Problem 11 - Chemistry and Chemical Reactivity

Question

You have a sample of gas in a flask with a volume of 250 mL. At \(25.5^{\circ} \mathrm{C},\) the pressure of the gas is \(360 \mathrm{mm} \mathrm{Hg} .\) 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 pressure at \(-5.0^{\circ}C\) is approximately 323.43 mm Hg.

Step 1: Identify the variables and constants

Let's identify the given quantities and what we need to find. We have an initial temperature \(T_1 = 25.5^{\circ}C\), initial pressure \(P_1 = 360 \, \text{mm Hg}\), and constant volume. We need to find the new pressure \(P_2\) at the lower temperature \(T_2 = -5.0^{\circ}C\).

Step 2: Convert temperatures from Celsius to Kelvin

To use the gas laws, convert the temperatures from Celsius to Kelvin by adding 273.15. \(T_1 = 25.5 + 273.15 = 298.65 \, \text{K}\) and \(T_2 = -5.0 + 273.15 = 268.15 \, \text{K}\).

Step 3: Apply the ideal gas law relation

Since the volume is constant, we can use the pressure and temperature relationship of gases: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). We need to find \(P_2\), so rearrange the formula to get \( P_2 = P_1 \times \frac{T_2}{T_1} \).

Step 4: Calculate the final pressure

Substitute the known values into the rearranged formula:\[ P_2 = 360 \, \text{mm Hg} \times \frac{268.15 \, \text{K}}{298.65 \, \text{K}} \approx 323.43 \, \text{mm Hg}.\]

Step 5: State the final answer

The pressure of the gas at \(-5.0^{\circ}C\) is approximately 323.43 mm Hg.

Key Concepts

Pressure-Temperature Relationship

The relationship between pressure and temperature in gases is beautifully illustrated by a principle known as Charles's Law. This law is part of the ideal gas law, which helps us predict how gases will behave under various conditions. At a constant volume, the pressure of a given amount of gas is directly proportional to its temperature when measured in Kelvin. This means that if the temperature of a gas increases, its pressure increases as well, provided the volume doesn't change. Conversely, when the temperature decreases, the pressure also decreases. This is why when we cool a gas, like in our exercise where the temperature drops from \(25.5^{\circ}C\) to \(-5.0^{\circ}C\), we expect the pressure to fall too. The mathematical expression for this relationship at constant volume is \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), where \(P\) refers to pressure and \(T\) is temperature in Kelvin. This formula allows us to calculate the new pressure once we know the initial conditions and how much the temperature has changed.

Kelvin Temperature Conversion

To use the gas laws effectively, we must always work with temperature in Kelvin. The Kelvin scale is an absolute temperature scale that is essential for gas law calculations because it starts at absolute zero, where theoretically molecular motion ceases. Temperature conversions from Celsius to Kelvin are straightforward: simply add 273.15 to the Celsius temperature. This conversion is crucial because the direct proportionality between pressure and temperature in the gas laws only holds when temperatures are measured in the absolute Kelvin scale. For example, in the exercise provided, the conversion was done as follows: \( T_1 = 25.5 + 273.15 = 298.65 \, \text{K} \) and \( T_2 = -5.0 + 273.15 = 268.15 \, \text{K} \). Proper conversion helps ensure accurate calculations and understanding of thermodynamic processes.

Constant Volume Gas Process

In a constant volume gas process, the volume of the gas does not change despite variations in other parameters like temperature and pressure. This scenario is often encountered in rigid containers like a sealed flask. Keeping the volume constant simplifies our calculations since we can directly relate changes in pressure to changes in temperature using the formula \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). This relationship derives from the ideal gas law by holding volume (\(V\)) constant. When dealing with a constant volume process, as in our exercise, the absence of volume changes means we don't have to worry about how volume might complicate the pressure-temperature calculations. Instead, we focus on how these two properties interact, allowing us to predict how one will respond to shifts in the other when all else is held steady. This type of analysis is very useful for understanding behavior of gases in closed systems.