Question

A gas has an initial volume of \(3.08 \mathrm{~L}\) and an initial temperature of \(-73^{\circ} \mathrm{C}\). What is its new volume if its temperature is changed to \(104^{\circ} \mathrm{C}\) ? Assume pressure and amount are held constant.

Short answer

The new volume is approximately 5.80 L.

Step-by-step solution

Understand the Problem

We need to find the new volume of a gas whose temperature changes while pressure and the amount of gas remain constant. This is a classic example of Charles's Law, which states that the volume of a gas is directly proportional to its temperature in Kelvin.

Convert Temperatures to Kelvin

Initial temperature: \( T_1 = -73^{\circ} \mathrm{C} + 273.15 = 200.15 \mathrm{~K} \)Final temperature: \( T_2 = 104^{\circ} \mathrm{C} + 273.15 = 377.15 \mathrm{~K} \)

Apply Charles's Law

Charles's Law is given by the formula: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), where\( V_1 = 3.08 \mathrm{~L} \)\( T_1 = 200.15 \mathrm{~K} \)\( T_2 = 377.15 \mathrm{~K} \)We need to find \( V_2 \).

Solve for New Volume

Rearrange Charles's Law equation to solve for \( V_2 \): \[ V_2 = V_1 \times \frac{T_2}{T_1} \]Substitute the known values into equation:\[ V_2 = 3.08 \mathrm{~L} \times \frac{377.15 \mathrm{~K}}{200.15 \mathrm{~K}} \]\[ V_2 = 3.08 \mathrm{~L} \times 1.884 \approx 5.80 \mathrm{~L} \]

Conclude

The new volume of the gas when the temperature is increased to \(104^{\circ} \mathrm{C}\) is approximately \(5.80 \mathrm{~L}\).

Key concepts

Gas Laws

Gas laws describe how gases behave under different conditions of temperature, pressure, and volume. These collective scientific principles are essential for understanding how gases expand and contract.

One of the core concepts governing gases is Charles's Law. Charles's Law states that for a fixed mass of gas at constant pressure, the volume of the gas is directly proportional to its temperature (in Kelvin). This means that when the temperature of a gas increases, its volume also increases, provided the pressure remains unchanged. Similarly, if the temperature decreases, the volume decreases.

In mathematical terms, Charles's Law can be represented as:\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
where:

• \( V_1 \) is the initial volume of the gas
• \( T_1 \) is the initial temperature of the gas in Kelvin
• \( V_2 \) is the final volume of the gas
• \( T_2 \) is the final temperature of the gas in Kelvin

By using Charles's Law, we can predict how the volume of a gas changes with its temperature when the pressure is constant.

Temperature Conversion

Before applying Charles's Law to a problem, it's crucial to convert temperatures from Celsius to Kelvin. Kelvin is the SI (International System of Units) unit of temperature used in scientific calculations involving gas laws.

The conversion from Celsius to Kelvin requires adding 273.15 to the Celsius temperature. This is because absolute zero, the point at which a theoretical zero particle motion occurs, is 0 Kelvin, which equals -273.15°C.

For example, converting \(-73^{\circ} \mathrm{C}\):

• Initial Temperature: \(-73^{\circ} \mathrm{C} + 273.15 = 200.15 \mathrm{~K}\)
• Final Temperature: \(104^{\circ} \mathrm{C} + 273.15 = 377.15 \mathrm{~K}\)

These conversions ensure that calculations using Charles's Law are accurate and conform with the temperature scales appropriate for gas law equations.

Volume Calculation

Calculating the new volume of a gas when its temperature changes involves solving Charles's Law’s equation for the variable representing the unknown volume. Takes as an example a gas with an initial volume of \(3.08 \mathrm{~L}\) and initial temperature \(-73^{\circ} \mathrm{C}\) and a final temperature of \(104^{\circ} \mathrm{C}\).

First, ensure temperatures are in Kelvin:

• Initial temperature: \(200.15 \mathrm{~K}\)
• Final temperature: \(377.15 \mathrm{~K}\)

Then use Charles's Law’s formula to find the new volume \(V_2\):\[ V_2 = V_1 \times \frac{T_2}{T_1} \]
Substituting the known values gives:\[ V_2 = 3.08 \mathrm{~L} \times \frac{377.15 \mathrm{~K}}{200.15 \mathrm{~K}} \approx 5.80 \mathrm{~L} \]
Through this calculation, you can determine that the new volume of the gas after its temperature increase will be approximately \(5.80 \mathrm{~L}\), illustrating the direct relationship between temperature and volume when pressure remains constant.