The mean kinetic energy of `1` kg-mol of nitrogen at `27^(@)C` is `600 J`. What will be its mean kinetic energy at `127^(@)C` ?

Answer

To solve the problem, we need to determine the mean kinetic energy of 1 kg-mol of nitrogen at a temperature of 127°C, given that its mean kinetic energy at 27°C is 600 J. The relationship between the mean kinetic energy and temperature is given by the formula:

\[ KE \propto T \]

Where \( KE \) is the mean kinetic energy and \( T \) is the absolute temperature in Kelvin.

### Step-by-Step Solution:

1. **Convert the given temperatures from Celsius to Kelvin:**
- The formula to convert Celsius to Kelvin is:
\[ T(K) = T(°C) + 273 \]
- For 27°C:
\[ T_1 = 27 + 273 = 300 \, K \]
- For 127°C:
\[ T_2 = 127 + 273 = 400 \, K \]

2. **Set up the ratio of mean kinetic energies:**
- Since the mean kinetic energy is directly proportional to the temperature, we can write:
\[ \frac{KE_1}{KE_2} = \frac{T_1}{T_2} \]
- Where \( KE_1 = 600 \, J \) at \( T_1 = 300 \, K \) and \( KE_2 \) is the unknown mean kinetic energy at \( T_2 = 400 \, K \).

3. **Substitute the values into the equation:**
- Plugging in the known values:
\[ \frac{600}{KE_2} = \frac{300}{400} \]

4. **Cross-multiply to solve for \( KE_2 \):**
- Cross-multiplying gives:
\[ 600 \times 400 = KE_2 \times 300 \]
- This simplifies to:
\[ 240000 = 300 \times KE_2 \]

5. **Isolate \( KE_2 \):**
- Dividing both sides by 300:
\[ KE_2 = \frac{240000}{300} = 800 \, J \]

### Final Answer:
The mean kinetic energy of 1 kg-mol of nitrogen at 127°C is **800 J**.

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