Problem 11 - Chemistry and Chemical Reactivity

Question

You have a sample of gas in a flask with a volume of \(250 \mathrm{mL}\). At \(25.5^{\circ} \mathrm{C}\), the pressure of the gas is \(360 \mathrm{mm}\) Hg. If you decrease the temperature to \(-5.0^{\circ} \mathrm{C},\) what is the gas pressure at the lower temperature?

Step-by-Step Solution

Answer

The gas pressure at \(-5.0^{\circ} \mathrm{C}\) is approximately \(322.5\) mm Hg.

Step 1: Understand the given

We have a sample of gas with an initial volume of \( V_1 = 250 \) mL, an initial temperature of \( T_1 = 25.5^{\circ} \mathrm{C} \), and an initial pressure of \( P_1 = 360 \) mm Hg. We're asked to find the pressure \( P_2 \) when the temperature drops to \( T_2 = -5.0^{\circ} \mathrm{C} \). The volume is constant here.

Step 2: Convert temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin by adding 273.15 to each:\[ T_1 = 25.5 + 273.15 = 298.65 \, \text{K} \]\[ T_2 = -5.0 + 273.15 = 268.15 \, \text{K} \]

Step 3: Use the Combined Gas Law

Since the volume is constant, use the equation for the relationship between pressure and temperature (Charles's Law for isochoric processes):\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]

Step 4: Solve for the desired variable

Solve for \( P_2 \) by rearranging the equation:\[ P_2 = P_1 \times \frac{T_2}{T_1} \]Substitute the known values:\[ P_2 = 360 \, \text{mm Hg} \times \frac{268.15 \, \text{K}}{298.65 \, \text{K}} \]

Step 5: Calculate the new pressure

Calculate the result using the formula:\[ P_2 = 360 \times \frac{268.15}{298.65} \approx 322.5 \, \text{mm Hg} \]

Key Concepts

Charles's Law and Isochoric Processes

Charles's Law focuses on the relationship between the volume and temperature of a gas, typically keeping the pressure constant. However, in isochoric processes, the volume remains unchanged, allowing us to explore how temperature affects pressure instead.
This concept is part of the Combined Gas Law, which shows the interconnection between pressure, volume, and temperature. In this exercise, since the flask's volume remains constant, we use Charles's Law as it applies to pressure and temperature:

• Charles's Law (Isochoric): \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), where \(P_1\) and \(T_1\) are the initial pressure and temperature, and \(P_2\) and \(T_2\) are the new pressure and temperature.
• This relationship shows that pressure increases or decreases as temperature rises or falls, respectively, when volume is constant.
• It emphasizes the direct proportion between pressure and temperature in Kelvin.

To find the new pressure at a lower temperature, we rearrange the formula to solve for \(P_2\). This allows us to predict how pressure changes when the temperature decreases under constant volume conditions.

Temperature Conversion from Celsius to Kelvin

Temperature conversion is essential when working with gas laws, as these laws use the Kelvin scale. Kelvin is an absolute temperature scale, starting at absolute zero, where all molecular motion ceases.
Celsius, on the other hand, is more common in everyday contexts but must be converted to Kelvin for scientific calculations. Here's why and how:

• Celsius to Kelvin Conversion: Add 273.15 to the Celsius temperature to convert it to Kelvin.
• Example: For our exercise, convert 25.5°C to Kelvin: \( T_1 = 25.5 + 273.15 = 298.65 \, \text{K} \).
• Similarly, convert -5.0°C to Kelvin: \( T_2 = -5.0 + 273.15 = 268.15 \, \text{K} \).

Using Kelvin provides a more accurate measure for the change in thermal energy, crucial for applying gas laws since they directly depend on absolute temperatures.

Gas Pressure Calculation Using Combined Gas Law

Calculating changes in gas pressure involves understanding how pressure, temperature, and volume interact within a gas sample. For this scenario, we utilize the Combined Gas Law, focusing on an isochoric process where volume remains constant. This means we are primarily concerned with how temperature variations affect pressure.

• Given values: Initial pressure \( P_1 = 360 \, \text{mm Hg} \), initial temperature \( T_1 = 298.65 \, \text{K} \), and new temperature \( T_2 = 268.15 \, \text{K} \).
• Using the formula: Rearrange Charles's Law for pressure, \( P_2 = P_1 \times \frac{T_2}{T_1} \).
• Substitute the values into the equation: \( P_2 = 360 \times \frac{268.15}{298.65} \).

This yields \( P_2 \approx 322.5 \, \text{mm Hg} \), showing the calculated decrease in pressure when the temperature drops from 25.5°C to -5.0°C. This result illustrates the direct relationship between temperature and pressure, holding volume constant, as outlined by the Combined Gas Law.