Air consists mainly of nitrogen and oxygen molecules. Relative molecular masses of nitrogen and oxygen are 28 and 32 respectiely. Calculate `:`
(i) the ratio of the rms speed of nitrogen molecules to that of the oxygen molecules in air and
(ii) the ratio of the r.m.s. speed of oxygen molecules in air at `0^(@)C` to that at `10^(@)C`.

Answer

To solve the problem, we need to calculate two ratios related to the root mean square (rms) speeds of nitrogen and oxygen molecules in air.

### Given Data:
- Molecular mass of Nitrogen (N₂) = 28 g/mol
- Molecular mass of Oxygen (O₂) = 32 g/mol
- Temperature at 0°C = 273 K
- Temperature at 10°C = 283 K

### (i) Ratio of the rms speed of nitrogen molecules to that of the oxygen molecules in air

1. **Formula for rms speed**: The rms speed (v_rms) of a gas is given by the formula:
\[ v_{rms} = \sqrt{\frac{5RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas.

2. **Calculate the rms speed for Nitrogen (v_rms(N₂))**:
\[ v_{rms(N₂)} = \sqrt{\frac{5RT}{M_{N₂}}} \]

3. **Calculate the rms speed for Oxygen (v_rms(O₂))**:
\[ v_{rms(O₂)} = \sqrt{\frac{5RT}{M_{O₂}}} \]

4. **Find the ratio of rms speeds**:
\[ \frac{v_{rms(N₂)}}{v_{rms(O₂)}} = \frac{\sqrt{\frac{5RT}{M_{N₂}}}}{\sqrt{\frac{5RT}{M_{O₂}}}} = \sqrt{\frac{M_{O₂}}{M_{N₂}}} \]
Since \( R \) and \( T \) are constants, they cancel out.

5. **Substituting the values**:
\[ \frac{v_{rms(N₂)}}{v_{rms(O₂)}} = \sqrt{\frac{32}{28}} = \sqrt{1.142857} \approx 1.067 \]

6. **Final ratio**:
\[ \frac{v_{rms(N₂)}}{v_{rms(O₂)}} \approx 1.06 : 1 \]

### (ii) Ratio of the rms speed of oxygen molecules in air at 0°C to that at 10°C

1. **Using the rms speed formula**:
\[ v_{rms} = \sqrt{\frac{5RT}{M}} \]

2. **Calculate rms speed at 0°C (273 K)**:
\[ v_{rms(O₂, 0°C)} = \sqrt{\frac{5R \cdot 273}{M_{O₂}}} \]

3. **Calculate rms speed at 10°C (283 K)**:
\[ v_{rms(O₂, 10°C)} = \sqrt{\frac{5R \cdot 283}{M_{O₂}}} \]

4. **Find the ratio of rms speeds**:
\[ \frac{v_{rms(O₂, 0°C)}}{v_{rms(O₂, 10°C)}} = \frac{\sqrt{\frac{5R \cdot 273}{M_{O₂}}}}{\sqrt{\frac{5R \cdot 283}{M_{O₂}}}} = \sqrt{\frac{273}{283}} \]
Again, \( M_{O₂} \) and \( 5R \) cancel out.

5. **Substituting the values**:
\[ \frac{v_{rms(O₂, 0°C)}}{v_{rms(O₂, 10°C)}} = \sqrt{\frac{273}{283}} \approx 0.982 \]

6. **Final ratio**:
\[ \frac{v_{rms(O₂, 0°C)}}{v_{rms(O₂, 10°C)}} \approx 0.982 : 1 \]

### Final Answers:
(i) The ratio of the rms speed of nitrogen molecules to that of the oxygen molecules in air is approximately **1.06 : 1**.

(ii) The ratio of the rms speed of oxygen molecules in air at 0°C to that at 10°C is approximately **0.982 : 1**.