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When dealing with direct variation problems, one of the key pieces of the puzzle is the proportionality constant. This constant, often denoted as "k," acts as the glue that connects the changing variables. In the context of our problem, the speed \(s\) of oxygen molecules in the air is directly proportional to the square root of the temperature \(T\). The formula capturing this relationship is \(s = k \sqrt{T}\). Here, the role of \(k\) is crucial because it allows us to translate how temperature impacts speed in a consistent way across different conditions.

To find \(k\), we need specific values that can be substituted into the equation. In our example, we know the speed \(s = 460\, \text{m/s}\) at a temperature \(T = 273\, \text{K}\). Plugging these values into the formula, we rearrange to solve for \(k\): \(k = \frac{460}{\sqrt{273}}\). Now, \(k\) becomes the cornerstone for predicting speeds at any other given temperatures.

In mathematics, the square root function is a unique function that forms a fundamental part of proportional and variation problems. This special function is used when one quantity varies as the square root of another, just like in our problem. When you see a square root, it implies that changes in the temperature have a gradual effect on speed.

Mathematically speaking, taking the square root of a number \(T\) provides us with a value \(\sqrt{T}\) that relates to how the molecule speeds will adjust as temperature changes. With this understanding, if you double the temperature, the effect on speed isn't a doubling but a more moderated increase captured by the square root relationship.

This function smooths the impact of temperature changes, which is why speed increases with temperature but not at the exact rate of the temperature increase.

The relationship between temperature and the speed of oxygen molecules is a classic example demonstrating direct variation. As the temperature of a gas increases, its molecules move faster. This is because higher temperatures provide energy to the molecules, causing them to increase in speed.

In the practical example given, we observe that at \(273\, \text{K}\), the speed of the molecules is \(460\, \text{m/s}\). We found earlier that the proportionality constant \(k\) is used to predict the speed at another temperature, such as \(300\, \text{K}\). Using the formula \(s = \frac{460}{\sqrt{273}} \times \sqrt{300}\), we compute the new speed and find it approximately \(480.36\, \text{m/s}\).

This logical link helps us model real-world phenomena where temperature changes influence molecule behavior, such as in processes like diffusion or reaction kinetics.