Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Doggiebounce

  • one year ago

What is the average kinetic energy of 1 mole of a gas at 125 Kelvin? (R = 8.314 J/K-mol) A. 1.04 x 10^3 J B. 3.12 x 10^3 J C. 1.56 x 10^3 J D. 1.77 x 10^3 J

  • This Question is Closed
  1. chmvijay
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    do you know equation for average kinetic energy

  2. .Sam.
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    From kinetic theory, \[E_K=\frac{1}{2}m<c^2>=\frac{3}{2}k_BT\] --------------------------------------- We can say that \[E_K=\frac{3}{2}k_BT\] \(k_B=Boltzmann~Constant=1.38 \times 10^{-23}m^2kgs^{-2}K^{-1}\) Substitute in \[E_K=\frac{3}{2}(1.38 \times 10^{-23})(125)=?\] Multiply that by avogadro number \(N_A\) gives average kinetic energy

  3. .Sam.
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @chmvijay do you have a different approach? I can't think of one relating R

  4. chmvijay
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ur right i is 3KT/2

  5. .Sam.
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah but the problem is the question includes (R = 8.314 J/K-mol)

  6. Pumpkin890
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Did anyone ever figure out how to do this?

  7. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.