Question

A sample of gas occupies \(135 \mathrm{mL}\) at \(22.5^{\circ} \mathrm{C} ;\) the pressure is \(165 \mathrm{mm}\) Hg. What is the pressure of the gas sample when it is placed in a 252 -mL. flask at a temperature of \(0.0^{\circ} \mathrm{C} ?\)

Short answer

The final pressure of the gas is approximately 81.37 mm Hg.

Step-by-step solution

Understand the Problem

We need to find the new pressure of the gas when it is transferred to a different container at a different temperature. Using the combined gas law, given initial and final volumes, temperatures and pressures, we need to find the final pressure.

Identify the Given Variables

Let's denote \( P_1 = 165 \text{ mm Hg} \), \( V_1 = 135 \text{ mL} \), \( T_1 = 22.5^{\circ} C \), \( V_2 = 252 \text{ mL} \), and \( T_2 = 0.0^{\circ} C \). Convert temperatures from Celsius to Kelvin for the gas law: \( T_1 = 22.5 + 273.15 = 295.65 \text{ K} \) and \( T_2 = 0 + 273.15 = 273.15 \text{ K} \).

Use the Combined Gas Law

The combined gas law relates two states of a gas sample and is given by \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \). We need to rearrange this formula to isolate \( P_2 \), the final pressure.

Rearrange the Formula

Rearranging the equation for \( P_2 \), we get: \( P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} \).

Substitute the Known Values

Substitute the known values into the equation: \( P_2 = 165 \times \frac{135}{252} \times \frac{273.15}{295.65} \).

Calculate the Final Pressure

First calculate component values: \( \frac{135}{252} \approx 0.5357 \) and \( \frac{273.15}{295.65} \approx 0.923 \). Then multiply everything: \( P_2 = 165 \times 0.5357 \times 0.923 \approx 81.37 \text{ mm Hg} \).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a cornerstone in understanding the behavior of gases. It is expressed as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the amount of substance (in moles), \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
Understanding this law helps to comprehend how changes in temperature, volume, and pressure influence a gas. It assumes an ideal scenario where gas molecules do not interact and occupy no space. Real gases can deviate from these assumptions under high pressure and low temperature.
The Ideal Gas Law is versatile and can be used in different forms to solve problems involving one or more gas state variables, like our case with the combined gas law, which simplifies scenarios involving two states.

Temperature Conversion

Temperature conversion, particularly from Celsius to Kelvin, is crucial when working with gas laws. The Kelvin scale is used because it starts at absolute zero, the theoretically lowest possible temperature, which is essential for accurate gas law calculations.
To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example:

• \( T_1 = 22.5^{\circ} C \rightarrow T_1 = 22.5 + 273.15 = 295.65 \, \text{K} \)
• \( T_2 = 0.0^{\circ} C \rightarrow T_2 = 0 + 273.15 = 273.15 \, \text{K} \)

Using Kelvin ensures mathematical consistency in gas law equations, as it prevents negative temperature values, which are not physically meaningful.

Pressure Calculation

Pressure calculation in the context of the combined gas law involves understanding how the pressure of a gas changes with volume and temperature changes.
The Combined Gas Law is expressed as \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \), where each variable represents the initial and final states of a gas sample.

• \( P_1 \) = initial pressure
• \( V_1 \) = initial volume
• \( T_1 \) = initial temperature (in Kelvin)
• \( P_2 \) = final pressure
• \( V_2 \) = final volume
• \( T_2 \) = final temperature (in Kelvin)

To find \( P_2 \), rearrange the formula: \( P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} \).
This approach was used to find the new pressure of the gas as 81.37 mm Hg, demonstrating the relationship between volume, temperature, and pressure in a gas sample.

Volume Change in Gases

Volume changes significantly affect gas pressure and temperature due to gas expansion or compression.
When a gas expands into a larger volume, its pressure decreases if the temperature is constant. Conversely, compressing a gas will increase its pressure. This principle is evident in the combined gas law, used in this problem to understand how shifting the gas to a larger flask affected its pressure.
The initial volume was 135 mL, and changed to 252 mL. The resulting pressure drop is calculated using the relationship in the combined gas law. Despite these volume changes, the gas stays in adherence to the ideal gas laws, showcasing their application in predicting the behavior of gases under various conditions.