Calculate the temperature at which the rms velocity of oxygen molecules will be `2 km//s-R = 8.31 J m^(-1) K^(-1)`.

Explanation

To calculate the temperature at which the root mean square (rms) velocity of oxygen molecules is 2 km/s, we can follow these steps:

### Step-by-Step Solution:

1. **Convert the rms velocity to SI units**:
\[ v_{\text{rms}} = 2 \text{ km/s} = 2 \times 10^3 \text{ m/s} \]

2. **Identify the molar mass of oxygen (O₂)**:
- The molar mass of oxygen (O) is 16 g/mol. Therefore, for O₂:
\[ M = 16 \text{ g/mol} + 16 \text{ g/mol} = 32 \text{ g/mol} = 32 \times 10^{-3} \text{ kg/mol} \]

3. **Use the formula for rms velocity**:
The formula for rms velocity is given by:
\[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the gas constant (8.31 J/(mol·K))
- \( T \) is the temperature in Kelvin
- \( M \) is the molar mass in kg/mol

4. **Square both sides of the equation**:
\[ v_{\text{rms}}^2 = \frac{3RT}{M} \]

5. **Rearrange the equation to solve for temperature \( T \)**:
\[ T = \frac{v_{\text{rms}}^2 \cdot M}{3R} \]

6. **Substitute the known values into the equation**:
\[ T = \frac{(2 \times 10^3)^2 \cdot (32 \times 10^{-3})}{3 \cdot 8.31} \]

7. **Calculate \( v_{\text{rms}}^2 \)**:
\[ (2 \times 10^3)^2 = 4 \times 10^6 \]

8. **Substitute this back into the equation**:
\[ T = \frac{4 \times 10^6 \cdot (32 \times 10^{-3})}{3 \cdot 8.31} \]

9. **Calculate the numerator**:
\[ 4 \times 10^6 \cdot 32 \times 10^{-3} = 128000 \text{ (or } 1.28 \times 10^5 \text{)} \]

10. **Calculate the denominator**:
\[ 3 \cdot 8.31 = 24.93 \]

11. **Now calculate \( T \)**:
\[ T = \frac{128000}{24.93} \approx 5134.4 \text{ K} \]

### Final Answer:
The temperature at which the rms velocity of oxygen molecules will be \( 5134.4 \) K.