Skip to main content
Ch 18: Thermal Properties of Matter

Chapter 18, Problem 18

Oxygen (O2) has a molar mass of 32.0 g>mol. What is (d) the momentum of an oxygen molecule traveling at this speed?

Verified Solution
Video duration:
3m
This video solution was recommended by our tutors as helpful for the problem above.
6960
views
Was this helpful?

Video transcript

Hey everyone in this problem. We have a nitrogen molecule in the atmosphere has a root mean square speed of 509 m per second. We're told the molar mass of nitrogen is 28.1 g per mole. We're asked to find its momentum as it moves at that speed. Alright, let's recall momentum P is equal to mass times velocity. Okay. Mv. Now in this problem we are given the root mean squared speed. Okay, so velocity will be V. RMS And so our momentum is going to be M the RMS. And so we have that V RMS given in the problem but we don't know the mass. Okay. We're given the molar mass but not the mass. So in order to find P the momentum we're looking for, we need to first find that mass. Now let's recall that we can relate the molar mass in the mass through avocados number. Using big M. The molar mass is equal to M. The mass times N A. Avocados number. This gives us the mass M is equal to big M over N A. The molar mass we're given is 28.1 g per mole which is going to be 0. 01 kg per mole. Okay. In our standard unit, in avocados number, you can look this up in your textbook or chart that your professor provided to you. Okay. And this is a good one to try to remember 6.22 times 10 to the 23. Okay. And the unit is one over all. Okay, so the unit of per mole will cancel we're just gonna be left with the mass is equal to 4.65 times 10 to the negative 26 kilograms. All right. So we have our mask now and we can get back to finding the momentum. Okay? So the momentum is equal to m times V V r M. S. And the mass is going to be 4.65 times 10 to the negative 26 kg Times the root mean square speed 500 and nine m/s. This is gonna give a momentum p of 2.37 times 10 to the negative 23. Our unit is going to be kilogram meter per second. Okay, So that's the momentum we were looking for of that nitrogen molecule. That's going to correspond with answer be That's it for this one. Thanks everyone for watching. See you in the next video.
Related Practice
Textbook Question
Martian Climate. The atmosphere of Mars is mostly CO2 (molar mass 44.0 g/mol) under a pressure of 650 Pa, which we shall assume remains constant. In many places the temperature varies from 0.0°C in summer to -100°C in winter. Over the course of a Martian year, what are the ranges of (a) the rms speeds of the CO2 molecules and
1069
views
1
rank
Textbook Question
Martian Climate. The atmosphere of Mars is mostly CO2 (molar mass 44.0 g/mol) under a pressure of 650 Pa, which we shall assume remains constant. In many places the temperature varies from 0.0°C in summer to -100°C in winter. Over the course of a Martian year, what are the ranges of (b) the density (in mol/m^3) of the atmosphere?
927
views
1
rank
Textbook Question
Oxygen (O2) has a molar mass of 32.0 g>mol. What is (a) the average translational kinetic energy of an oxygen molecule at a temperature of 300 K;
596
views
Textbook Question
At what temperature is the root-mean-square speed of nitrogen molecules equal to the root-mean-square speed of hydrogen molecules at 20.0°C? (Hint: Appendix D shows the molar mass (in g/mol) of each element under the chemical symbol for that element. The molar mass of H2 is twice the molar mass of hydrogen atoms, and similarly for N2.)
705
views
Textbook Question
Smoke particles in the air typically have masses of the order of 10-16 kg. The Brownian motion (rapid, irregular movement) of these particles, resulting from collisions with air molecules, can be observed with a microscope. (a) Find the rootmean-square speed of Brownian motion for a particle with a mass of 3.00 * 10-16 kg in air at 300 K.
1386
views
Textbook Question
For diatomic carbon dioxide gas (CO2, molar mass 44.0 g/mol) at T = 300 K, calculate (a) the most probable speed v_mp;
867
views
1
rank