Question

A bubble of air has a volume of \(0.150 \mathrm{~mL}\) at \(24.2^{\circ} \mathrm{C}\). If the pressure remains constant, what is the volume of the bubble at \(62.5^{\circ} \mathrm{C}\) ?

Short answer

The volume of the bubble at \(62.5^{\circ}C\) will be \(0.169 \mathrm{mL}\)

Step-by-step solution

Convert the given temperatures from Celsius to Kelvin

Temperature in Kelvin = Temperature in Celsius + 273.15. So, \(T1 = 24.2^{\circ} \mathrm{C} + 273.15 = 297.35 \mathrm{K}\) and \(T2 = 62.5^{\circ} \mathrm{C} + 273.15 = 335.65 \mathrm{K}\)

Apply Charles's law

We use the formula \( V1/T1 = V2/T2 \). Substituting the known values we get, \(0.150 \mathrm{mL}/297.35 \mathrm{K} = V2 / 335.65 \mathrm{K}\)

Solving for \(V2\)

Rearranging the above equation to solve for \(V2\), we get \(V2 = (0.150 \mathrm{mL} \times 335.65 \mathrm{K}) / 297.35 \mathrm{K} = 0.169 \mathrm{mL}\)

Key concepts

Gas Laws

Gas laws describe the behavior of gases in various conditions of pressure, volume, and temperature. These laws provide a theoretical framework to understand the relationships between these physical properties. For chemistry students, mastering gas laws is crucial for solving problems and predicting how gases will respond to changes in conditions.

One of the foundational laws is Charles's Law, which states that, at constant pressure, the volume of a gas is directly proportional to its temperature (in Kelvin). This means that if the temperature goes up, the volume goes up, and if the temperature goes down, the volume goes down, assuming pressure remains unchanged. Understanding Charles's law allows you to predict how the volume of a gas will change with temperature, a concept that underlies many real-world applications, such as weather balloons and engine efficiency.

Temperature Conversion

Temperature conversion between Celsius and Kelvin is a basic but vital skill in chemistry, especially when working with gas laws like Charles's Law, which requires temperature to be in Kelvin. To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature, because the Kelvin scale starts at absolute zero, rather than the freezing point of water.

To easily remember, you can use the formula:
\[ K = {}^\circ C + 273.15 \]
When solving chemistry problems, getting the right temperature scale will make sure that the subsequent calculations are correct. In the example problem, converting the given temperatures from Celsius to Kelvin ensures that Charles's Law can be applied correctly to determine how the volume of gas changes with temperature.

Volume-Temperature Relationship

The volume-temperature relationship, as described by Charles's Law, is an example of how gas volume responds to temperature changes at a constant pressure. The mathematical expression of this relationship is:
\[ V1/T1 = V2/T2 \]
where \(V1\) and \(V2\) are the initial and final volumes, and \(T1\) and \(T2\) are the initial and final temperatures in Kelvin, respectively. To solve for the unknown volume \(V2\), you rearrange the equation to:
\[ V2 = V1 \times \frac{T2}{T1} \]

This algebraic manipulation allows us to find the final volume of the gas when the initial volume and the temperatures at both initial and final states are known. This quantitative understanding of volume's responsiveness to temperature is fundamental in predicting gas behavior in various scientific and engineering processes.