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!

Ideal gas compression

  1. Feb 16, 2010 #1
    1. The problem statement, all variables and given/known data

    A compressed cylinder holds nitrogen gas at room temperature. How cold is the gas that is escaping from the cylinder?

    T(cylinder) = 298 K
    P(cylinder) = 1.034e7 Pa
    n = 1

    2. Relevant equations

    PV = nRT

    3. The attempt at a solution

    So since I was not given either the volume of the container or its final volume, I assumed that the initial and final volumes were approximately the same. This may be a fatal flaw to my line of reasoning.

    Assuming the above is correct, I said:

    P/V = constant , and therefore:

    P1/T1 = P2/T2 , where P1 and T1 are the pressure and temperature in the cylinder.

    I am not sure if I need the adiabatic exponent of f+2/2 in there somewhere or not. Also, I used P2 = 101.3 kPa and yes I put that into Pa before doing the algebra.

    Without the adiabatic exponent, I went through the algebra and found the temperature of the escaping gas is 2.94 K. This to me doesn't seem like a reasonable answer since that is really cold. However, I do not have a good understanding of what is reasonable since I do not know much about gas temperatures.

    Any comments on whether or not this is correct or flawed would be greatly appreciated.
     
  2. jcsd
  3. Feb 16, 2010 #2
    In addition to above, there are two equations that I know of but neither of them contain both pressure and volume. They are:

    VT^(f/2) = constant

    PV^(f+2/2) = constant

    I don't know how to get P and T together.
     
  4. Feb 16, 2010 #3

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Look at it as an adiabatic expansion. That'll give you a relationship between two of the variables. With that relationship along with the ideal gas law, you can solve for the final temp.
     
  5. Feb 16, 2010 #4
    Okay, I rearranged the equation for adiabatic expansion to get that P^(1-gamma)*T^(gamma) = constant.

    Going through the algebra for this, I get that the temperature of the gas being released is about 79 K.

    The only thing that is still throwing me off a bit is that the problem explicitly states that the calculation should be for one mole of gas. However, nowhere in my calculations is that value relevant. This makes me think I'm not including everything.

    Can anybody confirm this number as correct? Thanks for the advice vela.
     
    Last edited: Feb 16, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Ideal gas compression
  1. Ideal gas (Replies: 5)

  2. Ideal gas and piston (Replies: 2)

Loading...