Problem 26 - College Physics

Question

A husband buys a helium-filled anniversary balloon for his wife. The balloon has a volume of \(3.5 \mathrm{~L}\) in the warm store at \(74^{\circ} \mathrm{F}\). When he takes it outside, where the temperature is \(48^{\circ} \mathrm{F}\), he finds it has shrunk. By how much has the volume decreased?

Step-by-Step Solution

Answer

The volume decreases by 0.17 L.

Step 1: Understand the problem

We need to find how much the volume of the balloon decreases when it is taken from the warm store (74°F) to outside (48°F). We will use the concept of gas laws, specifically Charles' Law, since the pressure remains constant and we are dealing with the temperature and volume of the gas.

Step 2: Convert Fahrenheit to Kelvin

Charles' Law requires temperature in Kelvin. First, we convert the Fahrenheit temperatures to Celsius:- To Celsius: \( T(^{\circ}C) = \frac{5}{9} (T(^{\circ}F) - 32) \)- Store temperature: \( 74^{\circ}F = \frac{5}{9} (74 - 32) = 23.33^{\circ}C \)- Outside temperature: \( 48^{\circ}F = \frac{5}{9} (48 - 32) = 8.89^{\circ}C \)Now, convert Celsius to Kelvin:- Store temperature: \( 23.33^{\circ}C = 23.33 + 273.15 = 296.48 \, K \)- Outside temperature: \( 8.89^{\circ}C = 8.89 + 273.15 = 282.04 \, K \)

Step 3: Apply Charles' Law

Charles' Law states that \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where:\( V_1 \) and \( T_1 \) are the initial volume and temperature, \( V_2 \) and \( T_2 \) are the final volume and temperature. Given:- \( V_1 = 3.5 \, L \), \( T_1 = 296.48 \, K \)- \( T_2 = 282.04 \, K \)We need to find \( V_2 \):\[ V_2 = V_1 \times \frac{T_2}{T_1} = 3.5 \, L \times \frac{282.04 \, K}{296.48 \, K} \]

Step 4: Calculate the final volume

Using the Charles' Law formula from the previous step:\[ V_2 = 3.5 \, L \times \frac{282.04 \, K}{296.48 \, K} \approx 3.33 \, L \]

Step 5: Calculate the change in volume

To find the decrease in volume:\[ \text{Decrease in volume} = V_1 - V_2 = 3.5 \, L - 3.33 \, L = 0.17 \, L \]

Step 6: Conclusion: Write the final answer

The volume of the balloon decreases by 0.17 L when taken outside.

Key Concepts

Gas Laws

Understanding the behavior of gases is crucial, especially when we are dealing with changes in volume, pressure, and temperature. One of the fundamental principles that help in predicting these changes is the **Gas Laws**. These laws include Boyle's Law, Avogadro's Law, and Charles' Law, each addressing different aspects of gas behavior. Charles' Law is particularly important in our scenario, as it describes the direct relationship between volume and temperature. When pressure is kept constant, an increase in temperature results in an increase in volume and vice versa. This means that for a gas, if you heat it, it expands; if you cool it, it contracts. This relationship is depicted by the equation \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where \( V_1 \) and \( T_1 \) are the initial volume and temperature, and \( V_2 \) and \( T_2 \) are the final volume and temperature. Charles' Law applies perfectly in practical situations like our balloon example, explaining why a helium balloon shrinks when moved to a cooler environment.

Temperature Conversion

Temperature is a measure of how hot or cold something is, and it plays a vital role in how gases behave. In gas law calculations, temperatures must be in Kelvin, as this is the standard unit in the International System of Units (SI) for temperature. To convert temperatures from Fahrenheit to Kelvin, you need to pass through another temperature scale, Celsius, first. Here’s how it’s done:

• Convert Fahrenheit to Celsius using the formula: \( T(^\circ C) = \frac{5}{9} (T(^\circ F) - 32) \).
• Convert Celsius to Kelvin by adding 273.15: \( T(K) = T(^\circ C) + 273.15 \).

In our example, the store temperature is converted from \( 74 ^\circ F \) to about \( 296.48 \, K \) and the outside temperature from \( 48 ^\circ F \) to \( 282.04 \, K \). These conversions are crucial because they allow us to apply Charles' Law accurately.

Volume Change Calculation

Calculating changes in volume, especially for gases, often revolves around the relationship defined in Charles' Law when temperature changes are involved. Here's a simplified step-by-step guide using our balloon case:1. **Set the initial conditions**: You know the initial volume \( V_1 = 3.5 \, L \) and the initial temperature \( T_1 = 296.48 \, K \).2. **Identify the final condition**: The final temperature when the balloon is taken outside is \( T_2 = 282.04 \, K \).3. **Apply Charles' Law**: Rearrange the equation \[ V_2 = V_1 \times \frac{T_2}{T_1} \]. Substitute in the values: \[ V_2 = 3.5 \, L \times \frac{282.04 \, K}{296.48 \, K} \approx 3.33 \, L \].4. **Determine the change**: Calculate the decrease by subtracting the final volume from the initial volume: \( 3.5 \, L - 3.33 \, L = 0.17 \, L \).The balloon's volume decreases by 0.17 L, illustrating how precise calculations and understanding the principles can lead to accurate results.