Problem 71 - University Physics with Modern Physics

Question

You blow up a spherical balloon to a diameter of 50.0 \(\mathrm{cm}\) until the absolute pressure inside is 1.25 atm and the temperature is \(22.0^{\circ} \mathrm{C}\) . Assume that all the gas in \(\mathrm{N}_{2}\) is of molar mass 28.0 \(\mathrm{g} / \mathrm{mol}\) . (a) Find the mass of a single \(\mathrm{N}_{2}\) molecule. (b) How much translational kinetic energy does an average \(\mathrm{N}_{2}\) molecule have? (c) How many \(\mathrm{N}_{2}\) molecules are in this balloon? (d) What is the total translational kinetic energy of all the inolecules in the balloon?

Step-by-Step Solution

Answer

(a) Mass of one \( \mathrm{N}_{2} \) molecule is approximately \( 4.65 \times 10^{-26} \) kg. (b) Average translational kinetic energy is about \( 6.09 \times 10^{-21} \) J. (c) There are roughly \( 2.03 \times 10^{24} \) molecules in the balloon. (d) Total translational kinetic energy is approximately \( 1.24 \times 10^{4} \) J.

Step 1: Convert Units and Find Volume

First, convert the diameter of the balloon to meters to find the radius, which is 0.5 / 2 = 0.25 meters. The volume of a sphere is given by \( V = \frac{4}{3} \pi r^3 \). Using the radius calculated, the volume is \( V = \frac{4}{3} \pi (0.25)^3 \approx 0.06545 \) cubic meters.

Step 2: Convert Pressure to Pascals

Convert the given pressure from atm to pascals using the conversion 1 atm = 101325 Pa. Thus, \( 1.25 \text{ atm} = 1.25 \times 101325 \text{ Pa} = 126656.25 \text{ Pa} \).

Step 3: Calculate Number of Molecules (n)

Use the Ideal Gas Law \( PV = nRT \) to find the number of moles, \( n \). First, convert the temperature to Kelvin by adding 273.15 to 22.0°C, which gives 295.15 K. Using \( R = 8.314 \text{ J/mol·K} \), solve for \( n \): \ \( n = \frac{PV}{RT} = \frac{126656.25 \times 0.06545}{8.314 \times 295.15} \approx 3.373 \text{ moles} \).

Step 4: Calculate Mass of a Single Molecule

The molar mass of \( \mathrm{N}_{2} \) is 28 g/mol, which is equivalent to 0.028 kg/mol. The mass of a single molecule is found by \( \frac{0.028 \text{ kg/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 4.65 \times 10^{-26} \text{ kg} \).

Step 5: Calculate Average Translational Kinetic Energy

The average translational kinetic energy of a molecule in a gas is given by \( \frac{3}{2}kT \), where \( k = 1.38 \times 10^{-23} \text{ J/K} \) is Boltzmann's constant. \ \( \frac{3}{2} \times 1.38 \times 10^{-23} \times 295.15 \approx 6.09 \times 10^{-21} \text{ J} \).

Step 6: Number of Molecules in Balloon

Using the number of moles calculated, \( n = 3.373 \), find the number of molecules using Avogadro's number: \ \( \text{number of molecules} = 3.373 \times 6.022 \times 10^{23} \approx 2.03 \times 10^{24} \text{ molecules} \).

Step 7: Total Translational Kinetic Energy

The total translational kinetic energy is the average kinetic energy per molecule times the number of molecules: \ \( \text{Total energy} = 2.03 \times 10^{24} \times 6.09 \times 10^{-21} \approx 1.24 \times 10^{4} \text{ J} \).

Key Concepts

Translational Kinetic Energy

The translational kinetic energy of a gas molecule is a fundamental concept crucial for understanding the movement of particles within a gas. It refers to the energy associated with the motion of an object in space. In the context of gas molecules, translational kinetic energy specifically relates to the energy due to their motion along the three spatial dimensions. This type of energy is essentially the energy of movement that molecules possess due to their constant, random motion.To compute the average translational kinetic energy of a molecule in a gas, we use the formula:\[ KE_{avg} = \frac{3}{2} k T \]Here, \( k \) is the Boltzmann constant, and \( T \) is the temperature in Kelvin. Notice that this direct relationship with temperature indicates how crucial temperature is in determining the kinetic energy of molecules. At higher temperatures, molecules generally have more kinetic energy, moving faster and contributing to the overall pressure exerted by the gas.

Molar Mass

Molar mass is a measure of the mass of a given substance (usually in grams) per amount of substance (typically per mole). It serves as a bridge connecting macroscopic physical quantities with their atomic-scale counterparts.For nitrogen gas, \( \mathrm{N}_{2} \), the molar mass is given as 28 g/mol. This means that one mole of nitrogen molecules weighs 28 grams. This conversion is vital when calculating the mass of a single molecule or a given number of moles of gas. By knowing the molar mass, you can also find the mass of individual molecules. Using Avogadro's number, we derive the mass of a single molecule by dividing the molar mass by Avogadro's number:\[ \text{mass of a single molecule} = \frac{\text{molar mass}}{6.022 \times 10^{23}} \]

Volume of a Sphere

The volume of a sphere provides a relationship between the radius of a balloon and the space it occupies. When dealing with spherical objects, such as balloons, understanding how to calculate their volume is fundamental.The formula for the volume \( V \) of a sphere is:\[ V = \frac{4}{3} \pi r^3 \]Where \( r \) is the radius of the sphere. In this exercise, converting the diameter of the balloon to its radius involves dividing by two and then using this value in the volume calculation. This concept ties in with the ideal gas law, as knowing the volume helps determine the number of moles in a given space.

Boltzmann's Constant

Boltzmann's constant \( k \) is a physical constant that relates the average kinetic energy of particles in a gas with the temperature of the gas. Named after the Austrian physicist Ludwig Boltzmann, it plays a pivotal role in statistical mechanics.The value of Boltzmann's constant \( k \) is approximately \( 1.38 \times 10^{-23} \text{ J/K} \). It provides a way to calculate how energy at the molecular level depends on temperature, offering insight into molecular speeds and how they vary with temperature changes.In the formula for translational kinetic energy, \( KE_{avg} = \frac{3}{2} k T \), Boltzmann's constant directly connects microscopic thermal energy with macroscopic temperature. This allows scientists and engineers to anticipate the behavior of gases under different thermal conditions.