Question

If $m$ represents the mass of each molecule of a gas and $T$ its absolute temperature, then the root mean square speed of the gas molecule is proportional to

• A. $m^{-\frac{1}{2}}T^{\frac{1}{2}}$
• B. $mT$
• C. $m^{\frac{1}{2}}T^{-\frac{1}{2}}$
• D. $m^{\frac{1}{2}}T^{\frac{1}{2}}$

Hint:

Remember that lighter molecules move faster at a given temperature because kinetic energy depends on both mass and speed. Thus, $v_{\text{rms}}$ must be inversely proportional to the square root of mass ($m^{-\frac{1}{2}}$).

Answer 1

The Correct Option is A

Step 1: Identify what is being asked.
We want to know how the root mean square (rms) speed of a gas molecule depends on the molecular mass $m$ and the absolute temperature $T$. The answer will be a proportionality in $m$ and $T$.
Step 2: Recall the energy idea behind rms speed.
From kinetic theory, the average translational kinetic energy of one molecule depends only on temperature: $\frac{1}{2}mv_{rms}^2 = \frac{3}{2}k_BT$, where $k_B$ is the Boltzmann constant.
Step 3: Solve for the speed.
Multiplying both sides by $2$ and dividing by $m$ gives $v_{rms}^2 = \frac{3k_BT}{m}$.
Step 4: Take the square root.
\[ v_{rms} = \sqrt{\frac{3k_BT}{m}} \]
Step 5: Strip away the constants.
Since $3$ and $k_B$ are fixed numbers, only $T$ and $m$ control the speed, so $v_{rms} \propto \sqrt{\frac{T}{m}}$.
Step 6: Write using exponents.
A square root is a power of one half, so $\sqrt{T} = T^{1/2}$ and $\frac{1}{\sqrt{m}} = m^{-1/2}$. Therefore $v_{rms} \propto m^{-1/2}T^{1/2}$, which is option (1).
\[ \boxed{v_{rms} \propto m^{-\frac{1}{2}}T^{\frac{1}{2}}} \]