At 27°C, the average speed of `N_(2)` molecules is `x ms^(-1)` At 927°, the speed will be_______ `x ms^(-1)`

Answer

To solve the problem, we need to determine the average speed of nitrogen molecules at two different temperatures using the relationship between speed and temperature.

### Step-by-Step Solution:

1. **Convert the temperatures from Celsius to Kelvin:**
- The initial temperature \( T_1 \) is given as 27°C. To convert this to Kelvin:
\[ T_1 = 27 + 273 = 300 \, \text{K} \]
- The final temperature \( T_2 \) is given as 927°C. To convert this to Kelvin:
\[ T_2 = 927 + 273 = 1200 \, \text{K} \]

2. **Understand the relationship between average speed and temperature:**
- The average speed \( v \) of gas molecules is proportional to the square root of the temperature in Kelvin:
\[ v \propto \sqrt{T} \]
- Therefore, we can express the ratio of the average speeds at the two temperatures as:
\[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \]

3. **Substitute the temperatures into the ratio:**
- Using the temperatures calculated:
\[ \frac{v_2}{v_1} = \sqrt{\frac{1200}{300}} \]

4. **Simplify the fraction:**
- Simplifying \( \frac{1200}{300} \):
\[ \frac{1200}{300} = 4 \]
- Therefore:
\[ \frac{v_2}{v_1} = \sqrt{4} = 2 \]

5. **Express the final speed in terms of \( x \):**
- Since \( v_1 \) (the average speed at 27°C) is given as \( x \):
\[ v_2 = 2 \cdot v_1 = 2x \]

### Final Answer:
At 927°C, the average speed of \( N_2 \) molecules will be \( 2x \, \text{ms}^{-1} \).

---