Question

At what temperature is the root mean square (rms) speed of oxygen molecules ($O_2$) equal to that of helium molecules ($He$) at $27^\circ\text{C}$? (Given molar mass of $O_2 = 32\text{ g/mol}$, $He = 4\text{ g/mol}$)

• A. $2400\text{ K}$
• B. $2127\text{ K}$
• C. $2427^\circ\text{C}$
• D. $2127^\circ\text{C}$

Hint:

Since oxygen is 8 times heavier than helium ($32 / 4 = 8$), its absolute temperature must be exactly 8 times higher to maintain the same rms speed: $8 \times 300\text{ K} = 2400\text{ K}$.

Answer 1

The Correct Option is A

Step 1: Recall the rms speed.
The root mean square speed of gas molecules is \[ v_{rms} = \sqrt{\frac{3RT}{M}} \] where $T$ is temperature in kelvin and $M$ is the molar mass.

Step 2: Set the two speeds equal.
We want oxygen at temperature $T_1$ to have the same rms speed as helium at $300$ K. So \[ \sqrt{\frac{3RT_1}{M_1}} = \sqrt{\frac{3RT_2}{M_2}} \]

Step 3: Square and simplify.
Squaring both sides and cancelling $3R$ gives a neat rule. \[ \frac{T_1}{M_1} = \frac{T_2}{M_2} \]

Step 4: Put in the known values.
Oxygen has $M_1 = 32$, helium has $M_2 = 4$, and helium is at $27^{\circ}$C which is $300$ K. \[ \frac{T_1}{32} = \frac{300}{4} \]

Step 5: Solve for $T_1$.
Multiply both sides by $32$. \[ T_1 = \frac{32}{4}\times 300 = 8\times 300 = 2400 \text{ K} \]

Step 6: Read the meaning.
Oxygen is heavier, so it must be hotter to move as fast as the light helium. \[ \boxed{T_1 = 2400 \text{ K}} \]