At what temperarture, the kinetic energy of a gas molecule is half of the value at `27^(@)C.` ?

Options

A. `13.5^(@)C`
B. `150^(@)C`
C. 75 K
D. `-123^(@)C`

Answer

To solve the problem of finding the temperature at which the kinetic energy of a gas molecule is half of the value at 27°C, we can follow these steps:

### Step-by-Step Solution:

1. **Convert the initial temperature from Celsius to Kelvin**:
\[ T_1 = 27°C + 273 = 300 \, K \]

2. **Write the formula for kinetic energy (KE) of a gas molecule**:
The kinetic energy of a gas molecule is given by:
\[ KE = \frac{3}{2} R T \]
where \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin.

3. **Calculate the kinetic energy at the initial temperature**:
For the initial temperature \( T_1 \):
\[ KE_1 = \frac{3}{2} R T_1 = \frac{3}{2} R (300) \]

4. **Set up the equation for the new kinetic energy**:
We want to find the temperature \( T_2 \) at which the kinetic energy is half of \( KE_1 \):
\[ KE_2 = \frac{1}{2} KE_1 = \frac{1}{2} \left( \frac{3}{2} R (300) \right) = \frac{3}{4} R (300) \]

5. **Write the expression for kinetic energy at the new temperature**:
\[ KE_2 = \frac{3}{2} R T_2 \]

6. **Set the two expressions for kinetic energy equal**:
\[ \frac{3}{2} R T_2 = \frac{3}{4} R (300) \]

7. **Cancel \( R \) from both sides** (since it is a constant):
\[ \frac{3}{2} T_2 = \frac{3}{4} (300) \]

8. **Solve for \( T_2 \)**:
\[ T_2 = \frac{3}{4} (300) \times \frac{2}{3} = 150 \, K \]

9. **Convert the temperature back to Celsius**:
\[ T_2 = 150 \, K - 273 = -123 \, °C \]

### Final Answer:
The temperature at which the kinetic energy of a gas molecule is half of the value at 27°C is:
\[ T_2 = -123 \, °C \]