/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 \(\bullet$$\bullet\) You blow up... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\bullet$$\bullet\) 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 is \(\mathrm{N}_{2},\) 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 molecules in the balloon?

Short Answer

Expert verified
Mass of N2 molecule: \(4.65 \times 10^{-26}\) kg. Avg KE of N2: \(6.08 \times 10^{-21}\) J; N2 molecules: \(2.12 \times 10^{24}\); Total KE: \(1.28 \times 10^{4}\) J.

Step by step solution

01

Find the Mass of a Single N2 Molecule

To find the mass of a single nitrogen molecule (N2), first use the molar mass of N2. The molar mass of N2 is 28.0 g/mol, which means 1 mole of N2 weighs 28.0 grams. Since 1 mole contains Avogadro's number of molecules, which is approximately \(6.022 \times 10^{23}\), the mass of a single N2 molecule can be calculated using the formula:\[m_{\text{molecule}} = \frac{28.0 \text{ g/mol}}{6.022 \times 10^{23} \text{ molecules/mol}}\approx 4.65 \times 10^{-23} \text{ g} \]Convert the mass to kilograms:\[m_{\text{molecule}} = 4.65 \times 10^{-26} \text{ kg}\]
02

Calculate the Translational Kinetic Energy of an Average N2 Molecule

The average translational kinetic energy of a molecule can be calculated using the formula related to its temperature:\[\text{KE}_{\text{avg}} = \frac{3}{2}kT\]where \(k\) is the Boltzmann constant \( (1.38 \times 10^{-23} \text{ J/K}) \) and \(T\) is the absolute temperature in Kelvin. Convert \(22.0^{\circ} \text{C}\) to Kelvin:\[T = 22.0 + 273.15 = 295.15 \text{ K}\]Substitute \(T\) into the kinetic energy formula:\[\text{KE}_{\text{avg}} = \frac{3}{2}(1.38 \times 10^{-23})(295.15) \approx 6.08 \times 10^{-21} \text{ J}\]
03

Determine the Number of N2 Molecules in the Balloon

To find the number of N2 molecules, use the ideal gas law in the form \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is moles, \(R\) is the gas constant (8.314 J/mol·K), and \(T\) is temperature in Kelvin. First, calculate the volume of the balloon. The diameter is given as 50.0 cm, so:\[r = \frac{50.0}{2} = 25.0 \text{ cm} = 0.25 \text{ m}\]\[V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.25)^3 \approx 0.065 \text{ m}^3\]Convert the pressure to pascals:\[1.25 \text{ atm} = 1.25 \times 101325 \text{ Pa} = 1.26656 \times 10^5 \text{ Pa}\]Use \(PV = nRT\):\[1.26656 \times 10^5 \times 0.065 = n \times 8.314 \times 295.15\]Solving for \(n\):\[n = \frac{1.26656 \times 10^5 \times 0.065}{8.314 \times 295.15} \approx 3.52 \text{ mol}\]Convert moles to molecules:\[N = n \times 6.022 \times 10^{23} \approx 2.12 \times 10^{24} \text{ molecules}\]
04

Total Translational Kinetic Energy of All N2 Molecules

The total translational kinetic energy of all the N2 molecules in the balloon can be found by multiplying the number of molecules \(N\) by the average kinetic energy of each molecule \(\text{KE}_{\text{avg}}\) calculated in Step 2:\[\text{KE}_{\text{total}} = N \times \text{KE}_{\text{avg}} = 2.12 \times 10^{24} \times 6.08 \times 10^{-21} \approx 1.28 \times 10^{4} \text{ J}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
The concept of kinetic energy, which represents energy due to motion, plays a crucial role in the behavior of gases, especially at the molecular level. For a nitrogen molecule (N_2) inside a balloon, we are particularly interested in its translational kinetic energy. This form of kinetic energy is associated with the linear motion of molecules. The average translational kinetic energy of a molecule at a given temperature can be found using the formula \(\text{KE}_{\text{avg}} = \frac{3}{2}kT\), where \(k\) denotes the Boltzmann constant (1.38 \times 10^{-23} J/K) and \(T\) is the temperature in Kelvin. For example, when the temperature is 22^{\circ} \text{C}, converting to Kelvin gives 295.15 K. By substituting this into the energy formula, we get each nitrogen molecule's average translational kinetic energy as \(6.08 \times 10^{-21}\text{ J}\). This small value reflects the high-speed movement of very tiny molecules within the balloon, showing how temperature influences molecular motion.
Molar Mass
Molar mass is an essential property in chemistry that helps us to relate the mass of a substance to the amount of matter it contains, using moles as a measuring unit. For gases like nitrogen (N_2), molar mass is the mass of one mole of its molecules. In the exercise, N_2 has a molar mass of 28.0 \, g/mol. From this, the mass of a single N_2 molecule can be determined using Avogadro's number (6.022 \times 10^{23}\text{ molecules/mol}), since it tells us the number of molecules in one mole. The calculation follows as \(\frac{28.0 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{molecules/mol}}\), resulting in approximately \(4.65 \times 10^{-23} \, \text{g}\). Converting to kilograms, it becomes \(4.65 \times 10^{-26} \, \text{kg}\), which is quite minuscule, reflecting the small but significant mass of individual molecules.
Avogadro's Number
Understanding Avogadro's number is crucial when dealing with quantities of matter at the molecular scale. Named after the scientist Amedeo Avogadro, this constant (6.022 \times 10^{23} \text{ molecules/mol}) represents the number of particles in exactly one mole of a substance. It enables chemists to convert between the amount of substance (in moles) and the number of units each contains, such as atoms or molecules.In our exercise, after determining the number of moles (3.52) of nitrogen gas using the ideal gas law, Avogadro's number allows us to calculate the number of N_2 molecules inside the balloon. Multiplying the moles by Avogadro's number gives approximately \(2.12 \times 10^{24} \text{ molecules}\). This conversion demonstrates Avogadro's number's essential role in bridging the macroscopic and molecular worlds.
Translational Motion
Translational motion pertains to the movement of molecules through space, contributing significantly to the macroscopic properties of gases, such as pressure and temperature. For molecules inside a balloon, such as nitrogen (N_2) molecules, they constantly move in collaboration, colliding with each other and the balloon's walls. This motion leads to the phenomenon we recognize as gas pressure.In this exercise, understanding the translational motion of each nitrogen molecule helped to determine the average and total translational kinetic energy. Total energy can be found by recognizing that each molecule carries a kinetic energy component. By multiplying the number of molecules (2.12 \times 10^{24}) by the average translational kinetic energy (6.08 \times 10^{-21} \text{ J}), it results in a total of \(1.28 \times 10^{4} \text{ J}\). This showcases the energy held by all molecules in the balloon, juxtaposing their minute individual energies with the total effect in aggregate.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\bullet$$\bullet\) A player bounces a basketball on the floor, compressing it to 80.0\(\%\) of its original volume. The air (assume it is essentially \(\mathrm{N}_{2}\) gas) inside the ball is originally at a temperature of \(20.2^{\circ} \mathrm{C}\) and a pressure of 2.00 atm. The ball's diameter is 23.9 \(\mathrm{cm} .\) (a) What temperature does the air in the ball reach at its maximum compression? (b) By how much does the internal energy of the air change between the ball's original state and its maximum compression?

\(\bullet\) In the air we breathe at \(72^{\circ} \mathrm{F}\) and 1.0 atm pressure, how many molecules does a typical cubic centimeter contain, assuming that the air is all \(\mathrm{N}_{2} ?\)

\(\bullet\) An ideal gas at 4.00 atm and 350 \(\mathrm{K}\) is permitted to expand adiabatically to 1.50 times its initial volume. Find the final pressure and temperature if the gas is (a) monatomic with \(C_{p} / C_{V}=\frac{5}{3},\) (b) diatomic with \(C_{p} / C_{V}=\frac{7}{5} .\)

\(\bullet$$\bullet\) Starting with 2.50 \(\mathrm{mol}\) of \(\mathrm{N}_{2}\) gas (assumed to be ideal) in a cylinder at 1.00 atm and \(20.0^{\circ} \mathrm{C},\) a chemist first heats the gas at constant volume, adding \(1.52 \times 10^{4} \mathrm{J}\) of heat, then continues heating and allows the gas to expand at constant pressure to twice its original volume. (a) Calculate the final temperature of the gas. (b) Calculate the amount of work done by the gas. (c) Calculate the amount of heat added to the gas while it was expanding. (d) Calculate the change in internal energy of the gas for the whole process.

\(\bullet$$\bullet$$\bullet\) A hot-air balloon stays aloft because hot air at atmospheric pressure is less dense than cooler air at the same pressure. (The calculation of the buoyant force is discussed in Chapter \(13 .\) ) If the volume of the balloon is 500.0 \(\mathrm{m}^{3}\) and the surrounding air is at \(15.0^{\circ} \mathrm{C},\) what must the temperature of the air in the balloon be for it to lift a total load of 290 \(\mathrm{kg}\) (in addition to the mass of the hot air)? The density of air at \(15.0^{\circ} \mathrm{C}\) and atmospheric pressure is 1.23 \(\mathrm{kg} / \mathrm{m}^{3}.\) BIO Temperature and degrees of freedom. The internal energy of an ideal monatomic gas is simply the kinetic energy associated with the translational motion of its atoms as they move randomly in each of the three independent spatial dimensions. However, for a diatomic ideal gas we must also take into account the kinetic and potential energies associated with molecular vibration, and the kinetic energy associated with molecular rotation. Roughly speaking, each independent way that energy can be stored is known as a degree of freedom. Although translational motion can occur at any temperature, rotational and vibrational motions typically cannot occur at lower temperatures- thus, the number of available degrees of freedom can change as the temperature changes. For example, an ideal monatomic gas has three degrees of freedom (one for each of its independent directions of translational motion) at all temperatures. In contrast, diatomic hydrogen \(\left(\mathrm{H}_{2}\right)\) has five degrees of freedom near room temperature \((3\) translational and 2 rotational). However, at higher temperatures, where molecular vibrations can occur, diatomic hydrogen has seven degrees of freedom \((3\) translational, 2 rotational, and 2 vibrational). The equipartition theorem states that equilibrium each degree of freedom contributes \(\frac{1}{2} n R T\) to the internal energy of the gas. For example, the internal energy of a monatomic gas, which has 3 degrees of freedom, would be \(\frac{3}{2} n R T\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.