Problem 11
Question
The gas inside a balloon will always have a pressure nearly equal to atmospheric pressure, since that is the pressure applied to the outside of the balloon. You fill a balloon with helium (a nearly ideal gas) to a volume of 0.600 L at a temperature of \(19.0^{\circ} \mathrm{C} .\) What is the volume of the balloon if you cool it to the boiling point of liquid nitrogen \((77.3 \mathrm{K}) ?\)
Step-by-Step Solution
Verified Answer
The volume of the balloon is approximately 0.159 liters when cooled to 77.3 K.
1Step 1: Identify the Given Variables
We are given the initial volume of the balloon \( V_1 = 0.600 \) liters and initial temperature \( T_1 = 19.0^{\circ}C \). The final temperature \( T_2 \) is the boiling point of liquid nitrogen \( 77.3 \mathrm{K} \). First, convert \( T_1 \) from Celsius to Kelvin using the formula: \( T = \text{Celsius} + 273.15 \). Thus, \( T_1 = 19.0 + 273.15 = 292.15 \mathrm{K} \).
2Step 2: Apply Charles's Law
Charles's Law states that \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) for a constant pressure. Here, \( V_1 \) and \( T_1 \) are the initial conditions, and \( V_2 \) and \( T_2 \) are the final conditions.
3Step 3: Solve for the Final Volume \( V_2 \)
Re-arrange Charles's Law to solve for \( V_2 \): \[ V_2 = V_1 \times \frac{T_2}{T_1} \]Substitute the known values: \[ V_2 = 0.600 \times \frac{77.3}{292.15} \approx 0.159 \text{ liters} \]
4Step 4: Verify the Calculation
Double-check the calculations for any arithmetic errors. The calculation is correct, confirming that the final volume \( V_2 \) of the balloon is approximately \( 0.159 \) liters.
Key Concepts
Ideal Gas BehaviorTemperature ConversionGas LawsVolume Change
Ideal Gas Behavior
Gases that behave ideally follow a specific set of rules that make predicting their behavior simpler. An ideal gas is a theoretical gas composed of many randomly moving point particles that interact only through elastic collisions. It's an approximation used in many gas-related calculations, including Charles's Law.
- An ideal gas follows the ideal gas law equation, which is a combination of various simpler gas laws including Boyle's and Charles's Laws.
- Real gases can often be approximated as ideal gases when under conditions of high temperature and low pressure.
- Helium, the gas used in balloons, approximates ideal gas behavior closely under many conditions, making it suitable for use in theoretical and practical exercises.
Temperature Conversion
Temperature conversion is essential when working with gas laws as they often require temperatures to be in Kelvin.
The Kelvin scale is used because it begins at absolute zero, making it ideal for calculations in physics and chemistry.
To convert from Celsius to Kelvin, you use this simple formula:
- Add 273.15 to the Celsius temperature to obtain the temperature in Kelvin.
- The initial temperature was 19.0°C. By converting to Kelvin, it becomes 292.15 K.
- This ensures compatibility in calculations using gas laws, which require absolute temperature measurements.
Gas Laws
Gas laws are fundamental principles that describe how gases behave under various conditions of pressure, temperature, and volume.Charles’s Law is one of these laws, focusing specifically on the relationship between temperature and volume:
- For a given mass of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature (measured in Kelvin).
- This can be expressed mathematically as \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \).
- This means if you increase the temperature, the volume will increase, provided the pressure remains constant.
Volume Change
Volume change in gases is directly linked to changes in temperature when the pressure is held constant, as explained by Charles’s Law.In our example, as the temperature of the helium-filled balloon is lowered to the boiling point of liquid nitrogen, a significant decrease in volume occurs.The formula from Charles's Law allows us to predict how much the volume will change:
- Using the relationship \( V_2 = V_1 \times \frac{T_2}{T_1} \), you can calculate the final volume.
- For the balloon, when cooled from 292.15 K to 77.3 K, the volume reduces from 0.600 L to approximately 0.159 L.
- This illustrates the impact temperature changes have on the volume of a gas, crucial for applications and experiments involving gas expansion or compression.
Other exercises in this chapter
Problem 9
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