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 \(360 \mathrm{mm}\) 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°C is approximately 323.5 mm Hg.

Step 1: Identify the Known Values

First, we identify the initial conditions of the gas. We have an initial pressure ( P_1 ) of 360 mm Hg and an initial temperature ( T_1 ) of 25.5°C. To use the formula, temperature should be converted to Kelvin: T_1 = 25.5 + 273.15 = 298.65 K .

Step 2: Convert Final Temperature to Kelvin

Next, convert the final temperature to Kelvin. The final temperature ( T_2 ) is -5.0°C, which in Kelvin is T_2 = -5.0 + 273.15 = 268.15 K .

Step 3: Use Gay-Lussac's Law

We apply Gay-Lussac's Law, which relates pressure and temperature for a gas at constant volume: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). Substitute the known values into this formula to solve for P_2, the pressure at the lower temperature.

Step 4: Solve for the Final Pressure

Rearrange the formula to solve for P_2: \( P_2 = \frac{P_1 \cdot T_2}{T_1} \). Substitute P_1 = 360 \text{ mm Hg}, T_2 = 268.15 \text{ K}\, and T_1 = 298.65 \text{ K}: \( P_2 = \frac{360 \cdot 268.15}{298.65} \approx 323.5 \text{ mm Hg} \).

Step 5: State the Final Answer

The final pressure of the gas at -5.0°C is approximately 323.5 mm Hg .

Key Concepts

pressure and temperature relationship

Gay-Lussac's Law helps us understand how pressure and temperature are related in a gas. It states that if the volume of a gas is constant, the pressure is directly proportional to its absolute temperature. This means, when the temperature increases, the pressure also increases, provided the gas is not allowed to expand. Conversely, if the temperature decreases, the pressure also decreases.

• This relationship is mathematically expressed by the formula: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). Here, \( P_1 \) and \( P_2 \) represent the initial and final pressures, while \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.
• It is important that only absolute temperatures (Kelvin) are used in the equations to ensure the proportionate relationship holds true.
• With this law, changes in temperature can predictably affect the pressure within the same volume of gas.

conversion to Kelvin

Temperature conversion is crucial for applying gas laws correctly. In Gay-Lussac's Law, temperature must be in Kelvin, the SI unit of absolute temperature.

• Why Kelvin? The Kelvin scale starts at absolute zero, the point where all molecular movement stops. This makes it ideal for scientific calculations where proportional relationships require a zero baseline.
• How to convert: To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example, \( 25.5^{\circ} \mathrm{C} \) converts to \( 298.65 \mathrm{K} \), and \(-5.0^{\circ} \mathrm{C} \) converts to \( 268.15 \mathrm{K} \).
• This conversion is essential for ensuring calculations in gas laws are accurate and meaningful.

gas pressure calculation

Calculating gas pressure changes involves using the relationship expressed in Gay-Lussac's Law. By maintaining a constant volume, predicting how a change in temperature affects pressure becomes straightforward.

• Using the formula \( P_2 = \frac{P_1 \cdot T_2}{T_1} \) allows us to find the unknown final pressure when we know the initial pressure \( P_1 \) and both temperatures \( T_1 \) and \( T_2 \) in Kelvin.
• In the problem provided, the initial pressure \( P_1 \) was 360 mm Hg, with temperatures \( T_1 = 298.65 \mathrm{K} \) and \( T_2 = 268.15 \mathrm{K} \). Substituting these values into the equation gives a final pressure \( P_2 \) of approximately 323.5 mm Hg.
• This calculation confirms our understanding that a decrease in temperature leads to a decrease in pressure when the volume is constant.