Question

Calculate the final Celsius temperature when \(5.00 \mathrm{~L}\) of fluorine gas at \(50^{\circ} \mathrm{C}\) is cooled to give a volume of \(3.75 \mathrm{~L}\). Assume that the pressure remains constant.

Short answer

The final Celsius temperature is approximately -30.79°C.

Step-by-step solution

Understand the problem

We need to find the final temperature in Celsius when the volume of a gas changes, given the initial temperature and volume. We are given that the process occurs at constant pressure, indicating the use of Charles's Law, which is applicable to ideal gases.

Recall Charles's Law

Charles's Law states that the volume of an ideal gas is directly proportional to its absolute temperature (in Kelvin) when pressure is held constant. This can be expressed as \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) where \(V_1\) and \(T_1\) are the initial volume and temperature, and \(V_2\) and \(T_2\) are the final volume and temperature, respectively.

Convert initial temperature to Kelvin

The initial temperature is given as \(50^{\circ} \mathrm{C}\). To use Charles's Law, convert this to Kelvin by adding 273.15. The initial temperature in Kelvin is \(T_1 = 50 + 273.15 = 323.15 \: \text{K}\).

Set up the equation using Charles's Law

Using Charles's Law \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), we substitute \(V_1 = 5.00 \: \mathrm{L}\), \(T_1 = 323.15 \: \text{K}\), and \(V_2 = 3.75 \: \mathrm{L}\) to solve for \(T_2\): \[ \frac{5.00}{323.15} = \frac{3.75}{T_2} \].

Calculate the final temperature in Kelvin

Solve the equation from Step 4 for \(T_2\): Multiply both sides by \(T_2\) to get \(5.00 \, T_2 = 3.75 \, 323.15\). Divide both sides by 5.00 to isolate \(T_2\): \(T_2 = \frac{3.75 \, 323.15}{5.00} = 242.3625 \: \text{K}\).

Convert final temperature back to Celsius

Convert \(T_2\) back to Celsius by subtracting 273.15: \(T_2 \text{ in Celsius} = 242.3625 - 273.15 = -30.79^{\circ} \mathrm{C}\).

Key concepts

Ideal Gas Behavior

To understand how gases behave, we often refer to the concept known as ideal gas behavior. Ideal gases follow certain rules described by various gas laws, like Charles's Law. Although real gases do not perfectly align with these rules, many behave similarly enough under standard conditions to be approximated as ideal gases. This concept helps in making predictions about how gases react to changes in temperature, volume, and pressure.
The basic assumption is that the gas particles move randomly and that the size of these particles is much smaller than the distance between them. Furthermore, interactions between these particles are minimal. The predictability of ideal gases helps us employ relationships, like the one Charles described, to solve problems and understand gas behavior under constant pressure conditions.

Temperature Conversion

When solving problems involving gas laws, converting temperature to Kelvin is crucial. Gas laws, including Charles's Law, depend on absolute temperature, which is represented in Kelvin. The Kelvin scale starts at absolute zero, which is the theoretical lowest temperature possible, reflecting a point where particle motion would cease.
To convert from Celsius to Kelvin, simply add 273.15. For example, an initial temperature of \(50^{\circ}\mathrm{C}\) converts to \(323.15\, \mathrm{K}\). This step is vital because calculations require consistency with absolute temperature measures. Conversely, after a calculation, you might convert back to Celsius by subtracting 273.15 from the Kelvin temperature, helping interpret results in a commonly used temperature scale.

Volume-Temperature Relationship

Exploring the volume-temperature relationship is key to understanding Charles's Law. Charles's Law states that as long as the pressure remains constant, the volume of an ideal gas is directly proportional to its absolute temperature, measured in Kelvin.
This means if the volume increases, so does the temperature, and vice versa. We express this mathematically as \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where \(V\) is volume and \(T\) is temperature. Solving problems often involves rearranging this formula to find unknown values, such as the final temperature or volume, when the other variables are known.

• When the initial conditions are \(V_1 = 5.00 \, \mathrm{L}\) and \(T_1 = 323.15 \, \mathrm{K}\), and the final volume is \(V_2 = 3.75 \, \mathrm{L}\), plug these into the equation to find the final temperature \(T_2\).
• This process allows you to predict how volume changes affect temperature under constant pressure, offering insight into gas behavior.

Understanding this proportionality enables you to solve real-world problems and grasp how temperature changes in practical situations, like cooling a gas and observing volume contraction.