Thermodynamics Questions for Helium and Flourine

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 4K views
Tofuphysics
Messages
1
Reaction score
0
Hey all, I just discrovered these forums, thankoodness, because I'm taking physics over the summer (its a requirement at my school), and not being a science-y person, I really need help.

Homework Statement

[/B]

This is the problem:

Problem 1
a. What is the average translational kinetic energy of a helium atom at 300 K?
b. What is the average translational velocity of a helium atom at 300 K?
c. What is the total internal energy of a 4.0 g sample of He at 300 K?
d. What is the total internal energy of a 4.0 g sample of He at 301 K?
e. How much energy is needed to raise the temperature of 4.0 g of He by one degree K?

And then the next problem is he same cocept, but with Flourine (F2)

Problem 2
a. What is the average translational kinetic energy of a F2 molceule at 300 K?
b. What is the average translational velocity of a F2 moelcule at 300 K?
c. What is the total internal energy of a 38 g sample of F2 at 300 K?
d. What is the total internal energy of a 38 g sample of F2 at 301 K?
e. How much energy is needed to raise the temperature of 38 g of F2 by one degree K?

Homework Equations


KE=(3/2)(Kb)(T)
KE=1/2mv^2
Avogadroes number= 6.02 x 10^-23
1 MW= 1x wx^6 w (i don't know what this means but the teacher wrote it on the board)

The Attempt at a Solution


I think that you have to use the KE equations and the molar mass (which you can figure out using the periodic table?) but I'm really not sure how to put it together.


Thankyou so much if you had the patience to read all that!
-Tofuphysics
 
Last edited:
Physics news on Phys.org
Welcome to PF Tofuphysics,
Tofuphysics said:
KE=(3/2)(Kb)(T)
KE=1/2mv^2
Avogadroes number= 6.02 x 10^-23
Assuming of course that you are treating these gases as ideal, then those equations are correct and will certainly allow you to answer questions (a) and (b) in each case. However, a third equation would be infinitely more useful in answering the remaining parts.