At what temperature do the average speed of `CH_(4)(g)` molecules equal the average speed of `O_(2)` molecules at 300 K ?

Options

A. 150 K
B. 900 K
C. 600 K
D. 300 K

Explanation

To find the temperature at which the average speed of CH₄ (g) molecules equals the average speed of O₂ molecules at 300 K, we can use the formula for average speed of gas molecules:

\[ U_{\text{average}} = \sqrt{\frac{8kT}{\pi M}} \]

where:
- \( U_{\text{average}} \) is the average speed,
- \( k \) is the Boltzmann constant,
- \( T \) is the temperature in Kelvin,
- \( M \) is the molar mass of the gas.

### Step 1: Write the equations for average speeds of CH₄ and O₂

For CH₄ (methane):
\[ U_{\text{CH}_4} = \sqrt{\frac{8kT_{\text{CH}_4}}{\pi M_{\text{CH}_4}}} \]

For O₂ (oxygen) at 300 K:
\[ U_{\text{O}_2} = \sqrt{\frac{8k \cdot 300}{\pi M_{\text{O}_2}}} \]

### Step 2: Set the two average speeds equal to each other

We want to find \( T_{\text{CH}_4} \) such that:
\[ U_{\text{CH}_4} = U_{\text{O}_2} \]

This gives us:
\[ \sqrt{\frac{8kT_{\text{CH}_4}}{\pi M_{\text{CH}_4}}} = \sqrt{\frac{8k \cdot 300}{\pi M_{\text{O}_2}}} \]

### Step 3: Simplify the equation

We can cancel out the common terms \( \sqrt{\frac{8k}{\pi}} \) from both sides:
\[ \sqrt{\frac{T_{\text{CH}_4}}{M_{\text{CH}_4}}} = \sqrt{\frac{300}{M_{\text{O}_2}}} \]

### Step 4: Square both sides

Squaring both sides results in:
\[ \frac{T_{\text{CH}_4}}{M_{\text{CH}_4}} = \frac{300}{M_{\text{O}_2}} \]

### Step 5: Rearrange to find \( T_{\text{CH}_4} \)

Rearranging gives:
\[ T_{\text{CH}_4} = 300 \cdot \frac{M_{\text{CH}_4}}{M_{\text{O}_2}} \]

### Step 6: Find the molar masses

- Molar mass of CH₄ (methane):
- Carbon (C) = 12 g/mol
- Hydrogen (H) = 1 g/mol × 4 = 4 g/mol
- Total = 12 + 4 = 16 g/mol

- Molar mass of O₂ (oxygen):
- Oxygen (O) = 16 g/mol × 2 = 32 g/mol

### Step 7: Substitute the molar masses into the equation

Now substituting the values:
\[ T_{\text{CH}_4} = 300 \cdot \frac{16}{32} \]

### Step 8: Calculate \( T_{\text{CH}_4} \)

\[ T_{\text{CH}_4} = 300 \cdot \frac{1}{2} = 150 \text{ K} \]

### Final Answer

The temperature at which the average speed of CH₄ molecules equals the average speed of O₂ molecules at 300 K is **150 K**.

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