Question

A weather balloon has a volume of \(750 \mathrm{~L}\) when filled with helium at \(8{ }^{\circ} \mathrm{C}\) at a pressure of 380 torr. What is the new volume of the balloon, where the pressure is \(0.20 \mathrm{~atm}\) and the temperature is \(-45^{\circ} \mathrm{C}\) ?

Short answer

The new volume of the balloon is approximately 1524.64 L.

Step-by-step solution

Understanding the Problem

The problem involves an initial and a final state of a weather balloon filled with helium. The task is to find the new volume given changes in temperature and pressure.

Collect and Convert the Given Data

Initial volume, \( V_1 = 750 \, \mathrm{L} \), initial temperature, \( T_1 = 8^{\bullet} \mathrm{C} \). Convert \( T_1 \) to Kelvin: \( T_1 = 8 + 273.15 = 281.15 \mathrm{K} \). Initial pressure, \( P_1 = 380 \mathrm{torr} \). Convert \( P_1 \) to atm: \( P_1 = 380 / 760 = 0.5 \mathrm{atm} \). Final temperature, \( T_2 = -45^{\bullet} \mathrm{C} \). Convert \( T_2 \) to Kelvin: \( T_2 = -45 + 273.15 = 228.15 \mathrm{K} \). Final pressure, \( P_2 = 0.20 \mathrm{atm} \).

Use the Combined Gas Law

The combined gas law formula is: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]. Substitute the known values: \[ \frac{0.5 \, \mathrm{atm} \cdot 750 \, \mathrm{L}}{281.15 \, \mathrm{K}} = \frac{0.20 \, \mathrm{atm} \cdot V_2}{228.15 \, \mathrm{K}} \].

Solve for the Final Volume

Rearrange the equation to solve for \( V_2 \): \[ V_2 = \frac{0.5 \cdot 750 \cdot 228.15}{0.20 \cdot 281.15} \]. Simplify and calculate \( V_2 \): \[ V_2 \approx 1524.64 \, \mathrm{L} \].

Key concepts

initial state

When dealing with problems involving the combined gas law, it's crucial to understand the initial state of the gas system. In this example, the initial state of the weather balloon is provided by three key parameters:
Initial Volume: The volume of the balloon is given as 750 liters (L).
Initial Temperature: The temperature is given as 8°C. However, for calculations, temperatures must be converted to Kelvin by adding 273.15. So, 8°C converts to 281.15 K.
Initial Pressure: The pressure is initially provided as 380 torr. To use this value in the combined gas law formula, we convert torr to atm. Since 1 atm = 760 torr, we have 380 torr = 0.5 atm.
These initial parameters set the stage for understanding how changes in temperature and pressure will affect the volume of the balloon.

final state

Now, let's look at the final state of the balloon, which involves changes in temperature and pressure. These changes impact the new volume we need to determine.
Final Temperature: The temperature is given as -45°C. Again, we convert this to Kelvin by adding 273.15. Thus, -45°C becomes 228.15 K.
Final Pressure: The pressure is given as 0.20 atm.
Since we have both the initial and final states, the final volume can be determined using the combined gas law, taking into account these final conditions.

volume change

Understanding how volume changes with variations in temperature and pressure is essential for solving gas law problems. To determine the new volume, we use the combined gas law:
\ \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \).
Plugging in the values from the initial and final states provides the relationship between the changes.
Rearranging to solve for the final volume \(V_2\), we get:
\[ V_2 = \frac{P_1 \cdot V_1 \cdot T_2}{P_2 \cdot T_1} \]
By inserting the known quantities, we proceed with:
\( \frac{0.5 \cdot 750 \cdot 228.15}{0.20 \cdot 281.15} \).
Simplifying, we find that the new volume \(V_2\) is approximately 1524.64 liters (L). This indicates a significant increase in the balloon's volume due to the decrease in both temperature and pressure.

temperature conversion

One crucial step in solving gas law problems is converting temperatures to the correct unit, Kelvin. The gas law constants require temperature in Kelvin because the Kelvin scale starts at absolute zero, making it useful for direct proportional relationships in gas laws.
To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature.
For instance, the initial temperature of 8°C is converted as follows:

• \(8 + 273.15 = 281.15 K\)

Similarly, the final temperature of -45°C is converted:

• \(-45 + 273.15 = 228.15 K\)

Always ensure temperatures are accurately converted to avoid errors in calculations.

pressure conversion

Properly converting pressure units is also critical when applying the combined gas law. In this problem, pressures are given in torr and atm.
Understanding that 1 atm equals 760 torr is key.
For the initial pressure of 380 torr, the conversion is:
\(380 \div 760 = 0.5 atm\).
No conversion is necessary for the final pressure of 0.20 atm as it is already in the correct unit.
Accurate pressure conversion ensures that the units remain consistent throughout the calculations, allowing for proper application of the gas law formula and ensuring credible results.