Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

At \(20^{\circ} \mathrm{C}\) (approximately room temperature) the average velocity of \(\mathrm{N}_{2}\) molecules in air is \(1050 \mathrm{mph}\). (a) What is the average speed in \(\mathrm{m} / \mathrm{s}\) ? (b) What is the kinetic energy (in J) of an \(\mathrm{N}_{2}\) molecule moving at this speed? (c) What is the total kinetic energy of \(1 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) molecules moving at this speed?

Short Answer

Expert verified
(a) Average speed in m/s = \(1050 \times 0.44704 \approx 469.4 \,\text{m/s}\) (b) Kinetic energy of a single N2 molecule = \(\frac{1}{2}\times m \times (1050 \times 0.44704)^2\approx 4.71 \times 10^{-21} \mathrm{J}\) (c) Total kinetic energy of 1 mol of N2 molecules = \(KE \times NA \approx 2.83 \times 10^3 \,\text{J}\)

Step by step solution

01

Convert mph to m/s

To convert the given speed to m/s, we use the conversion factor: 1 mph = 0.44704 m/s So, average speed in m/s = \(1050 \times 0.44704\)
02

Find the kinetic energy of a single N2 molecule

The kinetic energy of a single molecule can be calculated using the formula: \(KE = \frac{1}{2}mv^2\) Where m is the mass of the molecule and v is its average velocity. But first, we need to find the mass of a single N2 molecule. The molecular weight of N2 is approximately 28 g/mol. To find the mass of a single molecule, we will use Avogadro's number, NA = \(6.022\times10^{23} \mathrm{mol}^{-1}\). We also need to convert the molecular mass to kg. Mass of a single N2 molecule in kg: \(m = \frac{28 \times 10^{-3} \, \text{kg/mol}}{6.022\times10^{23}\, \text{mol}^{-1}}\) Now we can find the kinetic energy: \(KE = \frac{1}{2}\times m \times \left(1050 \times 0.44704\right)^2\)
03

Find the total kinetic energy of 1 mol of N2 molecules

To find the total kinetic energy for 1 mol of N2 molecules, multiply the kinetic energy of a single molecule (found in step 2) by Avogadro's number: Total kinetic energy = \(KE \times NA\) Combining all the steps in the solution, we get: (a) Average speed in m/s = \(1050 \times 0.44704\) (b) Kinetic energy of a single N2 molecule = \( \frac{1}{2}\times m \times (1050 \times 0.44704)^2\) (c) Total kinetic energy of 1 mol of N2 molecules = \(KE \times NA\)

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 Vaia!

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

(a) When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises \(1.642^{\circ} \mathrm{C}\). When a 0.265-g sample of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{~N}_{4}\), is burned, the temperature rises \(1.525^{\circ} \mathrm{C}\). Using the value \(26.38 \mathrm{~kJ} / \mathrm{g}\) for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume. (b) Assuming that there is an uncertainty of \(0.002^{\circ} \mathrm{C}\) in each temperature reading and that the masses of samples are measured to \(0.001 \mathrm{~g}\), what is the estimated uncertainty in the value calculated for the heat of combustion per mole of caffeine?

(a) Under what condition will the enthalpy change of a process equal the amount of heat transferred into or out of the system? (b) During a constant- pressure process the system absorbs heat from the surroundings. Does the enthalpy of the system increase or decrease during the process?

The air bags that provide protection in autos in the event of an accident expand because of a rapid chemical reaction. From the viewpoint of the chemical reactants as the system, what do you expect for the signs of \(q\) and \(w\) in this process?

For the following processes, calculate the change in internal energy of the system and determine whether the process is endothermic or exothermic: (a) A balloon is heated by adding 850 J of heat. It expands, doing \(382 \mathrm{~J}\) of work on the atmosphere. (b) A \(50-g\) sample of water is cooled from \(30^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\), thereby losing approximately \(3140 \mathrm{~J}\) of heat. (c) A chemical reaction releases \(6.47 \mathrm{~kJ}\) of heat and does no work on the surroundings.

Calculate \(\Delta E\), and determine whether the process is endothermic or exothermic for the following cases: (a) A system absorbs \(105 \mathrm{~kJ}\) of heat from its surroundings while doing \(29 \mathrm{~kJ}\) of work on the surroundings; (b) \(q=1.50 \mathrm{~kJ}\) and \(w=-657 \mathrm{~J} ;\) (c) the system releases \(57.5 \mathrm{~kJ}\) of heat while doing \(22.5 \mathrm{~kJ}\) of work on the surroundings.

See all solutions

Recommended explanations on Chemistry 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.

Sign-up for free