Question

For each of the following sets of volume/temperature data, calculate the missing quantity. Assume that the pressure and the amount of gas remain constant. a. \(V=22.4 \mathrm{~L}\) at \(0 \quad \mathrm{C} ; V=44.4 \mathrm{~L}\) at \(? \mathrm{~K}\) b. \(V=1.0 \times 10^{-3} \mathrm{~mL}\) at \(-272^{\circ} \mathrm{C} ; V=?\) at \(25^{\circ} \mathrm{C}\) c. \(V=32.3 \mathrm{~L}\) at \(-40^{\circ} \mathrm{C} ; V=1000 . \mathrm{L}\) at \(?{ }^{\circ} \mathrm{C}\)

Short answer

a. \(T_2=544\mathrm{~K}\) b. \(V_2 \approx 2.6 \times 10^{-3} \mathrm{mL}\) c. \(T_2 \approx 449.24^{\circ} \mathrm{C}\)

Step-by-step solution

Write Charles's Law equation and conversion formula.

Charles's Law states that, for a given amount of gas at constant pressure, the ratio of volume and temperature remains constant: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] First we need to convert the temperature from Celsius to Kelvin using the following formula: \[ K = ^\circ C + 273.15 \]

Solve (a) \(V=22.4 \mathrm{~L}\) at \(0^{\circ} \mathrm{C} ; V=44.4 \mathrm{~L}\) at \(? \mathrm{~K}\).

Using Charles's Law and the conversion formula, we can write the equation for given data: \[ \frac{22.4 \mathrm{~L}}{0+273.15 \mathrm{~K}} =\frac{44.4 \mathrm{~L}}{T_2} \] Now, solve for \(T_2\): \[ T_2 =\frac{44.4 \mathrm{~L} \times 273.15 \mathrm{~K}}{22.4 \mathrm{~L}} \approx 544 \mathrm{~K} \]

Solve (b) \(V=1.0 \times 10^{-3} \mathrm{~mL}\) at \(-272^{\circ} \mathrm{C} ; V=?\) at \(25^{\circ} \mathrm{C}\) .

Using Charles's Law and the conversion formula: \[ \frac{1.0 \times 10^{-3} \mathrm{~mL}}{-272+273.15 \mathrm{~K}} =\frac{V_2}{25+273.15 \mathrm{~K}} \] Solve for \(V_2\): \[ V_2 =\frac{1.0 \times 10^{-3} \mathrm{~mL} \times (25+273.15) \mathrm{~K}}{1.15 \mathrm{~K}} \approx 2.6 \times 10^{-3} \mathrm{mL} \]

Solve (c) \(V=32.3 \mathrm{~L}\) at \(-40^{\circ} \mathrm{C} ; V=1000 . \mathrm{L}\) at \(?{ }^{\circ} \mathrm{C}\).

Using Charles's Law and the conversion formula: \[ \frac{32.3 \mathrm{~L}}{-40+273.15 \mathrm{~K}} =\frac{1000 \mathrm{~L}}{T_2} \] Solve for \(T_2\): \[ T_2 =\frac{1000 \mathrm{~L} \times (233.15) \mathrm{~K}}{32.3 \mathrm{~L}} \approx 722.39 \mathrm{~K} \] Convert \(T_2\) to Celsius: \[ 722.39 \mathrm{~K}-273.15 = 449.24^{\circ} \mathrm{C} \] So the final solutions are: a. \(T_2=544\mathrm{~K}\) b. \(V_2 \approx 2.6 \times 10^{-3} \mathrm{mL}\) c. \(T_2 \approx 449.24^{\circ} \mathrm{C}\)

Key concepts

Gas Laws

The behavior of gases under various conditions of temperature, volume, and pressure is described by gas laws. These laws are based on observations and experiments, and they are crucial for understanding how gases interact in different scenarios, including natural phenomena and industrial processes.

One of the fundamental gas laws is Charles's Law, which focuses on the volume-temperature relationship at constant pressure. It states that if the pressure of a gas remains unchanged, the volume of the gas is directly proportional to its temperature (measured in Kelvin). This relationship is depicted in the formula: \[\begin{equation}\frac{V_1}{T_1} = \frac{V_2}{T_2}\end{equation}\]
where:

• \(V_1\) and \(V_2\) are the initial and final volumes of the gas, respectively.
• \(T_1\) and \(T_2\) are the initial and final temperatures of the gas in Kelvin.

Solving for the unknown in this equation allows us to predict how a gas will behave when subjected to temperature changes, assuming the amount of gas and the pressure are held constant. This can be particularly useful in various scientific and practical applications, such as in meteorology, culinary arts (like bread baking where gases expand), and the automotive industry (where engine temperatures affect the air volume in tires).

Volume-Temperature Relationship

The volume-temperature relationship in gases, described by Charles's Law, is a linear relationship that indicates how gas volume will change as temperature varies, assuming pressure and the amount of gas are constant. When the temperature increases, the kinetic energy of gas particles also increases, causing them to move more vigorously and occupy more space – thus increasing the volume.

This phenomenon can be demonstrated by a simple experiment: inflating a balloon and then exposing it to different temperatures. You will notice that the balloon expands when heated and contracts when cooled. This is a visual representation of Charles's Law in action.

When solving problems related to the volume-temperature relationship of gases, it is essential to use the correct formula and to make sure temperatures are in the proper unit, Kelvin:

• To convert Celsius to Kelvin, the formula \(K = ^\circ C + 273.15\) is used.
• To solve for the unknown volume or temperature, we rearrange the Charles's Law equation accordingly.

When given a set of volume and temperature data, like in the provided exercise, students can calculate the missing quantity (volume or temperature) by inserting the known values into the equation and solving for the unknown.

Temperature Conversion

Understanding temperature conversion is crucial when working with gas laws, as the temperature must be expressed in Kelvin (K) and not in Celsius (°C) or Fahrenheit (°F). Kelvin is the SI unit for temperature and is used in scientific equations because it provides an absolute scale where 0 K represents absolute zero, the lowest theoretical temperature.

The process of converting from Celsius to Kelvin is straightforward but essential for accurate calculations. The conversion formula is:

• \(K = ^\circ C + 273.15\)

This means that to convert a temperature from Celsius to Kelvin, you simply add 273.15 to the Celsius temperature. Note that there is no degree (\(^\circ\)) used when talking about Kelvin, as it is an absolute measure of temperature.

Proper temperature conversion ensures that we apply Charles's Law correctly, leading to accurate predictions of a gas's behavior. In the exercise provided, temperature conversion was a key step in determining the missing variables, and it is a skill that students need to master when dealing with gas-related problems in chemistry and physics.