At what temperature the average value of the kinetic energy of the molecule of a gas will be `1//3` of the average value of kinetic energy at `27^(@)C`?

Answer

To solve the problem, we need to find the temperature at which the average kinetic energy of gas molecules is one-third of the average kinetic energy at 27°C.

### Step-by-Step Solution:

1. **Understand the Relationship**: The average kinetic energy (KE) of gas molecules is given by the formula:
\[ KE = \frac{3}{2} nRT \]
where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the absolute temperature in Kelvin.

2. **Convert the Given Temperature**: First, convert 27°C to Kelvin:
\[ T_{27} = 27 + 273 = 300 \, K \]

3. **Set Up the Equation**: We need to find the temperature \( T \) such that the average kinetic energy at this temperature is one-third of the average kinetic energy at 27°C:
\[ KE_T = \frac{1}{3} KE_{27} \]

4. **Substitute the Kinetic Energy Formula**:
\[ \frac{3}{2} nRT = \frac{1}{3} \left(\frac{3}{2} nR \cdot 300\right) \]

5. **Simplify the Equation**: Cancel out the common terms:
\[ T = \frac{1}{3} \cdot 300 \]

6. **Calculate the Temperature**:
\[ T = 100 \, K \]

### Final Answer:
The temperature at which the average value of the kinetic energy of the molecules of a gas will be one-third of the average value of kinetic energy at 27°C is **100 K**.