: $\sqrt{\frac{3RT}{M}} = \sqrt{2gR_e}$.
Squaring both sides: $\frac{3RT}{M} = 2gR_e$.
Solving for $T$: $T = \frac{2gR_e M}{3R}$.
Given $R = k_B N_A \approx 8.314 \, J/(mol \cdot K)$.
Substituting the values: $T = \frac{2 \times 10 \times 6.4 \times 10^6 \times 2 \times 10^{-3}}{3 \times 8.314} = \frac{25600}{24.942} \approx 1026 \, K$.
Among the given options,the value closest to $1026 \, K$ is $800 \, K$ (Option $A$).

(A) The root mean square $(rms)$ velocity of a gas molecule is related to its temperature by the kinetic theory of gases.
For any gas molecule,the average translational kinetic energy is given by the relation:
$\frac{1}{2}m\bar{v}^2 = \frac{3}{2}k_B T$
Here,$\bar{v}$ represents the $rms$ velocity of the molecule.
By simplifying the equation:
$m\bar{v}^2 = 3k_B T$
Solving for temperature $T$:
$T = \frac{m\bar{v}^2}{3k_B}$
Thus,the correct option is $A$.

(B) The root mean square $(rms)$ speed of gas molecules is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M_w}}$.
From this relation,it is clear that $V_{rms} \propto \sqrt{T}$.
Note that the $rms$ speed is independent of pressure.
Given: $T_1 = 127\,^oC = 127 + 273 = 400\,K$ and $V_1 = 200\,m/s$.
Given: $T_2 = 227\,^oC = 227 + 273 = 500\,K$.
Using the proportionality: $\frac{V_2}{V_1} = \sqrt{\frac{T_2}{T_1}}$.
$\frac{V_2}{200} = \sqrt{\frac{500}{400}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$.
$V_2 = 200 \times \frac{\sqrt{5}}{2} = 100\sqrt{5}\,m/s$.

(D) The $r.m.s.$ velocity of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $V_{rms} \propto \sqrt{T}$.
Initial temperature $T_1 = 30 + 273 = 303 \ K$.
Final temperature $T_2 = 90 + 273 = 363 \ K$.
The ratio of velocities is $\frac{V_2}{V_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{363}{303}} \approx \sqrt{1.198} \approx 1.0945$.
The percentage increase is given by $\left( \frac{V_2 - V_1}{V_1} \right) \times 100 = (1.0945 - 1) \times 100 \approx 9.45 \%$.
Rounding to the nearest provided option,the percentage increase is approximately $10 \%$.

(B) The root mean square $(V_{RMS})$ velocity is defined as the square root of the mean of the squares of the individual speeds.
Given speeds are $V_1 = 1 \, m/s$,$V_2 = 2 \, m/s$,$V_3 = 4 \, m/s$,$V_4 = 8 \, m/s$,and $V_5 = 16 \, m/s$.
The formula for $V_{RMS}$ is:
$V_{RMS} = \sqrt{\frac{V_1^2 + V_2^2 + V_3^2 + V_4^2 + V_5^2}{N}}$
Substituting the values:
$V_{RMS} = \sqrt{\frac{1^2 + 2^2 + 4^2 + 8^2 + 16^2}{5}}$
$V_{RMS} = \sqrt{\frac{1 + 4 + 16 + 64 + 256}{5}}$
$V_{RMS} = \sqrt{\frac{341}{5}}$
$V_{RMS} = \sqrt{68.2} \approx 8.258 \, m/s$.
Rounding to the nearest provided option,we get $8.2 \, m/s$.

(B) The root mean square velocity is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M_w}}$,where $R$ is the universal gas constant,$T$ is the temperature,and $M_w$ is the molar mass.
From this,we can see that $V_{rms} \propto \sqrt{\frac{T}{M_w}}$.
For $O_2$ molecules,the molar mass is $M_w = 32 \ g/mol$. Given $V_{rms} = V$ at temperature $T$.
For oxygen atoms $(O)$,the molar mass is $M_w' = 16 \ g/mol$. The new temperature is $T' = 2T$.
Now,calculating the new $V_{rms}'$:
$V_{rms}' = V \times \sqrt{\frac{T'}{T} \times \frac{M_w}{M_w'}}$
$V_{rms}' = V \times \sqrt{\frac{2T}{T} \times \frac{32}{16}}$
$V_{rms}' = V \times \sqrt{2 \times 2} = V \times \sqrt{4} = 2V$.

(C) For an ideal gas,the equation of state is $PV = \mu RT$. Since the mass is fixed,$\mu$ is constant. Thus,$T \propto PV$.
From the graph,we can choose points on the isotherms to find the product $PV$ for each temperature.
For isotherm $T_1$,at $V = 1 \text{ m}^3$,$P = 1 \times 10^5 \text{ Pa}$. So,$T_1 \propto (1 \times 1) = 1$.
For isotherm $T_2$,at $V = 2 \text{ m}^3$,$P = 2 \times 10^5 \text{ Pa}$. So,$T_2 \propto (2 \times 2) = 4$.
Therefore,the ratio of temperatures is $\frac{T_2}{T_1} = \frac{4}{1} = 4$.
The $r.m.s.$ speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.
Thus,the ratio of $r.m.s.$ speeds is $\frac{v_{rms,2}}{v_{rms,1}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{4} = 2$.

(A) The root mean square speed $(V_{rms})$ is defined as the square root of the mean of the squares of the individual speeds of the molecules.
The formula is given by $V_{rms} = \sqrt{\frac{v_{1}^{2} + v_{2}^{2} + v_{3}^{2} + v_{4}^{2}}{4}}$.
Given speeds are $v_{1} = 1 \ km/s$,$v_{2} = 2 \ km/s$,$v_{3} = 3 \ km/s$,and $v_{4} = 4 \ km/s$.
Substituting the values into the formula:
$V_{rms} = \sqrt{\frac{1^{2} + 2^{2} + 3^{2} + 4^{2}}{4}}$
$V_{rms} = \sqrt{\frac{1 + 4 + 9 + 16}{4}}$
$V_{rms} = \sqrt{\frac{30}{4}}$
$V_{rms} = \sqrt{\frac{15}{2}} \ km/s$.

(D) The most probable speed $v_p$ of gas molecules is given by the formula $v_p = \sqrt{\frac{2RT}{M}}$,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 $v_p \propto \frac{1}{\sqrt{M}}$.
This means that the gas with the smaller molar mass will have a higher most probable speed.
The molar masses of the given gases are: $M(O_2) = 32 \text{ g/mol}$,$M(Ne) = 20 \text{ g/mol}$,and $M(He) = 4 \text{ g/mol}$.
Comparing these,we have $M(O_2) > M(Ne) > M(He)$.
Therefore,the order of their most probable speeds will be $v_p(O_2) < v_p(Ne) < v_p(He)$.
Looking at the graph,the peak of the distribution shifts to the right as speed increases. Thus,the order of speeds corresponding to the peaks is $A < B < C$.
Matching these,we get $A \to O_2$,$B \to Ne$,and $C \to He$.

(C) The $rms$ 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.
For oxygen molecules $(O_2)$ at temperature $T$,the molar mass is $M$. Thus,$v_{rms} = \sqrt{\frac{3RT}{M}} = \upsilon$.
When the temperature is doubled $(T' = 2T)$ and the oxygen dissociates into atomic oxygen $(O)$,the molar mass becomes $M' = M/2$.
The new $rms$ speed is: $v'_{rms} = \sqrt{\frac{3R(2T)}{M/2}} = \sqrt{4 \cdot \frac{3RT}{M}} = 2 \sqrt{\frac{3RT}{M}} = 2\upsilon$.

(B) The root mean square speed $(v_{rms})$ of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$,we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Given that $v_2 = \frac{1}{2}v_1$ and $T_1 = 0^{\circ}C = 273 \, K$.
Substituting these values: $\frac{1}{2} = \sqrt{\frac{T_2}{273}}$.
Squaring both sides: $\frac{1}{4} = \frac{T_2}{273}$.
$T_2 = \frac{273}{4} = 68.25 \, K$.
To convert this temperature to Celsius: $T(^{\circ}C) = T(K) - 273 = 68.25 - 273 = -204.75^{\circ}C$.

(D) Since the separator is diathermic,the system is in thermal equilibrium,so the temperature $T$ is the same for both gases.
The $rms$ speed of molecules in the $L$ part is given by $v_{rms, L} = \sqrt{\frac{3kT}{m_L}}$,where $m_L$ is the mass of a molecule in the $L$ part.
The mean speed of molecules in the $R$ part is given by $v_{mean, R} = \sqrt{\frac{8kT}{\pi m_R}}$,where $m_R$ is the mass of a molecule in the $R$ part.
Given that $v_{rms, L} = v_{mean, R}$,we have:
$\sqrt{\frac{3kT}{m_L}} = \sqrt{\frac{8kT}{\pi m_R}}$
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 $\frac{m_L}{m_R}$:
$\frac{m_L}{m_R} = \frac{3\pi}{8}$

(A) The average speed of $O_2$ is given by $V_{av} = \sqrt{\frac{8RT}{\pi M_{O_2}}}$,where $M_{O_2} = 32 \times 10^{-3} \, kg/mol$.
The $rms$ speed of $N_2$ is given by $V_{rms} = \sqrt{\frac{3RT'}{M_{N_2}}}$,where $T' = 47 + 273 = 320 \, K$ and $M_{N_2} = 28 \times 10^{-3} \, kg/mol$.
According to the problem: $\sqrt{\frac{8RT}{\pi \times 32}} = \frac{7}{4} \sqrt{\frac{3R \times 320}{28}}$.
Squaring both sides: $\frac{8RT}{32\pi} = \frac{49}{16} \times \frac{3R \times 320}{28}$.
Simplifying the equation: $\frac{RT}{4\pi} = \frac{49 \times 3R \times 320}{16 \times 28}$.
Solving for $T$: $T = \frac{49 \times 3 \times 320 \times 4\pi}{16 \times 28} \approx 1320 \, K$ (taking $\pi \approx \frac{22}{7}$ for calculation).