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. `23^(@)C`
C. `-100^(@)C`
D. `0^(@)`

Explanation

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

### Step-by-Step Solution:

1. **Convert the Initial Temperature to Kelvin:**
The initial temperature \(T_1\) in Celsius is given as \(327^\circ C\). To convert this to Kelvin, we use the formula:
\[ T_1 = 327 + 273 = 600 \, K \]

2. **Understand the Relationship Between RMS Velocity and Temperature:**
The root mean square velocity (\(v_{rms}\)) of gas molecules is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \(R\) is the gas constant, \(T\) is the absolute temperature in Kelvin, and \(M\) is the molar mass of the gas.

3. **Set Up the Equation for the Final RMS Velocity:**
We want the final RMS velocity \(v_{rms,2}\) to be half of the initial RMS velocity \(v_{rms,1}\):
\[ v_{rms,2} = \frac{1}{2} v_{rms,1} \]
Substituting the expressions for RMS velocity:
\[ \sqrt{\frac{3RT_2}{M}} = \frac{1}{2} \sqrt{\frac{3RT_1}{M}} \]

4. **Square Both Sides to Eliminate the Square Root:**
Squaring both sides gives:
\[ \frac{3RT_2}{M} = \frac{1}{4} \cdot \frac{3RT_1}{M} \]
Since \(R\) and \(M\) are constants and cancel out, we simplify to:
\[ T_2 = \frac{1}{4} T_1 \]

5. **Calculate the Final Temperature:**
Now substituting \(T_1 = 600 \, K\):
\[ T_2 = \frac{1}{4} \times 600 = 150 \, K \]

6. **Convert the Final Temperature Back to Celsius:**
To convert \(T_2\) back to Celsius:
\[ T_2 = 150 - 273 = -123 \, ^\circ C \]

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