Question

A sample of gas in a balloon has an initial temperature of 18 ' \(\mathrm{C}\) and a volume of \(1340 \mathrm{~L}\). If the temperature changes to 87 and there is no overall change of pressure or amount of gas, what is the new volume of the gas?

Short answer

The initial temperature and volume of the gas are given as 18°C and 1340 L, respectively. When the temperature changes to 87°C, we can use Charles/Gay-Lussac Law to find the new volume. Convert the temperatures to Kelvin: \(T_1 = 291.15 K\) and \(T_2 = 360.15 K\). Apply the law \(\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\) and plug in the values: \(\frac{1340}{291.15} = \frac{V_{2}}{360.15}\). Calculate the new volume: \(V_2 \approx 1662.67 ~L\). Thus, the new volume of the gas is approximately \(1662.67~L\).

Step-by-step solution

Write down the Charles/Gay-Lussac Law

First, let's recall the law of Charles/Gay-Lussac. This law states that the ratio of volume to temperature is constant for a gas at constant pressure and amount of gas. Mathematically, we can write this as: \[\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\]

Convert temperatures to Kelvins

We are given the temperatures in Celsius. However, the Charles/Gay-Lussac Law requires temperatures to be in Kelvins. To convert Celsius to Kelvin, simply add 273.15 to the given temperature. Initial temperature T1 in Kelvin: \(T_1 = 18 + 273.15 = 291.15~K\) Final temperature T2 in Kelvin: \(T_2 = 87 + 273.15 = 360.15~K\)

Plug in the given values and solve for the new volume

Now, we have all the initial and final values we need to solve the equation. Let's plug in the values: \[\frac{1340}{291.15} = \frac{V_{2}}{360.15}\] To find the new volume \(V_2\), simply multiply both sides of the equation by 360.15: \(V_2 = \frac{1340}{291.15} \times 360.15\)

Calculate the new volume

Now calculate the new volume: \(V_2 \approx 1662.67 ~L\) Thus, the new volume of the gas is approximately \(1662.67~L\).

Key concepts

Volume-Temperature Relationship of Gases

Understanding the volume-temperature relationship of gases is crucial, particularly when dealing with situations where the pressure and the amount of gas remain constant. This concept is grounded in Charles's Law, which asserts that the volume of a gas is directly proportional to its temperature when measured in Kelvin.
When a gas's temperature increases, its particles gain more energy and move more rapidly. This increased movement results in the particles spreading out, causing the volume of the gas to expand.
Conversely, if the temperature of the gas decreases, the particles lose energy and slow down, resulting in a decrease in volume. This behavior is mathematically represented as:

• \[\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\]
• Where \(V_1\) and \(V_2\) are the initial and final volumes, and \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin.

Grasping this relationship helps in predicting and calculating the behavior of gases under different temperature conditions.

Gas Law Calculations: Applying Charles's Law

Applying mathematical principles such as Charles's Law to practical problems involves systematic steps. First, ensure all variables (volume and temperature) fit the requirements of the law, usually involving Kelvin for temperature.
Get familiar with the equation \(\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\), as it depicts the heart of Charles's Law calculations. When tackling a problem, like finding the new volume after a temperature change, these steps apply:

• Identify initial conditions \((V_1, T_1)\) based on given data.
• Determine final temperature \((T_2)\) using conversions if necessary.
• Plug in known values into the equation and solve for the unknown variable \((V_2)\).

Understanding this process ensures accuracy and makes the problem-solving approach more efficient, as each piece of the calculation relies on a clear understanding of the relationship between the variables.

Kelvin Temperature Conversion: Crucial for Gas Laws

In scientific equations dealing with gases, like Charles's Law, temperature must be measured using the Kelvin scale. The reason is that Kelvin begins at absolute zero, making it an absolute measure of thermal energy. Celsius, on the other hand, is a relative scale, which affects its suitability for direct use in gas law equations.
To convert a temperature from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example, an initial temperature of 18°C becomes \(18 + 273.15 = 291.15 \text{ K}\).
Similarly, a Celsius temperature of 87°C converts to \(87 + 273.15 = 360.15 \text{ K}\). This step is critical because any error in conversion will throw off the entire gas law calculation. Make conversion accuracy a priority to ensure measurements adhere to the requirements of gas laws.