Question

A balloon is filled to a volume of \(7.00 \times 10^{2} \mathrm{mL}\) at a temperature of \(20.0^{\circ} \mathrm{C}\) . The balloon is then cooled at constant pressure to a temperature of \(1.00 \times 10^{2} \mathrm{K} .\) What is the final volume of the balloon?

Short answer

The final volume of the balloon after being cooled to \(1.00 \times 10^2 \mathrm{K}\) is approximately \(238.59 \mathrm{mL}\).

Step-by-step solution

Convert temperatures to Kelvin

Since we are dealing with temperatures in the context of a gas law, we need to work with temperatures in Kelvin. The initial temperature is given in Celsius, so we need to convert it to Kelvin using the formula: \(T(K) = T(^{\circ}C) + 273.15\) Now, let's convert the given initial temperature in Celsius to Kelvin: \(T_1(K) = 20.0^{\circ} C + 273.15 = 293.15 K\) The final temperature is already given in Kelvin as \(T_2 = 1.00 \times 10^{2} K\).

Use Charles's Law formula to find the final volume

Charles's Law states that \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), where \(V_1\) is the initial volume, \(T_1\) is the initial temperature, \(V_2\) is the final volume, and \(T_2\) is the final temperature. We have all the necessary values, so let's plug them into the formula: \(\frac{7.00 \times 10^2 \mathrm{mL}}{293.15 \mathrm{K}} = \frac{V_2}{1.00 \times 10^2 \mathrm{K}}\)

Solve for the final volume

To solve for the final volume, \(V_2\), we need to isolate it in the equation by multiplying both sides by the final temperature (\(1.00 \times 10^2 \mathrm{K}\)): \(V_2 = \frac{7.00 \times 10^2 \mathrm{mL} \times 1.00 \times 10^2 \mathrm{K}}{293.15 \mathrm{K}}\) Now calculate the value of \(V_2\): \(V_2 = \frac{7.00 \times 10^2 \mathrm{mL} \times 1.00 \times 10^2 \mathrm{K}}{293.15 \mathrm{K}} \approx 238.59 \mathrm{mL}\) The final volume of the balloon after being cooled to \(1.00 \times 10^2 \mathrm{K}\) is approximately \(238.59 \mathrm{mL}\).

Key concepts

Gas Laws

Gas laws describe how gases behave in different conditions. They focus on the relationship between variables like volume, temperature, and pressure. One of the most common gas laws is Charles's Law. This law explains how gases expand or contract when the temperature changes, while the pressure remains constant.

Charles's Law states:

• The volume of a gas is directly proportional to its temperature when measured in Kelvin.
• As the temperature increases, the volume of the gas will also increase if the pressure is unchanged.
• Mathematically, Charles's Law is expressed as \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \).

This means that if you know the initial temperature and volume of a gas, you can calculate its final volume by applying this formula. It is essential to always work with temperatures in Kelvin when using any gas laws, as Kelvin is the absolute temperature scale used in scientific calculations.

Temperature Conversion

Temperature conversion is crucial when working with gas laws. In the sciences, temperatures must be in Kelvin, not Celsius or Fahrenheit. This is because Kelvin is an absolute temperature scale with a true zero point. It allows for direct proportional calculations without negative numbers.

To convert from Celsius to Kelvin, use the formula:

• \( T(K) = T(^{\circ}C) + 273.15 \)

As shown in the exercise, the temperature of \( 20.0^{\circ}C \) was converted to \( 293.15\, K \). This step ensures calculations are accurate. The final temperature was already in Kelvin, so no further conversion was needed. Always check the units before solving any gas law problems to avoid mistakes.

Volume Calculation

Volume calculation using Charles's Law involves determining how much space a gas occupies after a temperature change, keeping pressure constant. Once temperatures are converted to Kelvin, use Charles's Law formula for calculations:

The initial data given:

• Initial Volume \( V_1 = 700 \mathrm{mL} \)
• Initial Temperature \( T_1 = 293.15 \mathrm{K} \)
• Final Temperature \( T_2 = 100 \mathrm{K} \)

The goal is to find the final volume \( V_2 \). Rearrange the formula to solve for \( V_2 \):

• \( V_2 = \frac{V_1 \times T_2}{T_1} \)
• \( V_2 = \frac{700 \mathrm{mL} \times 100 \mathrm{K}}{293.15 \mathrm{K}} \approx 238.59 \mathrm{mL} \)

This calculation shows that the volume decreases when the temperature is lowered. The process is straightforward once you ensure temperature units are correct and apply Charles's Law accurately.