Question

Imagine that you have a huge helium balloon for a parade. Around noon, it is \(27^{\circ} \mathrm{C}\) when you fill the balloon with helium gas to a volume of \(25,000\) L. Later in the day, the temperature drops to \(15^{\circ} \mathrm{C}\) . a. What is the proportionality constant, \(k=V / T\) , at the beginning of the day? b. Calculate the volume of the balloon when the temperature has dropped to \(22^{\circ} \mathrm{C}\) . c. What will the voportionality constant, \(k=V / T,\) be at the end of the day when the temperature is \(15^{\circ} \mathrm{C}\) ? Explain your answer.

Short answer

a) The initial proportionality constant \( k \) is approximately 83.29 L/K. b) The volume at 22°C is approximately 24,593.7 L. c) The proportionality constant remains 83.29 L/K.

Step-by-step solution

Understand the Equation

The formula to use here is Charles's Law which states that the volume of a gas is directly proportional to its temperature, assuming pressure and amount of gas are constant. This is written as \( k = \frac{V}{T} \), where \( V \) is volume and \( T \) is temperature in Kelvin. The equation can also be expressed as \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \).

Convert Celsius to Kelvin

Temperatures must be converted from Celsius to Kelvin by adding 273.15. Initial temperature: \( T_1 = 27^{\circ} \text{C} + 273.15 = 300.15 \text{K} \). Final temperatures for points b and c: \( 22^{\circ} \text{C} + 273.15 = 295.15 \text{K} \), and \( 15^{\circ} \text{C} + 273.15 = 288.15 \text{K} \).

Calculate Initial Proportionality Constant \(k\)

Use the initial conditions, volume \( V_1 = 25,000 \) L and \( T_1 = 300.15 \text{K} \). Calculate \( k = \frac{V_1}{T_1} = \frac{25,000}{300.15} \approx 83.29 \text{ L/K} \).

Calculate Volume at 22°C

Using the equation \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), substitute \( V_1 = 25,000 \) L, \( T_1 = 300.15 \text{K} \), and \( T_2 = 295.15 \text{K} \). Solve for \( V_2 \): \( V_2 = \frac{25,000 \times 295.15}{300.15} \approx 24,593.7 \text{ L} \).

Calculate Final Proportionality Constant \(k\) at 15°C

Use the same formula for \( k \). Since \( k = \frac{V}{T} \), and assuming volume changes only due to temperature changes under constant pressure, \( k \) stays constant. Thus, \( k \) remains 83.29 L/K as it is dependent on factors that remained unchanged (same amount of gas and pressure conditions).

Key concepts

Gas Laws

When working with gases, particularly in physics and chemistry, gas laws are an essential concept to understand. These laws describe the behavior of gases in various conditions, connecting measurable properties such as volume, temperature, and pressure.
Charles's Law specifically relates to the interplay between volume and temperature when pressure and the amount of gas remain constant. It states that the volume of a gas is directly proportional to its absolute temperature. This means, as the temperature rises, the volume increases as well, and conversely, as temperature decreases, volume diminishes.
Charles's Law is articulated mathematically by the equation \( V \propto T \) or \( \frac{V}{T} = k \), where \( V \) is the volume, \( T \) is the temperature measured in Kelvin, and \( k \) is the proportionality constant. This law is critical in situations like determining the effect of temperature on the volume of gases in balloons, as explored in this exercise.

Temperature Conversion

When dealing with gas laws and equations, temperatures often need to be converted from Celsius to Kelvin. Kelvin is the SI unit of temperature used in scientific calculations because it starts at absolute zero, eliminating the possibility of negative numbers that could lead to mathematical inaccuracies.
To convert Celsius to Kelvin, add 273.15 to the Celsius temperature. For example:

• \(27^{\circ}\text{C} = 27 + 273.15 = 300.15 \text{ K}\)
• \(22^{\circ}\text{C} = 22 + 273.15 = 295.15 \text{ K}\)
• \(15^{\circ}\text{C} = 15 + 273.15 = 288.15 \text{ K}\)

In the exercise, converting Celsius to Kelvin allows for the application of Charles's Law, ensuring accurate calculation of volume changes as temperatures vary.

Volume-Temperature Relationship

At the core of Charles's Law is the volume-temperature relationship, which demonstrates how these two properties of a gas are interconnected when pressure and the amount of gas are constant.
According to the law, \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \). If you have initial conditions \( V_1 \text{ and } T_1 \) and final conditions \( V_2 \text{ and } T_2 \), you can predict one unknown property if the others are known. In the helium balloon exercise, this relationship helps calculate the balloon's new volume when the temperature drops later in the day.
An essential point here is that, despite the temperature change, if no gas is added or lost and the pressure stays the same, the proportionality constant \( k \) = \( \frac{V}{T} \) remains stable.

Helium Balloon

Helium balloons serve as practical examples for analyzing gas behaviors under different conditions. When a helium balloon is filled, the volume directly reflects the temperature, demonstrating Charles's Law in a tangible way.
In the exercise, as the day progresses and the temperature changes, so does the balloon's volume. Initial conditions set the stage: at noon, the balloon is filled at \(27^{\circ} \text{C}\) to a volume of \(25,000 \text{ L}\). As the temperature shifts to \(15^{\circ} \text{C}\), recalculating the volume demonstrates how the principles of gas laws apply in real-world scenarios.
This specificity is crucial not just for understanding theoretical principles but also for applications in industries where storing gases safely and effectively is essential, such as aerospace and weather forecasting.