Problem 6 - Chemistry Matter and Change

Question

Assume that the pressure and the amount of gas remain constant in the following problems. The Celsius temperature of a 3.00-\(\mathrm{L}\) sample of gas is lowered from \(80.0^{\circ} \mathrm{C}\) to \(30.0^{\circ} \mathrm{C}\). What will be the resulting volume of this gas?

Step-by-Step Solution

Answer

The resulting volume of the gas is 2.57 L.

Step 1: Understand the Problem

We are dealing with a gas whose temperature changes while its pressure and amount remain constant. We need to find the new volume of the gas.

Step 2: Identify the Gas Law

Given that pressure and amount of gas are constant, we can use Charles's Law to solve this problem. Charles's Law states that \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), where \(V\) is volume and \(T\) is temperature in Kelvin.

Step 3: Convert Celsius to Kelvin

We must convert the temperatures from Celsius to Kelvin. Use the formula \(T(\text{K}) = T(\degree \text{C}) + 273.15\). Thus, the initial temperature \(T_1 = 80.0 + 273.15 = 353.15\,\text{K}\) and the final temperature \(T_2 = 30.0 + 273.15 = 303.15\,\text{K}\).

Step 4: Substitute Values into Charles's Law

Substitute the known values into Charles's Law:\[\frac{V_1}{T_1} = \frac{V_2}{T_2} \Rightarrow \frac{3.00\,\text{L}}{353.15\,\text{K}} = \frac{V_2}{303.15\,\text{K}}\].

Step 5: Solve for the Unknown Volume

Rearrange the equation to solve for \(V_2\):\[V_2 = \frac{3.00\,\text{L} \times 303.15\,\text{K}}{353.15\,\text{K}} = 2.57\,\text{L}\].

Key Concepts

Gas Laws

Gas laws are fundamental in understanding the behavior of gases under various conditions. One of the most well-known principles under this umbrella is Charles's Law. This law is crucial when working with problems that involve changes in the temperature and volume of a gas while maintaining constant pressure. It states that the volume of an ideal gas is directly proportional to its temperature when the pressure is kept constant. Expressed formulaically, Charles's Law is given by:

• \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)

Here, \(V_1\) and \(V_2\) represent the initial and final volumes, respectively, while \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin. Gas laws like this one help predict how gases will behave when subjected to changes in their environment.

Temperature Conversion

Conversion between different temperature scales is critical in applying gas laws like Charles's Law. Most scientific calculations involving gases require temperatures to be expressed in Kelvin, the absolute scale.

To convert a temperature from Celsius to Kelvin, you can use a simple formula:

• \(T(\text{K}) = T(\degree \text{C}) + 273.15\)

For example, a temperature of \(80.0^{\circ} \text{C}\) converts to \(353.15\,\text{K}\), and \(30.0^{\circ} \text{C}\) converts to \(303.15\,\text{K}\). Converting temperatures ensures that calculations are accurate and consistent with the mathematical framework of the gas laws.

Volume Calculation

Calculating the volume of a gas after a temperature change involves understanding how different factors interact in Charles's Law.

When solving for the new volume (\(V_2\)), the formula \(V_2 = \frac{V_1 \times T_2}{T_1}\) is applied.

• Start by inputting the known values: \(V_1 = 3.00\,\text{L}\), \(T_1 = 353.15\,\text{K}\), and \(T_2 = 303.15\,\text{K}\).
• Substitute these values into the equation: \[V_2 = \frac{3.00\,\text{L} \times 303.15\,\text{K}}{353.15\,\text{K}} = 2.57\,\text{L}\]

The solution shows the effect of temperature decrease resulting in a decrease in the gas volume. Proper calculation is essential for accurately predicting gas behavior under specified conditions.

Kelvin to Celsius Conversion

In some situations, you need to convert temperatures from Kelvin back to Celsius, especially after calculations are complete. This can help provide an intuitive understanding of temperatures, as Celsius is a more familiar scale for everyday use.

To convert Kelvin to Celsius, use the formula:

• \(T(\degree \text{C}) = T(\text{K}) - 273.15\)

For instance, converting \(353.15\,\text{K}\) back to Celsius gives \(80.0^{\circ} \text{C}\). Likewise, converting \(303.15\,\text{K}\) results in \(30.0^{\circ} \text{C}\). This conversion is valuable when you need to relay temperature in a common format that is understandable to most people.