Question

A sample of xenon gas occupies \(500.0 \mathrm{~mL}\) at \(-185^{\circ} \mathrm{C}\) and \(75.0 \mathrm{~cm} \mathrm{Hg} .\) Calculate the Celsius temperature if the volume is \(225.0 \mathrm{~mL}\) at \(55.0 \mathrm{~cm} \mathrm{Hg}\).

Short answer

The Celsius temperature is approximately \(-214.85^{\circ}\text{C}\).

Step-by-step solution

Identify the Variables

First, let's identify the variables in the problem. You are given an initial volume \(V_1 = 500.0\, \text{mL}\), an initial temperature \(T_1 = -185^{\circ}\text{C}\), an initial pressure \(P_1 = 75.0\, \text{cm Hg}\). You're also given a new volume \(V_2 = 225.0\, \text{mL}\) and a new pressure \(P_2 = 55.0\, \text{cm Hg}\). The task is to find the new temperature \(T_2\) when these conditions are met.

Convert Celsius to Kelvin

Before using the gas laws, convert temperatures from Celsius to Kelvin as Kelvin is the standard unit for temperature in these calculations. Convert \(T_1 = -185^{\circ}\text{C}\) to Kelvin: \(T_1 = -185 + 273.15 = 88.15\, \text{K}\).

Use the Combined Gas Law

Use the combined gas law formula: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\). Substitute the known values: \(\frac{75.0 \times 500.0}{88.15} = \frac{55.0 \times 225.0}{T_2}\).

Solve for the Unknown Temperature

Rearrange the equation to solve for \(T_2\): \(T_2 = \frac{55.0 \times 225.0 \times 88.15}{75.0 \times 500.0}\). Calculate \(T_2\): \(T_2 \approx 58.3\, \text{K}\).

Convert Back to Celsius

Convert \(T_2\) from Kelvin to Celsius: \(T_2^{\circ}\text{C} = 58.3 - 273.15 = -214.85^{\circ}\text{C}\).

Key concepts

Combined Gas Law

The combined gas law is a useful equation that helps us understand the relationship between pressure, volume, and temperature of a gas. This equation is particularly handy when you need to solve problems where these three variables change under different conditions. The combined gas law is represented by the equation:\[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\]In this equation:

• \(P_1\) and \(P_2\) are the initial and final pressures, respectively.
• \(V_1\) and \(V_2\) are the initial and final volumes, respectively.
• \(T_1\) and \(T_2\) are the initial and final temperatures, respectively, but must be in Kelvin.

This means if you know five of the variables, you can find the sixth. It combines Boyle's Law (pressure and volume relationship) and Charles's Law (volume and temperature relationship), providing a comprehensive tool for gas calculations. When using the combined gas law, always remember to keep the units consistent, especially temperature in Kelvin.

Temperature Conversion

Changing temperature units is crucial in gas law problems. In the realm of gas laws, we use Kelvin, not Celsius, as the standard unit. This is because Kelvin measures absolute temperature, starting from absolute zero where particles stop moving.To convert from Celsius to Kelvin, use the simple formula:\[T(\text{K}) = T(\text{C}) + 273.15\]For example, if a temperature is \(-185^{\circ}\text{C}\), converting it to Kelvin is calculated as:\[-185 + 273.15 = 88.15 \, \text{K}\]Similarly, converting back from Kelvin to Celsius involves the reverse:\[T(\text{C}) = T(\text{K}) - 273.15\]For instance, if you have 58.3 K, it converts to Celsius as:\[58.3 - 273.15 = -214.85^{\circ}\text{C}\]Always convert temperatures to Kelvin when using the gas laws to ensure accurate calculations!

Pressure and Volume Relationship

Pressure and volume have an inverse relationship as described by Boyle's Law within the context of gas behavior. This principle means that as pressure increases, volume decreases if the temperature is held constant. Conversely, if volume increases, pressure decreases as long as the temperature does not change.In the combined gas law formula, this relationship is shown when other variables, like temperature, stay constant or when working through steps to isolate changes:\[P_1 V_1 = P_2 V_2\]In this equation:

• If \(P_1 > P_2\), then \(V_1 < V_2\), indicating pressure decreased while volume increased.
• If \(V_1 > V_2\), then \(P_1 < P_2\), showing volume decreased while pressure increased.

Keep in mind this relationship is only accurate under constant temperature conditions without changing the number of gas particles. Understanding this principle helps when solving problems involving changes to gas conditions.