At what temperature, pressure remaining constant will the r.m.s. speed of a gas molecules increase by 10% of the r.m.s speed at STP?

Options

A. `57.3K`
B. `57.3^(@)C`
C. `557.3K`
D. `-57.3^(@)C`

Answer

To solve the problem, we need to determine the temperature at which the root mean square (r.m.s.) speed of gas molecules increases by 10% from its value at standard temperature and pressure (STP).

### Step-by-Step Solution:

1. **Understand the formula for r.m.s. speed**:
The r.m.s. speed \( V_{rms} \) 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, and \( m \) is the molecular mass of the gas.

2. **Recognize the relationship between r.m.s. speed and temperature**:
Since \( R \) and \( m \) are constants for a given gas, we can say that:
\[ V_{rms} \propto \sqrt{T} \]
This means that the r.m.s. speed is directly proportional to the square root of the temperature.

3. **Set up the ratio of initial and final r.m.s. speeds**:
Let \( V_1 \) be the initial r.m.s. speed at STP and \( V_2 \) be the final r.m.s. speed after the increase. According to the problem, \( V_2 \) is 10% greater than \( V_1 \):
\[ V_2 = 1.1 V_1 \]

4. **Use the proportionality to relate temperatures**:
From the proportionality, we have:
\[ \frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}} \]
Substituting \( V_2 = 1.1 V_1 \):
\[ \frac{V_1}{1.1 V_1} = \sqrt{\frac{T_1}{T_2}} \]
This simplifies to:
\[ \frac{1}{1.1} = \sqrt{\frac{T_1}{T_2}} \]

5. **Square both sides to eliminate the square root**:
\[ \left(\frac{1}{1.1}\right)^2 = \frac{T_1}{T_2} \]
This gives:
\[ \frac{1}{1.21} = \frac{T_1}{T_2} \]

6. **Rearrange to find \( T_2 \)**:
\[ T_2 = 1.21 T_1 \]

7. **Substitute the value of \( T_1 \)**:
The standard temperature \( T_1 \) at STP is 273 K. Thus:
\[ T_2 = 1.21 \times 273 \text{ K} \]
Calculating this gives:
\[ T_2 = 330.93 \text{ K} \]

8. **Convert \( T_2 \) to Celsius**:
To convert Kelvin to Celsius, we use:
\[ T_2 (\text{°C}) = T_2 (\text{K}) - 273 \]
Therefore:
\[ T_2 = 330.93 - 273 \approx 57.93 \text{ °C} \]

9. **Final answer**:
Rounding off, we can state that the temperature at which the r.m.s. speed of gas molecules increases by 10% is approximately:
\[ T_2 \approx 57.3 \text{ °C} \]