Problem 166 - General Chemistry

Question

A weather balloon filled with helium has a diameter of \(3.50 \mathrm{ft}\). What is the mass in grams of the helium in the balloon at \(21^{\circ} \mathrm{C}\) and normal pressure? The density of helium under these conditions is \(0.166 \mathrm{~g} / \mathrm{L}\).

Step-by-Step Solution

Answer

The mass of helium in the balloon is approximately 105.53 grams.

Step 1: Calculate Balloon's Volume

The first step is to find the volume of the balloon. Since the balloon is spherical, we can use the formula for the volume of a sphere, which is \(V = \frac{4}{3} \pi r^3\). The radius \(r\) of the balloon is half of its diameter, so \(r = \frac{3.50}{2} = 1.75 \text{ ft}\). Now, calculate the volume: \[ V = \frac{4}{3} \pi (1.75)^3 \approx 22.45 \text{ cubic feet} \]

Step 2: Convert Volume to Liters

We need the volume in liters to use with the given density of helium. There are approximately 28.3168 liters in a cubic foot. So, convert the volume: \[ V = 22.45 \times 28.3168 \approx 635.73 \text{ liters} \]

Step 3: Calculate Mass of Helium

Now, use the density of helium to find the mass. Density is mass per unit volume, so we can use \(\text{mass} = \text{density} \times \text{volume}\). The density of helium is \(0.166 \text{ g/L}\), and we calculated the volume as approximately \(635.73\) liters: \[\text{mass} = 0.166 \times 635.73 \approx 105.53 \text{ grams}\]

Step 4: Final Check

Ensure that all calculations have been performed accurately and account for unit conversions. Double check that the values used match those provided in the problem description.

Key Concepts

Understanding Density

Density is a key property of matter that tells us how much mass is contained in a given volume. It's calculated as the mass divided by the volume, and its formula is \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). This concept helps in understanding how heavy or light a substance is relative to its size. For gases like helium, density is commonly expressed in grams per liter (g/L). Knowing the density is crucial because it allows us to relate the measurable volume of a gas to its mass, which is often the required calculation in real-world problems.
In this exercise, the density given for helium is \(0.166\,\text{g/L}\), an important piece of information that lets us complete our calculation of the mass once the volume is known.

Volume Conversion Basics

Volume conversion is an important step when dealing with measurements in different units. A common challenge in problems like these is the need to convert from one unit of volume to another. In our case, the balloon's volume was initially calculated in cubic feet.
However, to find the mass using density in g/L, we needed the volume in liters. There are approximately 28.3168 liters in a cubic foot. Therefore, we multiplied the volume in cubic feet by this conversion factor to obtain the volume in liters. Doing such conversions accurately ensures that subsequent calculations, like finding mass, are correct.

Calculating the Volume of a Sphere

Calculating the volume of a sphere is essential in many real-world applications, such as finding the capacity of a spherical object like a weather balloon. The formula for the volume of a sphere is given by: \[ V = \frac{4}{3} \pi r^3 \]
Here, \(r\) is the radius of the sphere. In the exercise, we determined the radius by halving the diameter of the balloon.

• Diameter was 3.50 feet.
• Thus, radius \( r = \frac{3.50}{2} = 1.75\) feet.

Then, substituting this radius into the formula, we can find the volume in cubic feet. Converting this spherical volume into liters allows for an application in further density-based calculations which relate to mass.

Calculating Mass from Volume and Density

Once we have the volume of the balloon in the correct units, calculating the mass becomes straightforward using the density formula. For gases in particular, finding the mass involves multiplying the volume by the density.
With the provided density of helium being 0.166 g/L and the converted volume of the balloon being approximately 635.73 liters, the calculation proceeds as follows:

• \(\text{Mass} = 0.166 \times 635.73 = 105.53 \text{ grams}\).

Such calculations help determine how much of a gas like helium is contained within a given volume, which is practical for a range of scientific and industrial applications.