Two monoatomic gases are at absolute temperatures 300 K and 350 K respectively. Ratio of average kinetic energy of their molecules is

Options

A. `7:6`
B. `6:7`
C. `36:49`
D. `49:36`

Explanation

To find the ratio of the average kinetic energy of the molecules of two monoatomic gases at absolute temperatures 300 K and 350 K, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Kinetic Energy**:
The average kinetic energy (KE) of a monoatomic gas molecule is given by the formula:
\[ KE = \frac{3}{2} k T \]
where \( k \) is the Boltzmann constant and \( T \) is the absolute temperature.

2. **Identify the Temperatures**:
We have two gases with temperatures:
- Gas 1: \( T_1 = 300 \, K \)
- Gas 2: \( T_2 = 350 \, K \)

3. **Express Kinetic Energies in Terms of Temperature**:
Using the formula for kinetic energy, we can express the kinetic energies of the two gases:
- For Gas 1:
\[ KE_1 = \frac{3}{2} k T_1 = \frac{3}{2} k (300) \]
- For Gas 2:
\[ KE_2 = \frac{3}{2} k T_2 = \frac{3}{2} k (350) \]

4. **Set Up the Ratio of Kinetic Energies**:
To find the ratio of the average kinetic energies \( KE_1 \) and \( KE_2 \), we can write:
\[ \frac{KE_1}{KE_2} = \frac{\frac{3}{2} k (300)}{\frac{3}{2} k (350)} \]

5. **Simplify the Ratio**:
The constants \( \frac{3}{2} k \) cancel out:
\[ \frac{KE_1}{KE_2} = \frac{300}{350} \]

6. **Reduce the Fraction**:
Simplifying \( \frac{300}{350} \):
\[ \frac{300}{350} = \frac{6}{7} \]

7. **Conclusion**:
Therefore, the ratio of the average kinetic energy of the two gases is:
\[ KE_1 : KE_2 = 6 : 7 \]

### Final Answer:
The ratio of average kinetic energy of the two gases is \( 6 : 7 \).

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