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Magazine Chapter 11: Problem 28
An air bubble has a volume of \(0.500 \mathrm{~L}\) at \(18{ }^{\circ} \mathrm{C}\). What is the final volume, in liters, of the gas when the temperature changes to each of the following, if \(n\) and \(P\) do not change? a. \(0^{\circ} \mathrm{C}\) b. \(425 \mathrm{~K}\) c. \(-12^{\circ} \mathrm{C}\) d. \(575 \mathrm{~K}\)
Short Answer
Expert verified
a. 0.469 L, b. 0.729 L, c. 0.448 L, d. 0.987 L
Step by step solution
01
Understand the Problem
We need to find the final volume of a gas when its temperature changes. We know the initial volume, initial temperature, and final temperature for each part, assuming the number of moles () and pressure () remain constant. This scenario applies Charles's Law for gases, which is given by: \[\frac{V_1}{T_1} = \frac{V_2}{T_2}\]
02
Convert Temperatures to Kelvin
Charles's Law requires temperature to be in Kelvin. Convert each temperature from Celsius to Kelvin using the formula \[ T(K) = T(掳C) + 273.15 \]
03
Substitute Initial Values
The initial volume is \( V_1 = 0.500 \text{ L} \) and the initial temperature is \( T_1 = 18掳C \). Converting 18掳C to Kelvin, \( T_1 = 18 + 273.15 = 291.15 \text{ K} \)
04
Step 4a: Calculate Final Volume for 0掳C
Convert the final temperature: \( 0掳C = 273.15 \text{ K} \). Apply Charles's Law: \[ \frac{0.500 \text{ L}}{291.15 \text{ K}} = \frac{V_2}{273.15 \text{ K}} \]. Solving for \( V_2 \): \[ V_2 = \frac{0.500 \text{ L} \times 273.15 \text{ K}}{291.15 \text{ K}} = 0.469 \text{ L} \]
05
Step 4b: Calculate Final Volume for 425 K
The final temperature is already in Kelvin: \( 425 \text{K} \). Apply Charles鈥檚 Law: \[ \frac{0.500 \text{ L}}{291.15 \text{ K}} = \frac{V_2}{425 \text{ K}} \]. Solving for \( V_2 \): \[ V_2 = \frac{0.500 \text{ L} \times 425 \text{ K}}{291.15 \text{ K}} = 0.729 \text{ L} \]
06
Step 4c: Calculate Final Volume for -12掳C
Convert the final temperature: \( -12掳C = 261.15 \text{ K} \). Apply Charles鈥檚 Law: \[ \frac{0.500 \text{ L}}{291.15 \text{ K}} = \frac{V_2}{261.15 \text{ K}} \]. Solving for \( V_2 \): \[ V_2 = \frac{0.500 \text{ L} \times 261.15 \text{ K}}{291.15 \text{ K}} = 0.448 \text{ L} \]
07
Step 4d: Calculate Final Volume for 575 K
The final temperature is already in Kelvin: \( 575 \text{ K} \). Apply Charles鈥檚 Law: \[ \frac{0.500 \text{ L}}{291.15 \text{ K}} = \frac{V_2}{575 \text{ K}} \]. Solving for \( V_2 \): \[ V_2 = \frac{0.500 \text{ L} \times 575 \text{ K}}{291.15 \text{ K}} = 0.987 \text{ L} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Charles's Law
Charles's Law describes the direct relationship between the volume of a gas and its temperature. To visualize it, imagine a balloon filled with air: if you heat the balloon, it expands; if you cool it, it contracts. Charles鈥檚 Law can be expressed as \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\). Here, \(V_1\) and \(T_1\) are the initial volume and temperature, respectively, and \(V_2\) and \(T_2\) are the final volume and temperature. It is crucial to remember that temperature must always be in Kelvin when using this law to ensure accuracy.
Temperature Conversion to Kelvin
To use Charles鈥檚 Law, you must convert temperatures from Celsius to Kelvin. The Kelvin scale begins at absolute zero, the lowest possible temperature. The formula to convert Celsius to Kelvin is: \(T(K) = T(掳C) + 273.15\). For example, converting 18掳C to Kelvin involves adding 273.15, resulting in 291.15 K. This step is essential because Charles鈥檚 Law, and other gas laws, rely on the absolute temperature scale to ensure proportional relationships.
Gas Laws
Gas laws, such as Charles鈥檚 Law, Boyle鈥檚 Law, and Avogadro鈥檚 Law, describe how gases behave under different conditions of temperature, volume, and pressure. Understanding these laws helps us predict and manipulate gas behavior in various scientific and real-world applications. Charles鈥檚 Law specifically focuses on the volume-temperature relationship at constant pressure and moles. Familiarizing yourself with these laws provides a foundational understanding of thermodynamics and gas behavior.
Volume and Temperature Relationship
The volume and temperature relationship in Charles鈥檚 Law is a direct one: if temperature increases, volume increases and vice versa. This relationship assumes constant pressure and moles of gas. For instance, if a gas鈥檚 volume at 291.15 K (18掳C) is 0.500 L, raising the temperature to 425 K will increase the volume to 0.729 L. Conversely, lowering the temperature to 273.15 K (0掳C) will decrease the volume to 0.469 L. This relationship is useful in various scientific fields, including chemistry and physics.
Constant Pressure and Moles
In Charles鈥檚 Law, the volume and temperature relationship is explored under the condition of constant pressure and moles of gas. This means that while temperature and volume may change, the pressure and the number of gas molecules remain unchanged. These conditions are crucial for accurate predictions using Charles鈥檚 Law. For example, in a balloon, as long as the pressure from the surroundings and the amount of gas inside stays the same, increasing the temperature will cause the balloon to expand proportionally.
