Question

A sample of gas occupies \(754 \mathrm{~mL}\) at \(22^{\circ} \mathrm{C}\) and a pressure of \(165 \mathrm{mmHg}\). Calculate its volume if the temperature is raised to \(42^{\circ} \mathrm{C}\) and the pressure is raised to \(265 \mathrm{mmHg}\). (The amount of gas does not change.)

Short answer

The final volume is approximately 501 mL.

Step-by-step solution

Convert Temperature to Kelvin

Convert the initial and final temperatures from Celsius to Kelvin using the formula: \( T(K) = T(\degree C) + 273.15 \). The initial temperature is \( T_1 = 22 + 273.15 = 295.15 \ K \). The final temperature is \( T_2 = 42 + 273.15 = 315.15 \ K \).

Write Down the Initial Conditions

Note the initial volume, temperature, and pressure: \( V_1 = 754 \ mL \), \( T_1 = 295.15 \ K \), and \( P_1 = 165 \ mmHg \).

Write Down the Final Conditions

Note the final temperature and pressure: \( T_2 = 315.15 \ K \) and \( P_2 = 265 \ mmHg \). We need to find the final volume \( V_2 \).

Apply the Combined Gas Law

The Combined Gas Law is \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \). Use it to find \( V_2 \) by rearranging to get \( V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \).

Substitute Values into the Equation

Substitute the known values into the equation: \( V_2 = \frac{165 \ mmHg \times 754 \ mL \times 315.15 \ K}{265 \ mmHg \times 295.15 \ K} \).

Calculate the Final Volume

Perform the arithmetic: \( V_2 = \frac{39248185.5}{78294.55} = 501.11 \ mL \). Round to three significant figures to get \( V_2 \approx 501 \ mL \).

Key concepts

Gas Laws

The Gas Laws are fundamental principles that help us understand the behavior of gases. They describe how gases behave under different conditions, specifically how pressure, volume, and temperature are interrelated. In this exercise, we focus on the Combined Gas Law, which brings together Boyle's Law, Charles's Law, and Gay-Lussac's Law. The Combined Gas Law can be expressed as \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \), where:

• \( P \) represents pressure.
• \( V \) represents volume.
• \( T \) represents temperature in Kelvin.

In this formula, we assume the amount of gas remains constant. By using the Combined Gas Law, we can predict how one of these variables will change if the others are altered, as shown in our exercise. The main point is that as temperature, pressure, or volume change, at least one of the other factors will adjust in response to maintain equilibrium.

Temperature Conversion

In the context of gases, temperature is a crucial factor because it affects both volume and pressure. When working with gas laws, it is vital to convert temperatures from Celsius to Kelvin. This is because gas laws require an absolute temperature scale, where Kelvin serves as the default. This is due to the Kelvin scale starting at absolute zero, an important reference point in thermodynamics.To convert Celsius to Kelvin, use the simple formula: \[ T(K) = T(\degree C) + 273.15 \].For instance, in our exercise:

• The initial temperature of \(22^{\circ}C\) becomes \(295.15\ K\).
• The final temperature of \(42^{\circ}C\) converts to \(315.15\ K\).

By converting temperatures into Kelvin, we ensure that our calculations align properly with the units required by the gas laws.

Pressure Measurement

Measuring pressure accurately is key when dealing with gases. Pressure is defined as the force exerted by gas particles on the walls of a container. In this exercise, pressure is given in \( mmHg \), a common unit used alongside others like atmospheres (atm) and Pascals (Pa).To use the Combined Gas Law properly, ensure the pressures are in the same unit for both initial and final states. In this case:

• The initial pressure \( P_1 \) is \( 165 \ mmHg \).
• The final pressure \( P_2 \) is \( 265 \ mmHg \).

Consistency in units allows us to plug the values directly into the Combined Gas Law equation without additional conversion, making calculations straightforward. Understanding the units and methods for measuring pressure ensures the accuracy and reliability of the results when applying the gas laws.