# Kinetic Energy of gas Molecules

by Dulcis21
Tags: energy, gas, kinetic, molecule
 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
HW Helper
P: 6,671
 Quote by Dulcis21 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 $U = \frac{3}{2}nRT$, 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:

$$v = \sqrt{2E/m}$$

AM
P: 3
 Quote by Andrew Mason Your method is right. Using $U = \frac{3}{2}nRT$, 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: $$v = \sqrt{2E/m}$$ 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...

P: 3
Kinetic Energy of gas Molecules

 Quote by Dulcis21 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.