The average K.E. of hydrogen molecules at 27 °C is E. The average K.E. at 327 °C is

Options

A. K.E
B. `sqrt2` K.E
C. 2 K.E
D. 4 K.E

Answer

To find the average kinetic energy of hydrogen molecules at 327 °C given that the average kinetic energy at 27 °C is E, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Kinetic Energy Formula**:
The average kinetic energy (K.E.) per molecule for a gas can be expressed as:
\[ KE = \frac{f}{2} k T \]
where \( f \) is the degrees of freedom, \( k \) is the Boltzmann constant, and \( T \) is the absolute temperature in Kelvin.

2. **Identify the Degrees of Freedom**:
For a diatomic gas like hydrogen (H₂), the degrees of freedom \( f \) is typically 5 (3 translational and 2 rotational).

3. **Convert Temperatures to Kelvin**:
- For 27 °C:
\[ T_1 = 27 + 273 = 300 \text{ K} \]
- For 327 °C:
\[ T_2 = 327 + 273 = 600 \text{ K} \]

4. **Calculate Kinetic Energy at 27 °C**:
Using the formula for kinetic energy:
\[ KE_1 = \frac{5}{2} k T_1 = \frac{5}{2} k \cdot 300 \]
We know that \( KE_1 = E \).

5. **Calculate Kinetic Energy at 327 °C**:
Similarly, for 327 °C:
\[ KE_2 = \frac{5}{2} k T_2 = \frac{5}{2} k \cdot 600 \]

6. **Find the Ratio of Kinetic Energies**:
To find the relationship between \( KE_2 \) and \( KE_1 \):
\[ \frac{KE_2}{KE_1} = \frac{\frac{5}{2} k \cdot 600}{\frac{5}{2} k \cdot 300} = \frac{600}{300} = 2 \]

7. **Express Kinetic Energy at 327 °C in Terms of E**:
Since \( KE_1 = E \):
\[ KE_2 = 2 \cdot KE_1 = 2E \]

### Final Answer:
The average kinetic energy of hydrogen molecules at 327 °C is \( 2E \).

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