Problem 145 - Chemistry An Atoms First Approach

Question

A certain flexible weather balloon contains helium gas at a volume of \(855\) \(\mathrm{L}\). Initially, the balloon is at sea level where the temperature is \(25^{\circ} \mathrm{C}\) and the barometric pressure is \(730\) torr. The balloon then rises to an altitude of \(6000 \mathrm{ft}\), where the pressure is \(605\) torr and the temperature is \(15^{\circ} \mathrm{C}\). What is the change in volume of the balloon as it ascends from sea level to \(6000 \mathrm{ft} ?\)

Step-by-Step Solution

Answer

The change in volume of the weather balloon as it ascends from sea level to 6000 ft is approximately \(112.74 \mathrm{L}\).

Step 1: Convert temperatures to Kelvin

To work with the ideal gas law, we need the temperatures in Kelvin. Convert the given temperatures from Celsius to Kelvin using the following formula: \[T(K) = T(^\circ\mathrm{C}) + 273.15\]

Step 2: Convert pressure units

Pressure units should be consistent. In this case, both initial and final pressures are given in torr. It is not necessary to convert them to other units.

Step 3: Apply the ideal gas law for the initial state

At sea level (initial state) where the temperature is \(25^\circ\mathrm{C}\) and the pressure is 730 torr, apply the ideal gas law to find the value of the constant k for the helium gas. \[P_1V_1 = k(T_1)\]

Step 4: Apply the ideal gas law for the final state

At an altitude of 6000 ft (final state) where the temperature is \(15^\circ\mathrm{C}\) and the pressure is 605 torr, apply the ideal gas law again to find the value of the constant k for the helium gas. \[P_2V_2 = k(T_2)\]

Step 5: Find the final volume

Since the values of the constant k found in both equations (initial and final states) are equal, we can equate the right sides of equations from Step 3 and Step 4, and solve for the final volume, \(V_2\).

Step 6: Find the change in volume

The change in volume of the balloon can be found by subtracting the initial volume from the final volume. This will give us the overall change in volume as the balloon rises from sea level to 6000 ft.

Key Concepts

Temperature Conversion to Kelvin

When you're working with the ideal gas law, it's crucial to use temperatures in Kelvin. This is because the Kelvin scale starts at absolute zero, which makes calculations accurate in science.

• To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.
• For example, convert 25°C to Kelvin: 25°C + 273.15 = 298.15 K.
• Similarly, for 15°C: 15°C + 273.15 = 288.15 K.

Using Kelvin ensures that our calculations using the ideal gas law are consistent and correct. Always remember that this conversion step is critical before applying any atmospheric gas calculations.

Pressure Units

Pressure is a key component in the ideal gas law. It's important to ensure the pressure units are consistent throughout your calculations.

• In the given problem, pressures are already in torr, so there is no conversion needed between different units like atm or Pa.
• Having consistent units helps avoid errors when calculating changes in gas volume or when solving for other variables.

It's a good practice to double-check units, especially if they start out in different forms, to maintain clarity and accuracy as you work through each step of a problem.

Volume Change Calculation

To calculate how the volume of the gas changes as the balloon rises, we apply the ideal gas law, expressed as \(PV = nRT\). Here, we focus on relationships between pressure, volume, and temperature. Initially at sea level:

• Use \(P_1V_1 = kT_1\), where \(P_1 = 730\ torr\), \(V_1 = 855\ L\), and \(T_1 = 298.15\ K\).

At 6000 ft:

• Use \(P_2V_2 = kT_2\), where \(P_2 = 605\ torr\), and \(T_2 = 288.15\ K\).

Since the constant \(k\) remains the same, we equate the two expressions and solve for the new volume \(V_2\):\[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\]Rearrange to solve for \(V_2\):\[V_2 = \frac{P_1V_1T_2}{P_2T_1}\]Then, find the change in volume by subtracting \(V_1\) from \(V_2\). This gives us the overall change as the balloon ascends.