Initially, the translational rms speed of a molecule of an ideal gas is 463 m/s. The pressure and volume of this gas are kept constant, while the number of molecules is doubled. What is the final translational rms speed of the molecules?

Options

A. 655 m/s
B. 463 m/s
C. 327 m/s
D. 926 m/s

Answer

To solve the problem step by step, we will use the relationship between the translational root mean square (rms) speed of gas molecules and the number of molecules, while keeping pressure and volume constant.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The initial translational rms speed of the gas molecules is given as \( V_{rms1} = 463 \, \text{m/s} \).
- The pressure and volume of the gas are constant.
- The number of molecules is doubled.

2. **Using the Formula for rms Speed**:
- The formula for the translational rms speed of gas molecules is given by:
\[ V_{rms} = \sqrt{\frac{3RT}{m}} \]
- Here, \( R \) is the gas constant, \( T \) is the absolute temperature, and \( m \) is the molecular mass of the gas.

3. **Analyzing the Effect of Doubling the Number of Molecules**:
- According to the ideal gas law, if the pressure (P) and volume (V) are constant, then:
\[ PV = nRT \]
- Where \( n \) is the number of moles. If the number of molecules is doubled, the number of moles \( n \) also doubles (since \( n = \frac{N}{N_A} \), where \( N \) is the number of molecules and \( N_A \) is Avogadro's number).

4. **Relationship Between rms Speed and Number of Molecules**:
- Since \( V_{rms} \) is inversely proportional to the square root of the number of molecules (when pressure and volume are constant):
\[ V_{rms} \propto \frac{1}{\sqrt{N}} \]
- If the number of molecules doubles, we can express this as:
\[ \frac{V_{rms2}}{V_{rms1}} = \sqrt{\frac{N_1}{N_2}} = \sqrt{\frac{N_1}{2N_1}} = \frac{1}{\sqrt{2}} \]

5. **Calculating the Final rms Speed**:
- Rearranging the above relationship gives:
\[ V_{rms2} = V_{rms1} \cdot \frac{1}{\sqrt{2}} \]
- Substituting the known values:
\[ V_{rms2} = 463 \cdot \frac{1}{\sqrt{2}} = \frac{463}{\sqrt{2}} \approx \frac{463}{1.414} \approx 327.39 \, \text{m/s} \]

6. **Final Answer**:
- The final translational rms speed of the molecules after doubling the number of molecules is approximately:
\[ V_{rms2} \approx 327.39 \, \text{m/s} \]