If the rms velocity of gas is v , then

Options

A. `V^(2)T=`constant
B. `(V^(2))/(T)=`constant
C. `VT^(2)`=constant
D. V and T are independent

Answer

To solve the problem, we need to derive the relationship between the root mean square (RMS) velocity of a gas and its temperature. The RMS velocity (V) is given by the formula:

1. **Start with the formula for RMS velocity:**
\[ V = \sqrt{\frac{3RT}{M}} \]
where:
- \( V \) = RMS velocity
- \( R \) = universal gas constant
- \( T \) = absolute temperature
- \( M \) = molecular weight of the gas

2. **Square both sides to eliminate the square root:**
\[ V^2 = \frac{3RT}{M} \]

3. **Rearrange the equation to express \( V^2 \) in terms of \( T \):**
\[ V^2 = \frac{3R}{M} \cdot T \]

4. **Divide both sides by \( T \) to isolate \( V^2/T \):**
\[ \frac{V^2}{T} = \frac{3R}{M} \]

5. **Recognize that \( \frac{3R}{M} \) is a constant for a given gas:**
- Since \( R \) (universal gas constant) and \( M \) (molecular weight) are constants for a specific gas, \( \frac{3R}{M} \) remains constant.

6. **Conclude that \( \frac{V^2}{T} \) is constant:**
\[ \frac{V^2}{T} = \text{constant} \]

Thus, we have derived that if the RMS velocity of a gas is \( V \), then the relationship \( \frac{V^2}{T} = \text{constant} \) holds true.

### Final Answer:
The correct relation is \( \frac{V^2}{T} = \text{constant} \).

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