The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at 300 K is: Atomic masses: He=4 u, 0 = 16u)

Options

A. 300 K
B. 600 K
C. 1200 K
D. 2400 K

Explanation

To solve the problem, we need to find the temperature at which the root mean square (RMS) speed of oxygen molecules (O2) is equal to the RMS speed of helium atoms (He) at 300 K.

### Step-by-Step Solution:

1. **Understand the RMS Speed Formula**:
The formula for the root mean square speed (v_rms) of a gas is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas in kg.

2. **Identify Molar Masses**:
- For Helium (He), the molar mass \( M_{He} = 4 \, \text{u} = 4 \times 10^{-3} \, \text{kg} \).
- For Oxygen (O2), the molar mass \( M_{O2} = 32 \, \text{u} = 32 \times 10^{-3} \, \text{kg} \).

3. **Calculate RMS Speed of Helium at 300 K**:
Using the formula for helium:
\[ v_{rms, He} = \sqrt{\frac{3R \cdot 300}{M_{He}}} = \sqrt{\frac{3R \cdot 300}{4 \times 10^{-3}}} \]

4. **Set Up the Equation for Oxygen**:
We need to find the temperature \( T \) for oxygen such that:
\[ v_{rms, O2} = v_{rms, He} \]
Therefore, we can write:
\[ \sqrt{\frac{3RT}{M_{O2}}} = \sqrt{\frac{3R \cdot 300}{M_{He}}} \]

5. **Square Both Sides**:
Squaring both sides eliminates the square root:
\[ \frac{3RT}{M_{O2}} = \frac{3R \cdot 300}{M_{He}} \]

6. **Cancel Common Terms**:
The \( 3R \) terms cancel out:
\[ \frac{T}{M_{O2}} = \frac{300}{M_{He}} \]

7. **Substituting Molar Masses**:
Substitute the values of \( M_{O2} \) and \( M_{He} \):
\[ \frac{T}{32 \times 10^{-3}} = \frac{300}{4 \times 10^{-3}} \]

8. **Cross-Multiply to Solve for T**:
\[ T = \frac{300 \cdot 32 \times 10^{-3}}{4 \times 10^{-3}} \]
Simplifying this gives:
\[ T = 300 \cdot 8 = 2400 \, \text{K} \]

### Final Answer:
The temperature at which oxygen molecules have the same root mean square speed as helium atoms at 300 K is **2400 K**.