The average kinetic energy of the molecules of an ideal gas at `10 ^@ C` has the value (E). The temperature at which the kinetic energy of the same gas becomes (2 E) is.

Options

A. `5^@ C`
B. `10^@ C`
C. `40^@ C`
D. None of these

Explanation

To solve the problem, we need to determine the temperature at which the average kinetic energy of the molecules of an ideal gas becomes twice the value of E, given that the average kinetic energy at 10°C is E.

### Step-by-Step Solution:

1. **Understanding Kinetic Energy Relation**:
The average kinetic energy (KE) of an ideal gas is directly proportional to its temperature. The formula for the average kinetic energy of a gas is given by:
\[ KE = \frac{f}{2} k T \]
where \( f \) is the degrees of freedom, \( k \) is the Boltzmann constant, and \( T \) is the absolute temperature in Kelvin.

2. **Setting Up the Initial Condition**:
At 10°C, we first convert this temperature to Kelvin:
\[ T_1 = 10 + 273 = 283 \text{ K} \]
The average kinetic energy at this temperature is given as \( E \).

3. **Setting Up the Final Condition**:
We want to find the temperature \( T_2 \) at which the kinetic energy becomes \( 2E \).

4. **Using the Proportionality**:
Since the kinetic energy is directly proportional to temperature, we can write:
\[ \frac{KE_1}{KE_2} = \frac{T_1}{T_2} \]
Substituting the known values:
\[ \frac{E}{2E} = \frac{283}{T_2} \]
Simplifying this gives:
\[ \frac{1}{2} = \frac{283}{T_2} \]

5. **Cross-Multiplying to Solve for \( T_2 \)**:
Cross-multiplying gives:
\[ T_2 = 2 \times 283 \]
Calculating this:
\[ T_2 = 566 \text{ K} \]

6. **Converting Back to Celsius**:
To convert \( T_2 \) back to Celsius:
\[ T_2 (\text{°C}) = T_2 (\text{K}) - 273 = 566 - 273 = 293 \text{°C} \]

### Final Answer:
The temperature at which the kinetic energy of the same gas becomes \( 2E \) is **293°C**.