Problem 55 - Chemistry: An Atoms-Focused Approach

Question

Balloons for a New Year's Eve party in Potsdam, New York, are filled to a volume of \(5.0 \mathrm{L}\) at a temperature of \(20^{\circ} \mathrm{C}\) and then hung outside where the temperature is \(-25^{\circ} \mathrm{C} .\) What is the volume of the balloons after they have cooled to the outside temperature? Assume that atmospheric pressure inside and outside the house is the same.

Step-by-Step Solution

Answer

Answer: The final volume of the balloon after cooling to -25°C is approximately 4.23L.

Step 1: Convert temperatures to Kelvin

First, we need to convert the initial and final temperatures from Celsius to Kelvin. Remember that \(K = °C + 273.15\). So, \(T_1 = 20°C + 273.15 = 293.15K\) \(T_2 = -25°C + 273.15 = 248.15K\)

Step 2: Write down the given values

Now we write down the given values, and what we want to solve for: \(V_1 = 5.0L\) \(T_1 = 293.15K\) \(T_2 = 248.15K\) \(V_2 = ?\)

Step 3: Use Charles' Law to find the final volume

We will use Charles' Law, which states that \(V_1/T_1 = V_2/T_2\). We're solving for \(V_2\), so we'll rearrange the equation to isolate \(V_2\): \(V_2 = \frac{V_1 * T_2}{T_1}\)

Step 4: Plug in the values and calculate the final volume

Now we just plug in the values for \(V_1\), \(T_1\), and \(T_2\): \(V_2 = \frac{5.0L * 248.15K}{293.15K}\) \(V_2 ≈ 4.23L\) The volume of the balloons after they have cooled to the outside temperature is approximately \(4.23L\).

Key Concepts

Temperature Conversion

When dealing with gas laws such as Charles' Law, it's crucial to convert temperatures into the Kelvin scale. This is because Kelvin is the absolute temperature scale used in thermodynamics. The Kelvin scale starts at absolute zero, making it ideal for scientific calculations.
To convert temperatures from Celsius to Kelvin, you add 273.15 to the Celsius temperature. For example, if you have an initial temperature of 20°C, converting it would result in:

• Initial Temperature: \( 20^{\circ}C + 273.15 = 293.15 K \).
• Final Temperature: \( -25^{\circ}C + 273.15 = 248.15 K \).

Why is this important? Using Kelvin ensures that all temperature values are positive and proportional to the energy content of a system. This way, you can apply formulas like Charles' Law accurately.

Volume Calculation

Volume calculation is a key factor in understanding how gases behave under different conditions. According to Charles' Law, if the pressure is constant, the volume of a gas is directly proportional to its temperature in Kelvin. This relationship allows us to foresee how the volume will change as the temperature changes.
To find the new volume (V_2), when the temperature changes, Charles' Law provides the equation: \[ V_1/T_1 = V_2/T_2 \]. By rearranging this equation, you find:

• \( V_2 = \frac{V_1 \times T_2}{T_1} \)
• Substitute the known values into this equation: \( V_2 = \frac{5.0L \times 248.15K}{293.15K} \)
• Solve to find: \( V_2 \approx 4.23L \).

This calculation shows that when temperature decreases, and pressure is constant, the volume will also decrease. So, when the balloon cools from 20°C to -25°C, the volume reduces as well.

Ideal Gas Law

The ideal gas law is a fundamental principle in chemistry and physics. It's expressed as \( PV = nRT \), where:

• P is pressure,
• V is volume,
• n is the number of moles of gas,
• R is the ideal gas constant,
• T is the temperature in Kelvin.

This law encompasses Boyle's, Charles', and Avogadro's laws, thereby combining the relationships between pressure, volume, and temperature.
While the exercise primarily uses Charles' Law (a component of the ideal gas law), understanding that it fits into the ideal gas law can provide a more comprehensive understanding of gas behavior. For the balloons, since the number of moles and pressure remain constant, Charles’ Law (\( V_1/T_1 = V_2/T_2 \)) effectively describes the situation.