Problem 18 - Chemistry and Chemical Reactivity

Question

A balloon holds \(30.0 \mathrm{kg}\) of helium. What is the volume of the balloon if its pressure is 1.20 atm and the temperature is \(22^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The volume of the balloon is approximately 151,338.21 liters.

Step 1: Convert the temperature to Kelvin

To use the Ideal Gas Law, the temperature must be in Kelvin. The formula to convert from Celsius to Kelvin is: \[ T(K) = T(°C) + 273.15 \] Plugging in the given temperature:\[ T(K) = 22 + 273.15 = 295.15 \, K \]

Step 2: Use the Ideal Gas Law equation

The Ideal Gas Law is expressed as \( PV = nRT \). We are looking for \( V \), the volume. Rearranging the equation gives:\[ V = \frac{{nRT}}{{P}} \]Where:- \( P \) is the pressure (1.20 atm)- \( R \) is the ideal gas constant (0.0821 L·atm/mol·K)- \( n \) is the number of moles of heliumFirst, we need to determine \( n \).

Step 3: Calculate the number of moles (n)

Determine the number of moles \( n \) using the molar mass of helium which is about 4.00 g/mol. First, we convert the mass of helium from kilograms to grams:\[ 30.0 \,\text{kg} = 30000 \, \text{g} \]Then, divide by the molar mass of helium:\[ n = \frac{{30000}}{4.00} = 7500 \, \text{mol} \]

Step 4: Calculate the volume of the balloon

Substitute the known values into the Ideal Gas Law equation:\[ V = \frac{{nRT}}{P} = \frac{{7500 \, \text{mol} \times 0.0821 \, \text{L·atm/mol·K} \times 295.15 \, K}}{1.20 \, \text{atm}} \]\[ V = \frac{{7500 imes 0.0821 imes 295.15}}{1.20} \approx 151338.21 \, \text{L} \]Thus, the volume of the balloon is approximately 151,338.21 liters.

Key Concepts

Mole Calculation

To understand gases like helium in a quantitative way, we use a concept called 'moles'. A mole is a unit that counts particles and it makes chemistry calculations much easier. Helium, a noble gas, is characterized by its light atomic weight. The molar mass of helium is about 4.00 grams per mole, which tells us that every mole of helium weighs 4 grams.

In this example, we need to convert 30.0 kilograms of helium into moles. Since kilograms are a large unit for gases, we switch to grams, which are more manageable for chemical calculations. We do this by remembering that 1 kilogram equals 1000 grams. So, 30 kilograms becomes 30,000 grams.

Next, we figure out how many moles this is by dividing the mass in grams by the molar mass. For helium:

• Convert mass: 30.0 kilograms to 30,000 grams
• Calculate moles: \( n = \frac{30000 \, \text{g}}{4.00 \, \text{g/mol}} = 7500 \, \text{mol} \)

This shows there are 7500 moles of helium in the balloon. Understanding this conversion helps us apply further calculations to predict the behavior of gases.

Temperature Conversion to Kelvin

When working with gases using the Ideal Gas Law, it's important to convert temperatures to Kelvin, the absolute temperature scale. Kelvin simplifies calculations and is the standard unit since it starts at absolute zero, the point at which particles have minimal thermal motion.

To convert from Celsius to Kelvin, you use the formula:

• \( T(K) = T(°C) + 273.15 \)

Take the temperature in Celsius, which is given as 22 degrees, and simply add 273.15 to convert it to Kelvin:

• \( T(K) = 22 + 273.15 = 295.15 \, K \)

This calculation adjusts the temperature to an absolute scale suitable for use in equations such as the Ideal Gas Law.

Pressure and Volume Relationship

The behavior of gases is heavily influenced by how pressure and volume interact. According to the Ideal Gas Law, these two properties maintain a specific relationship: if one changes, the other must also change if the temperature and number of moles remain constant.

The Ideal Gas Law is expressed as:

• \( PV = nRT \)

Where:

• \( P \) represents pressure
• \( V \) represents volume
• \( n \) is the number of moles
• \( R \) is the gas constant
• \( T \) is the temperature in Kelvin

By rearranging this equation to solve for volume \( V \), we see how volume varies directly with both the number of moles and temperature, and inversely with pressure:

• \( V = \frac{nRT}{P} \)

In essence, if you increase the pressure on a gas, its volume will decrease if other factors are held constant, and vice versa. This principle is a key component of many practical applications in physics and engineering.

Gas Constant

The gas constant, \( R \), is a key factor in the Ideal Gas Law, which provides a link between the various state properties of a gas. For the calculations in the Ideal Gas Law, the value of \( R \) must be consistent with the units used for pressure, volume, and temperature.

In our problem, we're using:

• Pressure in atm
• Volume in liters
• Temperature in Kelvin

Therefore, the gas constant \( R \) used is 0.0821 L·atm/mol·K. This particular value of \( R \) ensures consistency with the unit of measurement and simplifies calculations.

• \( R = 0.0821 \, \text{L·atm/mol·K} \)

Understanding \( R \) allows us to relate the quantities in the Ideal Gas Law and accurately predict the behavior of a gas under different conditions.