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Kinetic Energy of gas Molecules

by Dulcis21
Tags: energy, gas, kinetic, molecule
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Dulcis21
#1
Dec7-08, 10:39 AM
P: 3
Hi, the question, I'm having problems with is this:

A 1.0 mol sample of hydrogen gas has a temperature of 30 C. What is the total kinetic energy of all the gas molecules in the sample? How fast would a 75 kg person have to run to have the same kinetic energy?



Relevant equations:
K = (3/2)kT
or K = (3/2)nRT




Ok, so I tried using both formulas:

(3/2)(1.38*10^-23)(30+273) = 6.2721 * 10^-21

Then, since its one mole and the question asks for the total KE, I multiplied this answer by Avogadro's number (6.02*10^23) and got

= 3775.8042 Joules.

For the second part, I would just set this number equal to 0.5mv^2 and solve for v but since I got the first one wrong, it doesn't quite work :(


I really need help; I thought I had the right idea but it's not working out. Thanks
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Andrew Mason
#2
Dec7-08, 11:23 AM
Sci Advisor
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P: 6,653
Quote Quote by Dulcis21 View Post
A 1.0 mol sample of hydrogen gas has a temperature of 30 C. What is the total kinetic energy of all the gas molecules in the sample? How fast would a 75 kg person have to run to have the same kinetic energy?



Relevant equations:
K = (3/2)kT
or K = (3/2)nRT




[b]Ok, so I tried using both formulas:

(3/2)(1.38*10^-23)(30+273) = 6.2721 * 10^-21

Then, since its one mole and the question asks for the total KE, I multiplied this answer by Avogadro's number (6.02*10^23) and got

= 3775.8042 Joules.
Your method is right. Using [itex]U = \frac{3}{2}nRT[/itex], U = 1.5*8.314*303 = 3778.7 Joules.

For the second part, I would just set this number equal to 0.5mv^2 and solve for v but since I got the first one wrong, it doesn't quite work
Think of one mole or 1 gram of hydrogen possessing 3778.7 J of kinetic energy. The effective average (rms) velocity is:

[tex]v = \sqrt{2E/m}[/tex]

AM
Dulcis21
#3
Dec7-08, 11:35 AM
P: 3
Quote Quote by Andrew Mason View Post
Your method is right. Using [itex]U = \frac{3}{2}nRT[/itex], U = 1.5*8.314*303 = 3778.7 Joules.

Think of one mole or 1 gram of hydrogen possessing 3778.7 J of kinetic energy. The effective average (rms) velocity is:

[tex]v = \sqrt{2E/m}[/tex]

AM

I thought my method was right but my homework web-assign isn't accepting that answer :(
It asked for it in 2 sig figs so i put it in as 3.8 *10^3 but it still said no...

Dulcis21
#4
Dec7-08, 07:38 PM
P: 3
Kinetic Energy of gas Molecules

Quote Quote by Dulcis21 View Post
I thought my method was right but my homework web-assign isn't accepting that answer :(
It asked for it in 2 sig figs so i put it in as 3.8 *10^3 but it still said no...
Oh I figured it out lol, it needed (5/2)kT to work and then multiply by Avogadro's number.
Andrew Mason
#5
Dec8-08, 09:01 AM
Sci Advisor
HW Helper
P: 6,653
Quote Quote by Dulcis21 View Post
Oh I figured it out lol, it needed (5/2)kT to work and then multiply by Avogadro's number.
Of course. H2 is diatomic.

AM


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