At what temperature `25 dm ^3` of oxygen at 283 K is heated to make its volume `30 dm^3` ?

Explanation

To find the final temperature \( T_2 \) when \( 25 \, \text{dm}^3 \) of oxygen at \( 283 \, \text{K} \) is heated to a volume of \( 30 \, \text{dm}^3 \), we can use Charles's Law. Charles's Law states that the volume of a gas is directly proportional to its temperature when pressure is held constant.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial Volume \( V_1 = 25 \, \text{dm}^3 \)
- Initial Temperature \( T_1 = 283 \, \text{K} \)
- Final Volume \( V_2 = 30 \, \text{dm}^3 \)
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