At 20 °C (approximately room temperature) the average velocity of N 2 molecules in air is 1050 mph. (b) What is the kinetic energy (in J) of an N 2 molecule moving at this speed?

Welcome back everyone in this example, we need to calculate the kinetic energy of an oxygen molecule at 30 degrees Celsius. So the key word here is molecule for our oxygen species. And we're going to recall our kinetic energy formula for one mole of a gas is going to be equal to three halves, multiplied by the moles of our gas times r gas constant R and then times our temperature in kelvin. Where again, n would be the moles of our gas molecule. Now we're going to interpret this in terms of molecules of our gas. So in this case we would say that kinetic energy of our oxygen should be equal to three halves times R gas constant, R times r temperature in kelvin divided by avocados number. So that's representatives N. A. So we would recall that avocados number tells us that we have 6.022 times 10 to the 23rd power molecules. And sorry, that's molecules per mole of our gas. And we should recall that according to kinetic theory, molecules will undergo elastic collisions, just like gasses are stated to in our kinetic theory. And so that is according to our kinetic theory for ideal gasses. So now that we have our formula laid out, we're gonna plug in what we know but we also want to make sure we convert our temperature to kelvin. So we're gonna add to 73.152 R 30 degrees Celsius. And that's going to give us our kelvin temperature equal 2 to 73.15 kelvin. So now let's go go ahead and plug everything in. So we would say that our kinetic energy of our oxygen molecule is equal to three halves multiplied by r gas constant. R which we should recall is equal to a value of 8.314 with units of jewels divided by moles times kelvin. This is then multiplied by our temperature in kelvin which we converted 2 to 73. kelvin. And then in our denominator we have our avocados constant which we understand is equal to six point oh 22 times 10 to the 23rd power. And this unit of molecules which we want to divide by moles because we want moles to cancel out with moles in our numerator. So we'll get rid of moles here as well as here will also be able to cancel out our units of kelvin and we're left with units of joules per molecule as our final units. And so for our final answer, we understand the kinetic energy of an oxygen molecule is going to equal a value of 6.278 times 10 to the 21st power jewels per molecule. And so what's highlighted in yellow here is our final answer to complete this example, I hope that everything I reviewed was clear if you have any questions, leave them down below and I'll see everyone in the next practice video

Ch.5 - Thermochemistry

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Brown 14th Edition

Ch.5 - Thermochemistry

Problem 118b

Chapter 5, Problem 118b

At 20 °C (approximately room temperature) the average velocity of N₂ molecules in air is 1050 mph. (b) What is the kinetic energy (in J) of an N₂ molecule moving at this speed?

Verified step by step guidance

Convert the velocity of the N2 molecule from miles per hour (mph) to meters per second (m/s) using the conversion factor: 1 mph = 0.44704 m/s.

Calculate the mass of an N2 molecule. The molecular weight of N2 is approximately 28 g/mol. Convert this mass to kilograms (kg) by using the conversion factor: 1 g = 0.001 kg. Then, use Avogadro's number (6.022 \times 10^{23} molecules/mol) to find the mass of one molecule.

Use the kinetic energy formula: KE = \frac{1}{2}mv^2, where 'm' is the mass of the molecule in kg and 'v' is the velocity in m/s.

Plug the values of 'm' and 'v' into the kinetic energy formula to calculate the kinetic energy of the N2 molecule.

Ensure the final kinetic energy value is in Joules (J), as this is the standard unit for energy in the International System of Units (SI).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion, calculated using the formula KE = 0.5 * m * v², where m is the mass of the object and v is its velocity. For gas molecules like N₂, this concept is crucial for understanding how speed translates into energy at the molecular level.

Molecular Mass

The molecular mass of a substance is the mass of a single molecule, typically expressed in atomic mass units (amu). For nitrogen (N₂), the molecular mass is approximately 28 amu, which is essential for calculating the kinetic energy of an individual N₂ molecule when given its velocity.

Conversion of Units

In physics and chemistry, it is often necessary to convert units to ensure consistency in calculations. In this case, converting the speed from miles per hour (mph) to meters per second (m/s) is essential for using standard SI units in the kinetic energy formula, allowing for accurate energy calculations.

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Conversion Factors

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Textbook Question

A 201-lb man decides to add to his exercise routine by walking up three flights of stairs (45 ft) 20 times per day. Hefigures that theworkrequired to increasehis potential energy in this way will permit him to eat an extra order of French fries, at 245 Cal, without adding to his weight. Is he correct in this assumption?

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Textbook Question

Sucrose (C12H22O11) is produced by plants as follows: 12 CO2(g) + 11 H2O(l) → C12H22O11 + 12 O2(g) H = 5645 kJ About 4.8 g of sucrose is produced per day per square meter of the earth's surface. The energy for this endothermic reaction is supplied by the sunlight. About 0.1 % of the sunlight that reaches the earth is used to produce sucrose. Calculate the total energy the sun supplies for each square meter of surface area. Give your answer in kilowatts per square meter 1kW/m2 where 1W = 1 J/s2.

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Open Question

It is estimated that the net amount of carbon dioxide fixed by photosynthesis on the landmass of Earth is 5.5 * 10^16 g/yr of CO2. Assume that all this carbon is converted into glucose. (b) Calculate the average rate of conversion of solar energy into plant energy in megawatts, MW (1 W = 1 J/s). A large nuclear power plant produces about 10^3 MW. The energy of how many such nuclear power plants is equivalent to the solar energy conversion?

Textbook Question

Suppose an Olympic diver who weighs 52.0 kg executes a straight dive from a 10-m platform. At the apex of the dive, the diver is 10.8 m above the surface of the water. (a) What is the potential energy of the diver at the apex of the dive, relative to the surface of the water?

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Textbook Question

Suppose an Olympic diver who weighs 52.0 kg executes a straight dive from a 10-m platform. At the apex of the dive, the diver is 10.8 m above the surface of the water. (b) Assuming that all the potential energy of the diver is converted into kinetic energy at the surface of the water, at what speed, in m>s, will the diver enter the water?

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Textbook Question

Consider the following unbalanced oxidation-reduction reactions in aqueous solution:

Ag⁺(aq) + Li(s) → Ag(s) + Li⁺(aq)

Fe(s) + Na⁺(aq) → Fe²⁺(aq) + Na(s)

K(s) + H₂O(l) → KOH(aq) + H₂(g)

(a) Balance second reaction.

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At 20 °C (approximately room temperature) the average velocity of N2 molecules in air is 1050 mph. (b) What is the kinetic energy (in J) of an N2 molecule moving at this speed?

Verified Solution

Key Concepts

Kinetic Energy

Molecular Mass

Conversion of Units

At 20 °C (approximately room temperature) the average velocity of N₂ molecules in air is 1050 mph. (b) What is the kinetic energy (in J) of an N₂ molecule moving at this speed?