Calculate the temperature at which the average speed of oxygen equals that of hydrogen at 20 K.

Explanation

To solve the problem of calculating the temperature at which the average speed of oxygen equals that of hydrogen at 20 K, we can follow these steps:

### Step 1: Understand the formula for average speed of gas
The average speed (v_avg) of a gas can be calculated using the formula:
\[ v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}} \]
where:
- \( R \) is the universal gas constant (8.314 J/(mol·K)),
- \( T \) is the temperature in Kelvin,
- \( M \) is the molar mass of the gas in kg/mol.

### Step 2: Identify the molar masses of hydrogen and oxygen
- Molar mass of hydrogen (H₂) = 2 g/mol = 0.002 kg/mol
- Molar mass of oxygen (O₂) = 32 g/mol = 0.032 kg/mol

### Step 3: Calculate the average speed of hydrogen at 20 K
Using the formula for average speed, we can calculate the average speed of hydrogen at 20 K:
\[ v_{\text{H}_2} = \sqrt{\frac{8 \cdot R \cdot T}{\pi M_{\text{H}_2}}} \]
Substituting the values:
\[ v_{\text{H}_2} = \sqrt{\frac{8 \cdot 8.314 \cdot 20}{\pi \cdot 0.002}} \]

### Step 4: Calculate the average speed of hydrogen
Calculating the above expression:
\[ v_{\text{H}_2} = \sqrt{\frac{8 \cdot 8.314 \cdot 20}{\pi \cdot 0.002}} \approx \sqrt{\frac{1333.44}{0.006283}} \approx \sqrt{212,000} \approx 460.52 \text{ m/s} \]

### Step 5: Set the average speed of oxygen equal to that of hydrogen
Now we want to find the temperature at which the average speed of oxygen equals that of hydrogen:
\[ v_{\text{O}_2} = v_{\text{H}_2} \]
Using the average speed formula for oxygen:
\[ \sqrt{\frac{8RT_{\text{O}_2}}{\pi M_{\text{O}_2}}} = 460.52 \]

### Step 6: Substitute the values for oxygen
Substituting the values for oxygen:
\[ \sqrt{\frac{8 \cdot 8.314 \cdot T_{\text{O}_2}}{\pi \cdot 0.032}} = 460.52 \]

### Step 7: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \frac{8 \cdot 8.314 \cdot T_{\text{O}_2}}{\pi \cdot 0.032} = (460.52)^2 \]

### Step 8: Solve for \( T_{\text{O}_2} \)
Calculating \( (460.52)^2 \):
\[ (460.52)^2 \approx 212,000 \]
Now rearranging to solve for \( T_{\text{O}_2} \):
\[ T_{\text{O}_2} = \frac{(460.52)^2 \cdot \pi \cdot 0.032}{8 \cdot 8.314} \]
Calculating:
\[ T_{\text{O}_2} \approx \frac{212,000 \cdot \pi \cdot 0.032}{66.512} \approx \frac{21,200 \cdot 3.14}{66.512} \approx \frac{66,508.8}{66.512} \approx 995.5 \text{ K} \]

### Final Answer
The temperature at which the average speed of oxygen equals that of hydrogen at 20 K is approximately **995.5 K**.

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