When the temperature of an ideal gas is increased from `27^@C `to `927^@C`, the kinetic energy will be

Options

A. same
B. eight times
C. four times
D. twice.

Explanation

To solve the problem of how the kinetic energy of an ideal gas changes when the temperature is increased from \(27^\circ C\) to \(927^\circ C\), we can follow these steps:

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin**:
- The formula to convert Celsius to Kelvin is:
\[ K = °C + 273 \]
- For the initial temperature:
\[ T_1 = 27 + 273 = 300 \, K \]
- For the final temperature:
\[ T_2 = 927 + 273 = 1200 \, K \]

2. **Use the Kinetic Energy Formula**:
- The kinetic energy (KE) of an ideal gas is given by the formula:
\[ KE = \frac{3}{2} nRT \]
- Here, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.

3. **Calculate Initial and Final Kinetic Energies**:
- Initial kinetic energy at \(T_1\):
\[ KE_1 = \frac{3}{2} nR \cdot 300 \]
- Final kinetic energy at \(T_2\):
\[ KE_2 = \frac{3}{2} nR \cdot 1200 \]

4. **Find the Ratio of Final to Initial Kinetic Energy**:
- To find how the kinetic energy changes, we can take the ratio:
\[ \frac{KE_2}{KE_1} = \frac{\frac{3}{2} nR \cdot 1200}{\frac{3}{2} nR \cdot 300} \]
- The \(\frac{3}{2} nR\) cancels out:
\[ \frac{KE_2}{KE_1} = \frac{1200}{300} = 4 \]

5. **Conclusion**:
- This means that the final kinetic energy is 4 times the initial kinetic energy. Therefore, when the temperature of the ideal gas is increased from \(27^\circ C\) to \(927^\circ C\), the kinetic energy increases by a factor of 4.

### Final Answer:
The kinetic energy will be 4 times the initial kinetic energy.