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: Understanding the Question:
The question asks for the proportional relationship between the root mean square (rms) speed of a gas molecule, its individual molecular mass ($m$), and its absolute temperature ($T$).

Step 2: Key Formula or Approach:
The root mean square speed $v_{\text{rms}}$ of gas molecules is derived from the kinetic theory of gases and is expressed by the formula:
$$v_{\text{rms}} = \sqrt{\frac{3k_BT}{m}}$$ where $k_B$ is the Boltzmann constant, $T$ is the absolute temperature, and $m$ is the mass of a single gas molecule.

Step 3: Detailed Explanation:
From the formula, we can isolate the constants to see how $v_{\text{rms}}$ scales with temperature and mass:
$$v_{\text{rms}} \propto \sqrt{\frac{T}{m}}$$ We can rewrite this square root expression as fractional exponents:
$$v_{\text{rms}} \propto \frac{T^{\frac{1}{2}}}{m^{\frac{1}{2}}}$$ Bringing the mass term from the denominator to the numerator changes the sign of its exponent:
$$v_{\text{rms}} \propto m^{-\frac{1}{2}}T^{\frac{1}{2}}$$ This direct algebraic rearrangement matches the formatting given in option (A).

Step 4: Final Answer:
The root mean square speed of the gas molecule is proportional to $m^{-\frac{1}{2}}T^{\frac{1}{2}}$, corresponding to option (A).