Molecular speeds Questions in English

(D) The formula for $RMS$ velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Squaring both sides,we get $v_{rms}^2 = \frac{3RT}{M}$.
Comparing this with the equation of a straight line $y = mx + C$,where $y = v_{rms}^2$ and $x = T$:
The slope $m = \frac{3R}{M}$.
Given slope $m = 249 \ m^2 \ s^{-2} \ K^{-1}$ and $R = 8.3 \ J \ mol^{-1} \ K^{-1}$.
Substituting the values: $249 = \frac{3 \times 8.3}{M}$.
$M = \frac{3 \times 8.3}{249} = \frac{24.9}{249} = 0.1 \ kg \ mol^{-1}$.
Thus,$M = 1 \times 10^{-1} \ kg \ mol^{-1}$.

(A) According to the Maxwell-Boltzmann distribution of molecular speeds,as the temperature increases,the peak of the curve shifts towards higher velocities and the curve becomes broader and flatter.
In the given graph,the peak for $T_2$ is at the highest velocity,followed by $T_1$,and then $T_3$.
Therefore,the correct order of temperature is $T_2 > T_1 > T_3$.

(A) Given: $T = 133.33 \ K$,$M = 0.083 \ kg \ mol^{-1}$,$R = 8.3 \ J \ mol^{-1} \ K^{-1}$.
The formula for the root mean square $(RMS)$ velocity of an ideal gas is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Substituting the given values into the formula:
$v_{rms} = \sqrt{\frac{3 \times 8.3 \times 133.33}{0.083}}$
$v_{rms} = \sqrt{\frac{3323.917}{0.083}}$
$v_{rms} = \sqrt{40047.19} \approx 200.12 \ m \ s^{-1}$.
Thus,the $RMS$ velocity is approximately $200 \ m \ s^{-1}$.

(D) The kinetic energy $(K.E.)$ for one mole of an ideal gas is given by the formula: $K.E. = \frac{1}{2} M v_{rms}^2$
Given:
Molar mass $(M)$ $= 0.1 \ kg \ mol^{-1}$
$v_{rms} = 4 \times 10^2 \ ms^{-1}$
Substituting the values:
$K.E. = \frac{1}{2} \times (0.1 \ kg \ mol^{-1}) \times (4 \times 10^2 \ ms^{-1})^2$
$K.E. = 0.05 \times (16 \times 10^4)$
$K.E. = 0.8 \times 10^4 = 8 \times 10^3 \ J \ mol^{-1}$

(D) $RMS$ velocity $(u_{rms})$ is given by the formula: $u_{rms} = \sqrt{\frac{3 RT}{M}}$
Squaring both sides,we get: $(u_{rms})^2 = \frac{3 RT}{M}$
Comparing this with the equation of a straight line passing through the origin,$y = mx + C$,where $y = (u_{rms})^2$ and $x = T$:
$(u_{rms})^2 = (\frac{3 R}{M}) \cdot T + 0$
Here,the slope $(m)$ is equal to $\frac{3 R}{M}$.

(A) The root mean square velocity $(u_{rms})$ is given by the formula: $u_{rms} = \sqrt{\frac{3RT}{M}}$.
For $1$ mole of an ideal gas,the ideal gas equation is $PV = RT$.
Substituting $RT = PV$ into the $u_{rms}$ equation,we get:
$u_{rms} = \sqrt{\frac{3PV}{M}}$
Squaring both sides,we get:
$u_{rms}^2 = \frac{3PV}{M}$
Comparing this with the equation of a straight line $y = mx + C$,where $y = u_{rms}^2$ and $x = PV$:
$y = \left(\frac{3}{M}\right)x + 0$
Thus,the slope $(m)$ is $\frac{3}{M}$.

(B) The formulas for the velocities are:
$u_{rms} = \sqrt{\frac{3RT}{M}}$
$u_{av} = \sqrt{\frac{8RT}{\pi M}}$
$u_{mp} = \sqrt{\frac{2RT}{M}}$
Calculating the ratios:
$1$. $\frac{u_{rms}}{u_{av}} = \frac{\sqrt{3}}{\sqrt{8/\pi}} = \sqrt{\frac{3\pi}{8}} \approx 1.085$
$2$. $\frac{u_{av}}{u_{mp}} = \frac{\sqrt{8/\pi}}{\sqrt{2}} = \sqrt{\frac{4}{\pi}} \approx 1.128 \approx 1.12$
$3$. $\frac{u_{rms}}{u_{mp}} = \frac{\sqrt{3}}{\sqrt{2}} = \sqrt{1.5} \approx 1.225$
Comparing these with the given options,the ratio $\frac{u_{av}}{u_{mp}} = 1.12$ is correct.

(A) The root mean square speed $(u_{rms})$ is directly proportional to the square root of the absolute temperature $(T)$:
$u_{rms} = \sqrt{\frac{3RT}{M}}$
For the same gas,$u_{rms} \propto \sqrt{T}$.
Initial temperature $T_{1} = 30 + 273 = 303 \ K$.
Final temperature $T_{2} = 930 + 273 = 1203 \ K$.
Taking the ratio:
$\frac{u_{2}}{u_{1}} = \sqrt{\frac{T_{2}}{T_{1}}} = \sqrt{\frac{1203}{303}} = \sqrt{3.97} \approx \sqrt{4} = 2$.
Therefore,$u_{2} = 2u_{1}$.
Thus,the root mean square speed of the gas gets doubled.

(C) The average molecular speed $(v_{avg})$ is given by the formula: $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.
This shows that $v_{avg} \propto \sqrt{\frac{T}{M}}$.
Comparing the ratios of $\frac{T}{M}$ for each gas:
$A: \frac{500}{32} = 15.625$
$B: \frac{400}{44} \approx 9.09$
$C: \frac{200}{4} = 50.0$
$D: \frac{300}{17} \approx 17.65$
Since the ratio $\frac{T}{M}$ is highest for $He$ $(50.0)$,it will have the greatest average speed.

(B) The formula for rms velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
For oxygen $(O_2)$,$T_1 = 27 + 273 = 300 \ K$ and $M_1 = 32 \ g/mol$. Given $v_{rms,1} = 800 \ m/s$.
For methane $(CH_4)$,$T_2 = 600 \ K$ and $M_2 = 16 \ g/mol$. Let $v_{rms,2} = x$.
Taking the ratio: $\frac{v_{rms,1}}{v_{rms,2}} = \sqrt{\frac{T_1}{M_1} \times \frac{M_2}{T_2}}$.
Substituting the values: $\frac{800}{x} = \sqrt{\frac{300}{32} \times \frac{16}{600}} = \sqrt{\frac{1}{2} \times \frac{1}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,$x = 800 \times 2 = 1600 \ m/s$.
Therefore,the correct option is $B$.

(C) The kinetic energy $(KE)$ of an ideal gas is given by $KE = \frac{3}{2} RT$.
The root mean square speed $(v_{rms})$ is given by $v_{rms} = \sqrt{\frac{3 RT}{M}}$.
Substituting $\frac{3}{2} RT = KE$,we get $v_{rms} = \sqrt{\frac{2 KE}{M}}$.
Given $KE = 4.0 \ kJ \ mol^{-1} = 4000 \ J \ mol^{-1}$ and molar mass $M$ of $O_2 = 32 \times 10^{-3} \ kg \ mol^{-1}$.
$v_{rms} = \sqrt{\frac{2 \times 4000 \ J \ mol^{-1}}{32 \times 10^{-3} \ kg \ mol^{-1}}} = \sqrt{\frac{8000}{32 \times 10^{-3}}} = \sqrt{250000} = 500 \ m \ s^{-1}$.
Converting to $cm \ s^{-1}$: $500 \ m \ s^{-1} \times 100 \ cm \ m^{-1} = 5.0 \times 10^4 \ cm \ s^{-1}$.

(C) Root mean square speed $(v_{rms}) = \sqrt{\frac{3RT}{M}}$
Given:
$M_1 = \text{molar mass of } H_2 = 2 \ g/mol$
$M_2 = \text{molar mass of } O_2 = 32 \ g/mol$
$T_1 = 50 \ K, T_2 = 800 \ K$
Taking the ratio of $v_{rms,1}$ and $v_{rms,2}$:
$\frac{v_{rms,1}}{v_{rms,2}} = \sqrt{\frac{3RT_1}{M_1}} \times \sqrt{\frac{M_2}{3RT_2}} = \sqrt{\frac{M_2 \times T_1}{M_1 \times T_2}}$
$\frac{v_{rms,1}}{v_{rms,2}} = \sqrt{\frac{32}{2} \times \frac{50}{800}} = \sqrt{16 \times \frac{1}{16}} = \sqrt{1} = 1$
Thus,the ratio is $1 : 1$.
Hence,option $(C)$ is correct.

(B) The kinetic energy $(KE)$ of one mole of an ideal gas is given by the formula: $KE = \frac{1}{2} M v_{rms}^2$,where $M$ is the molar mass in $kg \ mol^{-1}$ and $v_{rms}$ is the root mean square speed in $m \ s^{-1}$.
Given: $KE = 4.0 \ kJ \ mol^{-1} = 4000 \ J \ mol^{-1}$ and $v_{rms} = 5.0 \times 10^4 \ cm \ s^{-1} = 500 \ m \ s^{-1}$.
Substituting the values into the equation: $4000 = \frac{1}{2} \times M \times (500)^2$.
$4000 = \frac{1}{2} \times M \times 250000$.
$4000 = M \times 125000$.
$M = \frac{4000}{125000} = 0.032 \ kg \ mol^{-1}$.
Converting to grams: $M = 0.032 \times 1000 \ g \ mol^{-1} = 32 \ g \ mol^{-1}$.

(B) Given,\\ Temperature $(T)$ is the same for most probable velocity $(v_{mp})$ and root mean square velocity $(v_{rms})$. \\ The formulas are: \\ $v_{mp} = \sqrt{\frac{2RT}{M}}$ and $v_{rms} = \sqrt{\frac{3RT}{M}}$. \\ Taking the ratio: \\ $\frac{v_{rms}}{v_{mp}} = \sqrt{\frac{3RT/M}{2RT/M}} = \sqrt{\frac{3}{2}} = \sqrt{1.5} \approx 1.2247$. \\ Given $v_{mp} = 400 \ ms^{-1}$,then: \\ $v_{rms} = 1.2247 \times 400 \ ms^{-1} \approx 489.88 \ ms^{-1} \approx 490 \ ms^{-1}$. \\ Thus,option $(B)$ is the correct answer.

(C) The most probable speed $(v_{mp})$ is given by the formula $v_{mp} = \sqrt{\frac{2RT}{M}}$.
At a constant temperature,$v_{mp} \propto \frac{1}{\sqrt{M}}$.
From the given graph,the order of most probable speeds is $r_2 < r_1 < r_3$.
Since $v_{mp}$ is inversely proportional to the square root of the molar mass,a lower speed corresponds to a higher molar mass.
Therefore,the order of molar masses is $M_2 > M_1 > M_3$.

(B) Given,$U_{rms} = 3000 \ ms^{-1}$.
The formula for $RMS$ speed is $U_{rms} = \sqrt{\frac{3RT}{M}}$.
Squaring both sides,we get $U_{rms}^2 = \frac{3RT}{M}$.
For nitrogen $(N_2)$,the molar mass $M = 28 \ g/mol = 28 \times 10^{-3} \ kg/mol$.
Substituting the values: $(3000)^2 = \frac{3RT}{28 \times 10^{-3}}$.
$9 \times 10^6 = \frac{3RT}{28 \times 10^{-3}}$.
$3RT = 9 \times 10^6 \times 28 \times 10^{-3} = 252 \times 10^3 \ J/mol$.
The kinetic energy $(K.E.)$ for $1$ mole of an ideal gas is given by $K.E. = \frac{3}{2} RT$.
$K.E. = \frac{1}{2} \times (3RT) = \frac{1}{2} \times 252 \times 10^3 \ J/mol = 126 \times 10^3 \ J/mol = 126 \ kJ$.

(D) The most probable speed $(u_{mp})$ is given by the formula: $u_{mp} = \sqrt{\frac{2RT}{M}}$.
Given $u_{mp} = 400 \ ms^{-1}$ and molar mass of methane $(M)$ = $16 \ g \ mol^{-1} = 0.016 \ kg \ mol^{-1}$.
Squaring both sides: $u_{mp}^2 = \frac{2RT}{M} \Rightarrow \frac{RT}{M} = \frac{u_{mp}^2}{2}$.
Substituting the values: $\frac{RT}{0.016} = \frac{400^2}{2} = \frac{160000}{2} = 80000$.
Thus,$RT = 80000 \times 0.016 = 1280 \ J \ mol^{-1}$.
The kinetic energy $(KE)$ of one mole of an ideal gas is given by: $KE = \frac{3}{2} RT$.
$KE = \frac{3}{2} \times 1280 = 3 \times 640 = 1920 \ J$.

(D) The formula for $RMS$ velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that the $RMS$ velocity of $He$ at $T \ K$ is equal to the $RMS$ velocity of $SO_2$ at $127^{\circ} C$ $(400 \ K)$:
$\sqrt{\frac{3RT}{M_{He}}} = \sqrt{\frac{3R(400)}{M_{SO_2}}}$
$\frac{T}{M_{He}} = \frac{400}{M_{SO_2}}$
Given $M_{He} = 4 \ g/mol$ and $M_{SO_2} = 64 \ g/mol$:
$\frac{T}{4} = \frac{400}{64}$
$T = \frac{400 \times 4}{64} = \frac{1600}{64} = 25 \ K$.

(A) According to the Maxwell-Boltzmann distribution,the most probable velocity $(v_{mp})$ is given by the formula $v_{mp} = \sqrt{\frac{2RT}{M}}$.
This shows that $v_{mp} \propto \frac{1}{\sqrt{M}}$.
From the given graph,the order of most probable velocities is $v_{mp}(M_2) < v_{mp}(M_1) < v_{mp}(M_3)$.
Since the velocity is inversely proportional to the square root of the molar mass,the order of molar masses will be $M_2 > M_1 > M_3$.

(C) The root mean square $(RMS)$ velocity of an ideal gas is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$,we can write: $\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$.
Given $T_1 = 127^{\circ} C = 127 + 273 = 400 \ K$ and $v_1 = v$.
We want to find $T_2$ such that $v_2 = 2v$.
Substituting the values: $\frac{v}{2v} = \sqrt{\frac{400}{T_2}}$.
$\frac{1}{2} = \sqrt{\frac{400}{T_2}}$.
Squaring both sides: $\frac{1}{4} = \frac{400}{T_2}$.
$T_2 = 400 \times 4 = 1600 \ K$.

(C) The formula for $RMS$ velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given: $v_{H_2} = \sqrt{7} \times v_{N_2}$.
Substituting the formula: $\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{7} \times \sqrt{\frac{3RT_{N_2}}{M_{N_2}}}$.
Squaring both sides: $\frac{3RT_{H_2}}{M_{H_2}} = 7 \times \frac{3RT_{N_2}}{M_{N_2}}$.
Given molar masses: $M_{H_2} = 2 \ g/mol$ and $M_{N_2} = 28 \ g/mol$.
Substituting these values: $\frac{T_{H_2}}{2} = 7 \times \frac{T_{N_2}}{28}$.
Simplifying: $\frac{T_{H_2}}{2} = \frac{T_{N_2}}{4}$.
Therefore,$T_{H_2} = \frac{T_{N_2}}{2}$.

(A) The root mean square velocity $(u_{rms})$ is given by the formula: $u_{rms} = \sqrt{\frac{3RT}{M}}$.
Given $u_{rms} = 20 \ ms^{-1}$ and $M = 40 \ g \ mol^{-1} = 0.04 \ kg \ mol^{-1}$.
Squaring both sides: $u_{rms}^2 = \frac{3RT}{M} \implies 20^2 = \frac{3RT}{0.04}$.
$400 = \frac{3RT}{0.04} \implies 3RT = 400 \times 0.04 = 16$.
$RT = \frac{16}{3} \ J \ mol^{-1}$.
The average kinetic energy $(KE_{avg})$ for one mole of an ideal gas is given by: $KE_{avg} = \frac{3}{2}RT$.
Substituting the value of $RT$: $KE_{avg} = \frac{3}{2} \times \frac{16}{3} = 8 \ J \ mol^{-1}$.
Thus,the correct option is $A$.

(D) The formula for $RMS$ velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$ ...$(I)$
The formula for most probable velocity is $v_{mp} = \sqrt{\frac{2RT}{M}}$ ...$(II)$
Given:
For $SO_2$: $T_1 = 400 \ K$,$M_1 = 64 \ g/mol$
For $O_2$: $T_2 = ?$,$M_2 = 32 \ g/mol$
Equating the two velocities:
$\sqrt{\frac{3R \times 400}{64}} = \sqrt{\frac{2R \times T_2}{32}}$
Squaring both sides:
$\frac{1200}{64} = \frac{2T_2}{32}$
$\frac{1200}{64} = \frac{T_2}{16}$
$T_2 = \frac{1200 \times 16}{64} = \frac{1200}{4} = 300 \ K$

(A) The formula for average velocity is $\mu_{av} = \sqrt{\frac{8RT}{\pi M}}$.
Since $\mu_{av} \propto \sqrt{T}$,we have the ratio $\frac{\mu_{av_1}}{\mu_{av_2}} = \sqrt{\frac{T_1}{T_2}}$.
Given $T_1 = 300 \ K$,$T_2 = 1200 \ K$,and $\mu_{av_1} = 3 \times 10^2 \ cm / s$.
Substituting the values: $\frac{3 \times 10^2}{\mu_{av_2}} = \sqrt{\frac{300}{1200}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,$\mu_{av_2} = 2 \times 3 \times 10^2 = 6 \times 10^2 \ cm / s$.

(B) Given,root mean square $(rms)$ speed of $O_2$ is $500 \ m/s$ at a constant temperature.
Root mean square speed is given by the expression $u_{rms} = \sqrt{\frac{3RT}{Mw}}$.
For $H_2$ gas,$u_{rms(H_2)} = \sqrt{\frac{3RT}{Mw_{H_2}}}$ where $Mw_{H_2} = 2 \ g/mol$.
For $O_2$ gas,$u_{rms(O_2)} = \sqrt{\frac{3RT}{Mw_{O_2}}}$ where $Mw_{O_2} = 32 \ g/mol$.
Taking the ratio: $\frac{u_{rms(H_2)}}{u_{rms(O_2)}} = \sqrt{\frac{Mw_{O_2}}{Mw_{H_2}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
Thus,$u_{rms(H_2)} = 4 \times 500 \ m/s = 2000 \ m/s$.
Average kinetic energy per mole is given by $KE_{avg} = \frac{3}{2} RT$.
First,find $T$ from $u_{rms(O_2)} = \sqrt{\frac{3RT}{Mw_{O_2}}}$.
$500 = \sqrt{\frac{3 \times 8.33 \times T}{0.032 \ kg/mol}}$.
$250000 = \frac{24.99 \times T}{0.032} \implies T = \frac{250000 \times 0.032}{24.99} \approx 320 \ K$.
$KE_{avg} = \frac{3}{2} \times 8.33 \times 320 \approx 4000 \ J/mol = 4 \ kJ/mol$.

(C) The kinetic energy $(KE)$ of one mole of an ideal gas is given by $KE = \frac{3}{2} RT$.
From this,we get $RT = \frac{2}{3} KE$.
The most probable velocity $(v_{mp})$ is given by $v_{mp} = \sqrt{\frac{2RT}{M}}$.
Here,$M$ is the molar mass of oxygen $(O_2)$,which is $32 \ g \ mol^{-1} = 0.032 \ kg \ mol^{-1}$.
Given $KE = 3741.3 \ J \ mol^{-1}$.
Substituting $RT = \frac{2}{3} \times 3741.3$ into the velocity formula:
$v_{mp} = \sqrt{\frac{2 \times (\frac{2}{3} \times 3741.3)}{0.032}}$
$v_{mp} = \sqrt{\frac{4 \times 3741.3}{3 \times 0.032}} = \sqrt{\frac{14965.2}{0.096}} = \sqrt{155887.5} \ m \ s^{-1}$.
Thus,the approximate most probable velocity is $\sqrt{155887} \ m \ s^{-1}$.
Therefore,option $(C)$ is the correct answer.

(C) The formula for the most probable speed $(v_{mp})$ is $v_{mp} = \sqrt{\frac{2RT}{M}}$.
Given:
Molar mass of $N_2$ $(M_1)$ = $28 \ g \ mol^{-1}$,Temperature $(T_1)$ = $400 \ K$.
Molar mass of $CO$ $(M_2)$ = $28 \ g \ mol^{-1}$,Temperature $(T_2)$ = $800 \ K$.
The ratio of the most probable speed of $N_2$ to $CO$ is:
$\frac{v_{mp(N_2)}}{v_{mp(CO)}} = \frac{\sqrt{\frac{2RT_1}{M_1}}}{\sqrt{\frac{2RT_2}{M_2}}} = \sqrt{\frac{T_1}{T_2} \times \frac{M_2}{M_1}}$
Substituting the values:
$\frac{v_{mp(N_2)}}{v_{mp(CO)}} = \sqrt{\frac{400}{800} \times \frac{28}{28}} = \sqrt{\frac{1}{2}} = \sqrt{0.5} \approx 0.707$.

(D) The formula for $RMS$ velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Here,$R$ is the molar gas constant,$T$ is the temperature in Kelvin,and $M$ is the molar mass.
Given $M_{(N_2)} = 28 \ g \ mol^{-1}$ and $M_{(CO)} = 28 \ g \ mol^{-1}$.
Since the molar masses are equal,the ratio of $RMS$ velocities depends only on the square root of the temperatures:
$\frac{v_{rms}(N_2)}{v_{rms}(CO)} = \sqrt{\frac{T_{(N_2)}}{T_{(CO)}}} = \sqrt{\frac{200}{800}}$.
$\frac{v_{rms}(N_2)}{v_{rms}(CO)} = \sqrt{\frac{1}{4}} = \frac{1}{2} = 0.5$.

(D) According to the Maxwell-Boltzmann distribution,as the temperature of a gas increases,the most probable speed $(u_{mp})$ increases,and the peak of the distribution curve shifts to the right and flattens.
In the given figure,the peak for $T_2$ is at the highest speed,followed by $T_3$,and then $T_1$.
Therefore,the most probable speed follows the order $u_{mp}(T_2) > u_{mp}(T_3) > u_{mp}(T_1)$.
Since $u_{mp} = \sqrt{\frac{2RT}{M}}$,it implies that $u_{mp} \propto \sqrt{T}$.
Thus,the correct order of temperature is $T_2 > T_3 > T_1$.

(C) The formula for the most probable speed $(v_{mp})$ is given by $v_{mp} = \sqrt{\frac{2 R T}{M}}$.
For $O_2$ gas,the molar mass $M = 32 \ g \ mol^{-1}$.
Substituting the value of $M$ into the formula: $v_{mp} = \sqrt{\frac{2 R T}{32}}$.
Simplifying the expression,we get $v_{mp} = \sqrt{\frac{R T}{16}}$.

(C) Given velocities: $C_1 = 100 \ ms^{-1}, C_2 = 200 \ ms^{-1}, C_3 = 500 \ ms^{-1}$.
Root mean square (rms) velocity is given by the formula: $C_{rms} = \sqrt{\frac{C_1^2 + C_2^2 + C_3^2}{n}}$.
Substituting the values: $C_{rms} = \sqrt{\frac{100^2 + 200^2 + 500^2}{3}}$.
$C_{rms} = \sqrt{\frac{10000 + 40000 + 250000}{3}}$.
$C_{rms} = \sqrt{\frac{300000}{3}} = \sqrt{100000}$.
$C_{rms} = 100 \sqrt{10} \ ms^{-1}$.

(A) The formula for r.m.s. velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
For $CO_2$: $M_{CO_2} = 12 + (2 \times 16) = 44 \ g \ mol^{-1}$. Given $v_{rms} = x$ at temperature $T$.
For $N_2O$: $M_{N_2O} = (2 \times 14) + 16 = 44 \ g \ mol^{-1}$. Let the temperature be $T'$. Given $v_{rms} = 4x$.
Since $v_{rms} \propto \sqrt{T/M}$ and $M_{CO_2} = M_{N_2O} = 44$,we have $\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$.
Substituting the values: $\frac{x}{4x} = \sqrt{\frac{T}{T'}}$.
$\frac{1}{4} = \sqrt{\frac{T}{T'}}$.
Squaring both sides: $\frac{1}{16} = \frac{T}{T'}$.
Therefore,$T' = 16T$.

(D) The root mean square velocity of a gas is given by $u = \sqrt{\frac{3RT}{M}}$.
Since the temperature $T$ is the same for both gases,$u \propto \frac{1}{\sqrt{m}}$.
Therefore,$\frac{u_{1}}{u_{2}} = \sqrt{\frac{m_{2}}{m_{1}}}$.
Squaring both sides,we get $\frac{u_{1}^{2}}{u_{2}^{2}} = \frac{m_{2}}{m_{1}}$.
Cross-multiplying gives $m_{1} u_{1}^{2} = m_{2} u_{2}^{2}$.

(C) The formula for root mean square speed is $C_{rms} = \sqrt{\frac{3RT}{M}}$.
Initially,for $X_2$ gas at temperature $T$,$C_1 = x = \sqrt{\frac{3RT}{M_{X_2}}}$.
When temperature is doubled $(T_2 = 2T)$ and $X_2$ dissociates into $2X$,the molar mass becomes $M_2 = \frac{M_{X_2}}{2}$.
The new rms speed $C_2$ is given by $C_2 = \sqrt{\frac{3R(2T)}{M_{X_2}/2}} = \sqrt{4 \times \frac{3RT}{M_{X_2}}} = 2 \times \sqrt{\frac{3RT}{M_{X_2}}}$.
Substituting $x$ for the initial speed,we get $C_2 = 2x \ m/s$.

(B) The average kinetic energy $(KE_{avg})$ of an ideal gas is given by $\frac{3}{2}kT$. Since the temperature $T$ is halved in Case $-2$,the average kinetic energy is also halved.
The average speed $(C_{avg})$ is given by $\sqrt{\frac{8RT}{\pi M}}$.
For Case $-1$: $C_1 \propto \sqrt{\frac{T}{M}}$.
For Case $-2$: $C_2 \propto \sqrt{\frac{T/2}{2M}} = \sqrt{\frac{T}{4M}} = \frac{1}{2} \sqrt{\frac{T}{M}}$.
Thus,$C_2 = \frac{1}{2} C_1$.
Therefore,both the average kinetic energy and the average speed are halved in the second case.

(D) The formula for average speed is $C_{av} = \sqrt{\frac{8RT}{\pi M}}$.
Given that $(C_{av})_{H_2} = (C_{av})_{O_2}$,we have:
$\sqrt{\frac{8RT_1}{\pi M_{H_2}}} = \sqrt{\frac{8RT_2}{\pi M_{O_2}}}$
Squaring both sides:
$\frac{T_1}{M_{H_2}} = \frac{T_2}{M_{O_2}}$
$\frac{T_1}{T_2} = \frac{M_{H_2}}{M_{O_2}}$
Substituting the molar masses ($M_{H_2} = 2 \ g/mol$ and $M_{O_2} = 32 \ g/mol$):
$\frac{T_1}{T_2} = \frac{2}{32} = \frac{1}{16}$
Thus,the ratio $T_1: T_2$ is $1: 16$.

(B) The formula for r.m.s. speed is $(C_{rms}) = \sqrt{\frac{3RT}{M}}$ and for most probable speed is $(C_{mp}) = \sqrt{\frac{2RT}{M}}$.
Given $(C_{rms})_{H_2} = \sqrt{\frac{3 \times R \times 150}{2}}$.
According to the problem,$(C_{mp})_{He} = \frac{1}{2} (C_{rms})_{H_2}$.
Substituting the values: $\sqrt{\frac{2RT}{4}} = \frac{1}{2} \sqrt{\frac{3 \times R \times 150}{2}}$.
Squaring both sides: $\frac{2RT}{4} = \frac{1}{4} \times \frac{3 \times R \times 150}{2}$.
$\frac{RT}{2} = \frac{450R}{8}$.
$T = \frac{450}{4} = 112.5 \ K$.

(D) The $rms$ velocity $(v_{rms})$ is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{\frac{1}{M}}$,the gas with the smallest molar mass $(M)$ will have the highest $rms$ speed at a constant temperature.
The molar masses are: $SO_{2} = 64 \ g/mol$,$CO_{2} = 44 \ g/mol$,$O_{2} = 32 \ g/mol$,and $H_{2} = 2 \ g/mol$.
Since $H_{2}$ has the minimum molar mass,it has the highest $rms$ speed.

(B) The formula for $RMS$ velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$,we have $\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$.
Given $T_1 = 27^{\circ} C = 300 \ K$ and $v_1 = 1000 \ m / s$.
Given $T_2 = 600 \ K$.
Substituting the values: $\frac{1000}{v_2} = \sqrt{\frac{300}{600}} = \sqrt{\frac{1}{2}} = \frac{1}{1.414}$.
Therefore,$v_2 = 1000 \times 1.414 = 1414 \ m / s$.