The temperature at which the kinetic energy of a gas molecule is doubled to its value at 27 °C is

Options

A. `54^(@)C`
B. 300 K
C. `327^(@)C`
D. `108^(@)C`

Answer

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

### Step-by-Step Solution:

1. **Understand the relationship between temperature and kinetic energy**:
The kinetic energy (KE) of a gas molecule is directly proportional to its absolute temperature (in Kelvin). The formula for the kinetic energy of a gas with F degrees of freedom is given by:
\[ KE = \frac{F}{2} k T \]
where \( k \) is the Boltzmann constant and \( T \) is the absolute temperature in Kelvin.

2. **Determine the initial temperature in Kelvin**:
The initial temperature \( T_1 \) is given as 27 °C. To convert this to Kelvin:
\[ T_1 = 27 + 273 = 300 \, \text{K} \]

3. **Set up the equation for doubled kinetic energy**:
We want to find the temperature \( T_2 \) at which the kinetic energy is doubled:
\[ KE(T_2) = 2 \times KE(T_1) \]
This can be expressed as:
\[ \frac{F}{2} k T_2 = 2 \left( \frac{F}{2} k T_1 \right) \]

4. **Simplify the equation**:
Cancel out the common terms \( \frac{F}{2} k \) from both sides:
\[ T_2 = 2 T_1 \]

5. **Calculate the new temperature**:
Substitute \( T_1 = 300 \, \text{K} \):
\[ T_2 = 2 \times 300 = 600 \, \text{K} \]

6. **Convert back to Celsius**:
To convert \( T_2 \) back to Celsius:
\[ T_2 = 600 - 273 = 327 \, \text{°C} \]

### Final Answer:
The temperature at which the kinetic energy of a gas molecule is doubled to its value at 27 °C is **327 °C**.

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