Question

An ideal gas having molar mass $M_0$, has r.m.s. velocity $V$ at temperature $T$. Then

• A. $VT^2 = \text{constant}$
• B. $V^2T = \text{constant}$
• C. $\frac{V^2}{T} = \text{constant}$
• D. $V$ is independent of $T$

Hint:

Remember that the r.m.s. velocity of gas molecules is always directly proportional to the square root of the absolute temperature ($V \propto \sqrt{T}$). Squaring this proportional relationship instantly yields $V^2 \propto T \implies \frac{V^2}{T} = \text{constant}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the correct functional relationship between the root-mean-square (r.m.s.) velocity $V$ of an ideal gas and its absolute temperature $T$, given a fixed molar mass $M_0$.

Step 2: Key Formula or Approach:
The root-mean-square velocity of molecules in an ideal gas sample is defined by the kinetic theory expression: $$V = \sqrt{\frac{3RT}{M_0}}$$ where $R$ is the universal gas constant, $T$ is the absolute temperature, and $M_0$ is the molar mass. By squaring both sides, we can isolate the variable parameters from the absolute constants.

Step 3: Detailed Explanation:
Let's square the fundamental r.m.s. velocity formula to eliminate the radical sign: $$V^2 = \frac{3RT}{M_0}$$ Now, rearrange the equation by dividing both sides by the temperature variable $T$: $$\frac{V^2}{T} = \frac{3R}{M_0}$$ Since $3$, $R$, and $M_0$ are entirely fixed constant values for a specific gas sample, the right-hand side is a constant: $$\frac{V^2}{T} = \text{constant}$$ This matches the relationship shown in option (C).

Step 4: Final Answer:
The correct relationship is $\frac{V^2}{T} = \text{constant}$, which corresponds to option (C).