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 solve the problem of finding 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 the average speed of gas molecules. Here’s a step-by-step solution:

### Step 1: Write the formula for average speed
The average speed (U) of gas molecules is given by the formula:
\[ U = \sqrt{\frac{8RT}{\pi M}} \]
where:
- \( R \) is the gas constant,
- \( T \) is the temperature in Kelvin,
- \( M \) is the molar mass of the gas.

### Step 2: Set up the equation
Since we want the average speed of CH₄ to equal the average speed of O₂, we can set up the equation:
\[ U_{CH_4} = U_{O_2} \]
Substituting the formula for average speed, we have:
\[ \sqrt{\frac{8R T_{CH_4}}{\pi M_{CH_4}}} = \sqrt{\frac{8R T_{O_2}}{\pi M_{O_2}}} \]

### Step 3: Simplify the equation
We can simplify this equation by squaring both sides:
\[ \frac{8R T_{CH_4}}{\pi M_{CH_4}} = \frac{8R T_{O_2}}{\pi M_{O_2}} \]
Since \( \frac{8R}{\pi} \) is a constant, we can cancel it out:
\[ \frac{T_{CH_4}}{M_{CH_4}} = \frac{T_{O_2}}{M_{O_2}} \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ T_{CH_4} = T_{O_2} \cdot \frac{M_{CH_4}}{M_{O_2}} \]

### Step 5: Insert known values
We know:
- \( T_{O_2} = 300 \, K \)
- Molar mass of CH₄ (\( M_{CH_4} \)) = 16 g/mol
- Molar mass of O₂ (\( M_{O_2} \)) = 32 g/mol

Substituting these values into the equation:
\[ T_{CH_4} = 300 \cdot \frac{16}{32} \]

### Step 6: Calculate \( T_{CH_4} \)
Calculating the right side:
\[ T_{CH_4} = 300 \cdot \frac{1}{2} = 150 \, K \]

### Final Answer
The temperature at which the average speed of CH₄ molecules equals the average speed of O₂ molecules at 300 K is:
\[ \boxed{150 \, K} \]