Question

1. consider separate 1.0 - l gaseous samples of h₂, xe, cl₂, and o₂ all at stp. a. rank the gases in order of increasing average kinetic energy. b. rank the gases in order of increasing average velocity. c. how can separate 1.0 - l samples of o₂ and h₂ each have the same average velocity?

Explanation:

Step1: Recall kinetic - molecular theory for average kinetic energy

The average kinetic energy of a gas is given by $KE_{avg}=\frac{3}{2}RT$. At the same temperature (STP: $T = 273\ K$ and $P= 1\ atm$), all gases have the same average kinetic energy. So, for part a, the ranking is $H_2 = Xe=Cl_2 = O_2$.

Step2: Recall the formula for average velocity

The average velocity of a gas molecule is $\overline{v}=\sqrt{\frac{8RT}{\pi M}}$, where $M$ is the molar - mass of the gas. The molar masses are $M_{H_2}=2.02\ g/mol$, $M_{Xe} = 131.29\ g/mol$, $M_{Cl_2}=70.90\ g/mol$, and $M_{O_2}=32.00\ g/mol$. Since $\overline{v}\propto\frac{1}{\sqrt{M}}$, as $M$ increases, $\overline{v}$ decreases.

Step3: Rank the gases based on molar - mass for average velocity

The order of increasing molar - mass is $H_2

Step4: Analyze the condition for equal average velocities

Since $\overline{v}=\sqrt{\frac{8RT}{\pi M}}$, for $H_2$ and $O_2$ to have the same average velocity, from the formula $\sqrt{\frac{T_{H_2}}{M_{H_2}}}=\sqrt{\frac{T_{O_2}}{M_{O_2}}}$, or $\frac{T_{H_2}}{M_{H_2}}=\frac{T_{O_2}}{M_{O_2}}$. To make the average velocities equal, the temperature ratio should be $\frac{T_{H_2}}{T_{O_2}}=\frac{M_{H_2}}{M_{O_2}}$. So, the temperature of $H_2$ must be lower than the temperature of $O_2$ in the ratio of their molar - masses.

Answer:

a. $H_2 = Xe=Cl_2 = O_2$
b. $Xe < Cl_2c. The temperature of $H_2$ must be lower than the temperature of $O_2$ in the ratio of their molar - masses ($\frac{T_{H_2}}{T_{O_2}}=\frac{M_{H_2}}{M_{O_2}}$).