1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Calculate a values given equation of state for gases

  1. Aug 31, 2011 #1
    Calculate "a" values given equation of state for gases

    1. The problem statement, all variables and given/known data

    For my equation of state:
    Vm = (RT/P) - (a/T)
    what kind of sign for the constant a (positive or
    negative) do you think the following gases will have and why:
    a. H2 b. He c. CH4 d. CO2
    Hint: There is almost no math involved.

    2. Relevant equations
    Vm = (RT/P) - (a/T) (obviously, lol)
    I'm thinking critical values also might play a role, but since my professor stated that nearly no math is involved, I'm thinking that the answers are mostly derived from the given equation.

    3. The attempt at a solution
    Honestly I have no idea where to start with this. I know in the standard Van der Waal's equation that the a values represent the error intermolecular attractions, and that the values for hydrogen and helium are very small because they basically have no attraction. I'm guessing negative values would represent intermolecular repulsion...any help is appreciated
  2. jcsd
  3. Aug 31, 2011 #2
    Re: Calculate "a" values given equation of state for gases

    As an educated guess,

    I think the answer should relate to the number of atoms in a mole.

    where 2,1,5,3 respectfully
  4. Aug 31, 2011 #3
    Re: Calculate "a" values given equation of state for gases

    I think the answer is going to be something like CH4 and H2 are negative because they possess essentially no dipoles (especially H2, since it has no intermolecular interactions as H+ ions, much less as inert H2) while He would be weakly positive due to possessing some Van der Waals force and CO2 would be high positive due to having strong dipoles. However, this isn't based off the equation at all, but maybe it's a leap in the right direction.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook