Problem 21 - Chemistry The Molecular Science

Question

Suppose you have a sample of \(\mathrm{CO}_{2}\) in a gas-tight syringe with a movable piston. The gas volume is \(25.0 \mathrm{~mL}\) at a room temperature of \(20 .{ }^{\circ} \mathrm{C} .\) Calculate the final volume of the gas if you hold the syringe in your hand to raise the gas temperature to \(37^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The final volume of the gas is approximately 26.5 mL.

Step 1: Understand the Problem

We need to calculate the final volume of a gas when the temperature changes, while assuming that the pressure and the amount of gas remain constant. This is a classic application of Charles's Law in gas behavior.

Step 2: Convert Temperatures to Kelvin

Charles's Law requires temperatures to be in Kelvin. Convert the initial temperature of \(20^{\circ} \mathrm{C}\) to Kelvin by adding 273.15: \(20 + 273.15 = 293.15 \mathrm{~K}\). Similarly, convert \(37^{\circ} \mathrm{C}\) to Kelvin: \(37 + 273.15 = 310.15 \mathrm{~K}\).

Step 3: State Charles's Law

Charles's Law states that \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), where \(V_1\) and \(V_2\) are the initial and final volumes, and \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin.

Step 4: Substitute Known Values

We know that \(V_1 = 25.0 \mathrm{~mL}\), \(T_1 = 293.15 \mathrm{~K}\), and \(T_2 = 310.15 \mathrm{~K}\). Substitute these values into Charles's Law: \(\frac{25.0}{293.15} = \frac{V_2}{310.15}\).

Step 5: Solve for Final Volume \(V_2\)

Rearrange the equation to solve for \(V_2\): \(V_2 = \frac{25.0 \times 310.15}{293.15}\). Calculate the result: \(V_2 \approx 26.5 \mathrm{~mL}\).

Step 6: Conclusion

Based on the calculations, when the temperature of the gas is increased from \(20^{\circ} \mathrm{C}\) to \(37^{\circ} \mathrm{C}\), the final volume of the gas is approximately \(26.5 \mathrm{~mL}\).

Key Concepts

Gas Laws

Gas laws are essential principles that help us understand the behavior of gases under different conditions. One of these important laws is Charles's Law. Charles's Law shows the relationship between the volume and temperature of a gas at constant pressure. It explains that the volume of a gas is directly proportional to its temperature when measured in Kelvin. This means if the temperature of a gas goes up, its volume increases, given that pressure remains unchanged.

• Charles's Law formula: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
• \( V_1 \) and \( V_2 \) stand for initial and final volumes.
• \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.

Understanding gas laws like Charles's Law aids in predicting how gases will react to different environments. This is particularly useful in various fields like chemistry, physics, and engineering.

Temperature Conversion

When working with gas laws, it’s crucial to convert temperatures to the Kelvin scale. Kelvin is the standard unit of temperature in the scientific community, as it offers a direct correlation with energy and gas behavior.
To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. This conversion is necessary because Kelvin does not have negative numbers, making calculations straightforward and avoiding errors.
For example:

• To convert \(20^{\circ} \mathrm{C} \) to Kelvin: \( 20 + 273.15 = 293.15 \mathrm{~K} \)
• To convert \(37^{\circ} \mathrm{C} \) to Kelvin: \( 37 + 273.15 = 310.15 \mathrm{~K} \)

Always remember this step when dealing with gas problems aligned with Charles's Law or any other gas law.

Volume Calculation

Volume calculation is a pivotal step in determining the changes a gas undergoes under different temperatures as dictated by Charles's Law. Once the temperatures are correctly converted to Kelvin, we apply these to the formula \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \).
In this example, you calculate the final volume \( V_2 \) of the gas:

• Given: \( V_1 = 25.0 \mathrm{~mL}, \ T_1 = 293.15 \mathrm{~K}, \ T_2 = 310.15 \mathrm{~K} \)
• Substitute into the formula: \( \frac{25.0}{293.15} = \frac{V_2}{310.15} \)
• Solve for \( V_2 \): \( V_2 = \frac{25.0 \times 310.15}{293.15} \approx 26.5 \mathrm{~mL} \)

Using this step-by-step method guarantees accuracy in predicting how a gas's volume changes with a temperature increase, while ensuring that the pressure and amount of gas remain unchanged.