To what temperature should the hydrogen at `327^(@)C` be cooled at constant pressure, so that the root mean square velocity of its molecules become half of its previous value?

Options

A. `-123^@C`
B. `123^@C`
C. `-100^@C`
D. `0^@C`

Answer

To solve the problem, we need to find the temperature to which hydrogen at \(327^\circ C\) should be cooled so that the root mean square (RMS) velocity of its molecules becomes half of its previous value.

### Step-by-Step Solution:

1. **Convert the initial temperature to Kelvin:**
\[ T_i = 327^\circ C + 273 = 600 \, K \]

2. **Write the formula for RMS velocity:**
The root mean square velocity \(V_{rms}\) is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
where \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass of the gas.

3. **Set up the relationship for initial and final RMS velocities:**
We know that the final RMS velocity \(V_{rms, final}\) is half of the initial RMS velocity \(V_{rms, initial}\):
\[ V_{rms, final} = \frac{1}{2} V_{rms, initial} \]

4. **Express the initial and final RMS velocities using the formula:**
\[ V_{rms, initial} = \sqrt{\frac{3R T_i}{M}} = \sqrt{\frac{3R \cdot 600}{M}} \]
\[ V_{rms, final} = \sqrt{\frac{3R T_f}{M}} \]

5. **Equate the final RMS velocity to half of the initial RMS velocity:**
\[ \sqrt{\frac{3R T_f}{M}} = \frac{1}{2} \sqrt{\frac{3R \cdot 600}{M}} \]

6. **Square both sides to eliminate the square root:**
\[ \frac{3R T_f}{M} = \frac{1}{4} \cdot \frac{3R \cdot 600}{M} \]

7. **Cancel out common terms (3R/M):**
\[ T_f = \frac{600}{4} = 150 \, K \]

8. **Convert the final temperature back to Celsius:**
\[ T_f = 150 \, K - 273 = -123^\circ C \]

### Final Answer:
The temperature to which the hydrogen should be cooled is \(-123^\circ C\).