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!

Free expansion of a gas problem

  1. Jan 2, 2010 #1
    Hello. This is my first post in this forum, so bear with me on any flaw or mistake please.

    1. The problem statement, all variables and given/known data
    Imagine a gas confined within an insulated container as shown in the figure below. The gas is initially confined to a volume Vi at pressure Pi and temperature Ti. The gas then is allowed to expand into another insulated chamber with volume V2 that is initially evacuated. Show that, for small differences in volume [tex]\Delta[/tex]V= Vf - Vi, the difference of temperature [tex]\Delta[/tex]T before and after the expansion is given by:
    [tex]\Delta T = (\frac{P}{Nc_{v}} - \frac{T\alpha}{Nc_{v}K_{t}})\Delta V[/tex]
    P - pressure
    T - temperature
    [tex]\alpha[/tex] - coefficient of thermal expansion
    [tex]c_{v}[/tex] - molar heat capacity at constant volume
    [tex]K_{t}[/tex] - coefficient of isothermal compressibility

    2. Relevant equations
    None.


    3. The attempt at a solution
    It's known that there is no work done on the system, so [tex]\Delta[/tex]U = 0. Since it's an adiabatic process, there's no exchange of heat with the exterior, so [tex]\Delta[/tex]Q = 0.

    With the help of my "Concepts in Thermal Physics" and of course, Wikipedia, I managed to write down some promising equations that looked like they were gonna solve the problem on their own, but I just wound up with a huge headache. Here they are:
    [tex]\alpha = \frac{1}{V}(\frac{\delta V}{\delta T})_{P}[/tex]

    [tex]K_{t} = - \frac{1}{V} (\frac{\delta V}{\delta P})_{T}[/tex]

    [tex]dU = 0 = (\frac{\delta U}{\delta T})_{V} dT + (\frac{\delta U}{\delta V})_{T} dV[/tex]

    [tex]dU = T dS - P dV[/tex]

    [tex]\frac{P}{T} = (\frac{\delta S}{\delta V})_{U}[/tex]

    I kinda suck at differential equations, but I'm sure I am missing something, because I just couldn't relate all the variables the way they wanted me to. I've searched the internet, but I just end up reading a problem about Joule's Expansion (which differs on the fact that in this case we are not told that it's an ideal gas, and the gas doesn't expand to twice it's original volume - otherwise it would be fine) or a free gas expansion with constant temperature, which is clearly not the case.

    I know that the change of entropy will be a very useful tool in this problem, but I still can't see how. I am going crazy with this.


    EDIT: With the equations in this site http://scienceworld.wolfram.com/physics/Heat.html I managed to get an interesting equation: [tex]dT = (\frac{-\alpha K_{T} T}{C_{V}}) dV[/tex], but it still isn't good enough... =/
     
    Last edited: Jan 2, 2010
  2. jcsd
  3. Jan 2, 2010 #2
    With these sorts of problems we've always got to keep an eye on the expression we are trying to get to; in particular we want dT in terms of dV.

    Consequently it makes sense to consider temperature T as a function of volume V and another independent variable; here the only sensible choices seem to be pressure P and entropy S. (The entropy creeps in through the specific heat).

    Now the fundamental thermodynamic relation (which you listed):

    [tex] dU = T dS - P dV [/tex]

    gives us a relation between dS and dV since we know what dU is. Thus to get dT in terms of dV we should take T as a function of....
    giving dT = ( ... ) dV.

    You'll get some weird partial derivatives in this expression: you'll need to juggle some chain rules and Maxwell relations until you can get things in terms of heat capacity at constant volume, coefficient of isothermal compressibility and the coefficient of thermal expansion.

    Can you work it out?
     
  4. Jan 3, 2010 #3
    That's the thing, I think I'm missing some of those chain rules you mentioned. I understand perfectly what you said, and I've been over this for days trying to get the equation from it's variables, and also the reverse path. I think I'm actually missing some ground rules on differentials, because I just can't see the answer. I can't see how dS eventually turns into dT in the expression... All my attempts have been laid out for you in the original post.
     
  5. Jan 3, 2010 #4
    I just need help in a little thing so I can wrap this up, please...

    [tex]T (\frac{\delta S}{\delta V})_{T} = ??[/tex]
     
  6. Jan 3, 2010 #5
    Alright, the basic chain rules for partial derivatives are
    [tex]\left(\frac{\partial z}{\partial x}\right)_{y} = \left(\frac{\partial x}{\partial z}\right)^{-1}_{y}[/tex]

    and

    [tex]\left(\frac{\partial z}{\partial x}\right)_{y} = - \left(\frac{\partial z}{\partial y}\right)_{x} \left(\frac{\partial y}{\partial x}\right)_{z}[/tex]

    (you should be able to obtain these given a function z=z(x,y) from [tex] dz = \left(\frac{\partial z}{\partial x}\right)_{y} dx + \left(\frac{\partial z}{\partial y}\right)_{x} dy [/tex] by differentiating with respect to a certain variable keeping another variable fixed).

    The other important equations come from the fact mixed derivatives commute: see http://en.wikipedia.org/wiki/Maxwell_relations.

    Now you should have everything you need: are there any maxwell relations relating to the term you are trying to simplify? Keep applying the maxwell relations followed by appropriate chain rules until you get partial derivatives proportional to the specific heat, isothermal compressibility and coefficient of thermal expansion.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Free expansion of a gas problem
  1. Gas expansion (Replies: 3)

  2. Adiabatic Gas expansion (Replies: 29)

Loading...