One mole of oxygen at 273 K and one mole of sulphur dioxide at 546 K are taken in two separate containers, then

Options

A. kinetic energy of `O_2 gt` kinetic energy of `SO_2`
B. kinetic energy `O_2 lt` kinetic energy of `SO_2`
C. kinetic energy of both are equal
D. none of the above

Explanation

To solve the problem of comparing the kinetic energies of one mole of oxygen (O₂) at 273 K and one mole of sulfur dioxide (SO₂) at 546 K, we will use the formula for the average kinetic energy of a gas, which is given by:

\[ KE = \frac{3}{2} nRT \]

where:
- \( KE \) is the kinetic energy,
- \( n \) is the number of moles,
- \( R \) is the universal gas constant (approximately 8.314 J/(mol·K)),
- \( T \) is the absolute temperature in Kelvin.

### Step-by-Step Solution:

1. **Calculate the Kinetic Energy of Oxygen (O₂):**
- For oxygen, we have:
- Number of moles, \( n = 1 \)
- Temperature, \( T = 273 \, K \)
- Using the kinetic energy formula:
\[ KE_{O_2} = \frac{3}{2} nRT = \frac{3}{2} \times 1 \times R \times 273 \]
- This simplifies to:
\[ KE_{O_2} = \frac{3}{2} R \times 273 \]

2. **Calculate the Kinetic Energy of Sulfur Dioxide (SO₂):**
- For sulfur dioxide, we have:
- Number of moles, \( n = 1 \)
- Temperature, \( T = 546 \, K \)
- Using the kinetic energy formula:
\[ KE_{SO_2} = \frac{3}{2} nRT = \frac{3}{2} \times 1 \times R \times 546 \]
- This simplifies to:
\[ KE_{SO_2} = \frac{3}{2} R \times 546 \]

3. **Find the Ratio of Kinetic Energies:**
- To compare the two kinetic energies, we can take the ratio:
\[ \frac{KE_{SO_2}}{KE_{O_2}} = \frac{\frac{3}{2} R \times 546}{\frac{3}{2} R \times 273} \]
- The \( \frac{3}{2} R \) terms cancel out:
\[ \frac{KE_{SO_2}}{KE_{O_2}} = \frac{546}{273} = 2 \]

4. **Conclusion:**
- This means that:
\[ KE_{SO_2} = 2 \times KE_{O_2} \]
- Therefore, the kinetic energy of sulfur dioxide (SO₂) is twice that of oxygen (O₂).

### Final Answer:
The kinetic energy of O₂ is less than the kinetic energy of SO₂.