Question

A balloon for long-distance flying contains \(1.2 \times 10^{7} \mathrm{L}\) of helium. If the helium pressure is \(737 \mathrm{mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C},\) what mass of helium (in grams) does the balloon contain?

Short answer

The balloon contains approximately \(1.90 \times 10^6\) grams of helium.

Step-by-step solution

Identify the known quantities

We know the volume of helium is \(1.2 \times 10^7\) L, the pressure is \(737\) mmHg, and the temperature is \(25^\circ\)C. We need to find the mass of helium.

Convert the units

Convert pressure from mmHg to atm by using the conversion: \(1 \text{ atm} = 760 \text{ mmHg}\).\[P = \frac{737}{760} \text{ atm}\approx 0.97 \text{ atm}\]Convert temperature from Celsius to Kelvin by adding 273.15: \[T = 25 + 273.15 = 298.15 \, \text{K}\]

Use the ideal gas law to find moles of helium

The ideal gas law is given by \(PV = nRT\). Rearrange it to solve for \(n\), the number of moles.\[n = \frac{PV}{RT}\]Where \(R = 0.0821 \, \text{L} \, \text{atm} / \text{mol} \cdot \text{K}\). Substitute the known values:\[n = \frac{(0.97 \text{ atm})(1.2 \times 10^7 \text{ L})}{(0.0821 \text{ L atm/mol K})(298.15 \text{ K})}\]Calculate \(n\).

Calculate the mass of helium

After calculating the number of moles \(n\), we find the mass using the molar mass of helium, \(4.00 \, \text{g/mol}\).\[\text{mass} = n \times 4.00 \, \text{g/mol}\]

Perform the calculations to find the final mass

Substitute \(n\) from Step 3 into the equation from Step 4 to obtain the mass:1. Calculate \(n\): \[n \approx \frac{(0.97)(1.2 \times 10^7)}{(0.0821)(298.15)} \approx 4.746 \times 10^5 \, \text{mol} \]2. Calculate the mass: \[\text{mass} = 4.746 \times 10^5 \times 4.00 \, \text{g/mol} \approx 1.90 \times 10^6 \, \text{g} \]

Verify results

Ensure the calculated mass, \(1.90 \times 10^6\) g, is suitable considering the units used and matches typical calculations for this kind of problem.

Key concepts

Helium Properties

Helium is one of the most fascinating elements in the periodic table. It is a noble gas, which means it is extremely stable and non-reactive. This property makes helium an ideal choice for use in balloons, as it won't undergo any chemical changes that could affect its behavior.
Helium is the second lightest element, having an atomic number of 2 and an atomic mass of around 4 atomic mass units (amu). Because of its lightness, it provides excellent buoyancy, which is why it is used in airships and balloons.
Another significant property of helium is that it is colorless, odorless, and tasteless. This makes it safe for various applications, especially in environments where coloration or odors might be problematic.

• Non-reactivity: Remains stable.
• Light: Provides buoyancy.
• Colorless and odorless: Safe and non-intrusive.

Pressure Conversion

Converting pressure units is a crucial step in solving problems involving gases. In this exercise, the original pressure given is in millimeters of mercury (mmHg). This is a common metric used in many scenarios, like weather forecasting or medical measurements.
However, to use the ideal gas law, pressures typically need to be in atmospheres (atm). Fortunately, there is a straightforward conversion between these units: 1 atm = 760 mmHg. This means you can convert mmHg to atm by dividing the mmHg value by 760. For example, a pressure of 737 mmHg would be subsequently converted as follows: \[ P = \frac{737}{760} \text{ atm} \approx 0.97 \text{ atm} \]
Using correct units ensures that calculations in formulas like the Ideal Gas Law are accurate and that the result is meaningful.

• Use conversion factor: 1 atm = 760 mmHg.
• Ensures consistency in equation.

Molar Mass Calculation

The molar mass of an element or compound is essential for converting moles into mass, which is particularly useful in chemistry. For helium, the molar mass is approximately 4 grams per mole (g/mol). This is because helium's atomic structure comprises 2 protons and most commonly 2 neutrons, leading to a mass number of about 4.
Calculating molar mass allows for easy determination of the quantity of a substance in grams when the amount in moles is known. In the given exercise, having found the number of moles of helium using the ideal gas law, the next step was to multiply by the molar mass to find the mass.
For helium: \[\text{Mass} = n \times 4.00 \text{ g/mol} \] where \(n\) represents the number of moles of helium determined from earlier calculations.

• 2 protons and 2 neutrons: Atomic mass of ~4.
• Conversion factor from moles to grams.
• Necessary for final mass calculation.

Volume Measurement

Measuring the volume of gases like helium is a critical part of many scientific and industrial processes. Gases tend to fill their containers, and their volume can change dramatically with pressure and temperature changes.
In gas-related calculations, volume is typically given in liters (L) due to its compatibility with other metric units commonly used in chemistry equations, like the ideal gas law. For the balloon example in our problem, the volume was specified as \(1.2 \times 10^{7} \text{ L}\). This is quite a large volume, illustrating helium's application in lifting heavy loads or extensive balloons.
Understanding how volume works in tandem with pressure and temperature is crucial for predicting how gases will behave under different conditions.

• Gaseous adaptation: Fills container shape.
• Standard unit for gas volume: Liters (L).
• Part of the Ideal Gas Law: PV = nRT.