At what temperature is the R.M.S. speed of gas molecules half the value at N.T.P.?

Options

A. 68.25 K
B. 273 K
C. 345 K
D. 0 K

Answer

To solve the problem of finding the temperature at which the R.M.S. speed of gas molecules is half the value at N.T.P. (Normal Temperature and Pressure), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the R.M.S. Speed Formula**:
The formula for the root mean square (R.M.S.) speed of gas molecules is given by:
\[ V_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the absolute temperature in Kelvin, and \( M \) is the molar mass of the gas.

2. **Calculate R.M.S. Speed at N.T.P.**:
At N.T.P., the temperature is 0 degrees Celsius, which is equivalent to 273 Kelvin. Thus, the R.M.S. speed at N.T.P. can be expressed as:
\[ V_{\text{rms, STP}} = \sqrt{\frac{3R \cdot 273}{M}} \]

3. **Set Up the Equation for Half the R.M.S. Speed**:
We need to find the temperature \( T \) at which the R.M.S. speed \( V'_{\text{rms}} \) is half of the R.M.S. speed at N.T.P.:
\[ V'_{\text{rms}} = \frac{1}{2} V_{\text{rms, STP}} \]

4. **Express \( V'_{\text{rms}} \) Using the Formula**:
Using the R.M.S. speed formula for the new temperature \( T \):
\[ V'_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]

5. **Set Up the Equation**:
Now, we can equate the two expressions:
\[ \sqrt{\frac{3RT}{M}} = \frac{1}{2} \sqrt{\frac{3R \cdot 273}{M}} \]

6. **Simplify the Equation**:
We can cancel \( \sqrt{\frac{3R}{M}} \) from both sides:
\[ \sqrt{T} = \frac{1}{2} \sqrt{273} \]

7. **Square Both Sides**:
Squaring both sides gives:
\[ T = \left(\frac{1}{2}\right)^2 \cdot 273 = \frac{273}{4} \]

8. **Calculate the Temperature**:
Now, calculate \( T \):
\[ T = 68.25 \text{ Kelvin} \]

### Final Answer:
The temperature at which the R.M.S. speed of gas molecules is half the value at N.T.P. is **68.25 Kelvin**.