Question

A sample of argon gas occupies \(0.500 \mathrm{~L}\) at \(-125^{\circ} \mathrm{C}\) and \(225 \mathrm{~mm}\) Hg. Calculate the pressure in \(\mathrm{mm} \mathrm{Hg}\) if the gas occupies \(0.375 \mathrm{~L}\) at \(100^{\circ} \mathrm{C}\).

Short answer

The final pressure is approximately 755 mmHg.

Step-by-step solution

Convert Temperatures to Kelvin

First, we need to convert the given Celsius temperatures to Kelvin. The formula to convert Celsius to Kelvin is: \[ K = C + 273.15 \]For the initial temperature,\[ T_1 = -125^{\circ}C + 273.15 = 148.15 \ K \]For the final temperature,\[ T_2 = 100^{\circ}C + 273.15 = 373.15 \ K \]

Write Down the Ideal Gas Law Expression

The relationship between pressure, volume, and temperature under changing conditions is given by the combined gas law:\[ \frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2} \]Where:- \(P_1\) and \(V_1\) are the initial pressure and volume.- \(T_1\) is the initial temperature in Kelvin.- \(P_2\) and \(V_2\) are the final pressure and volume.- \(T_2\) is the final temperature in Kelvin.Plug in the known values:\[ \frac{225 \cdot 0.500}{148.15} = \frac{P_2 \cdot 0.375}{373.15} \]

Rearrange the Equation to Solve for Final Pressure

Rearrange the equation to solve for \(P_2\):\[ P_2 = \frac{225 \times 0.500 \times 373.15}{148.15 \times 0.375} \]

Substitute Values and Compute

Calculate the pressure \(P_2\):\[ P_2 = \frac{225 \times 0.500 \times 373.15}{148.15 \times 0.375} \approx \frac{41968.125}{55.55625} \approx 755.47 \ mmHg \]

Round the Answer

Round \(P_2\) to the correct number of significant figures based on the given data (which is three significant figures):\[ P_2 \approx 755 \ mmHg \]

Key concepts

Kelvin Temperature Conversion

When working with gases, it's crucial to convert temperatures from Celsius to Kelvin, as gas laws use absolute temperatures. This conversion is essential because Kelvin starts at absolute zero, the temperature where all molecular motion theoretically stops. In contrast, Celsius zero is based on the freezing point of water. The conversion formula is:

• \( K = C + 273.15 \)

For example, if you have a temperature of \(-125^{\circ}C\), converting this to Kelvin gives:

• \(-125^{\circ}C + 273.15 = 148.15 \, K \)

Similarly, a temperature of \(100^{\circ}C\) converts to:

• \(100^{\circ}C + 273.15 = 373.15 \, K \)

Using Kelvin ensures that calculations regarding gas laws are more accurate and reflect the actual behavior of gases under different conditions.

Combined Gas Law

The combined gas law is essential for understanding how gases respond to changes in pressure, volume, and temperature all at once. It is represented by the formula:

• \( \frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2} \)

This formula combines Charles’s Law, which relates volume and temperature, and Boyle’s Law, which relates pressure and volume. It shows the interconnected nature of these properties in a gas system, providing a powerful tool for solving real-world problems where multiple conditions change simultaneously.Let's consider an example where the initial conditions of a gas are detailed by a pressure \( P_1 = 225 \, mmHg \), a volume \( V_1 = 0.500 \, L \), and a temperature \( T_1 = 148.15 \, K \). Suppose we want to know the new pressure, \( P_2 \), when the gas volume changes to \( 0.375 \, L \) at a temperature \( T_2 = 373.15 \, K \). By rearranging and solving the combined gas law equation, we can accurately find \( P_2 \).

Pressure-Volume-Temperature Relationship

The relationship between pressure, volume, and temperature is a fundamental concept in thermodynamics and the study of gases. These three properties are interdependent:

• **Pressure (P)**: The force exerted by gas particles colliding with the walls of their container.
• **Volume (V)**: The space that the gas occupies.
• **Temperature (T)**: A measure of the kinetic energy of gas particles.

According to the ideal gas law, any change in one property will affect the others if not independently controlled. For instance, if you increase the temperature of a gas while keeping the volume constant, the pressure will increase due to more energetic particle collisions. Alternatively, decreasing the volume while keeping the temperature constant will also increase the pressure as particles encounter the container walls more frequently. Understanding these relationships helps predict how gases will behave under different conditions, which is crucial in fields ranging from chemistry to meteorology. In practice, using known values to find unknowns, like calculating the pressure change when volume and temperature shift, illustrates this deep connection between pressure, volume, and temperature.