Results 1 to 2 of 2

Math Help - integration word problem help

  1. #1
    Super Member
    Joined
    Sep 2008
    Posts
    607

    integration word problem help

    The reversible work per mol done, Wrev, to expand a fuid from a molar volume
    V1 to a molar volume V2 is given by

     W_{rev} = - \int^{V_{2}}_V_{{1}}   P  dV

    Where P the the pressure of the fluid at a volume V.

    Calculate Wrev when

    a)  PV = RT - - i.e. an ideal gas. Assume that the temperature, T, is constant
    (i.e. an isothermal expansion).

    b)  P = KV^{- \gamma} and K,  \gamma are constants. Adiabatic expansion of an ideal gas. For information's sake  \gamma = \frac{C_{p}}{C_{v}} where where Cp and Cv are the heat capacities at constant pressure and volume respectively.


    c)  P = \frac{RT}{(V-b)} - \frac{a}{V^{2}} - van der Waals equation of state with a and b constants,
    assume that the temperature T is constant.
    I think I am able to do part 'a', can someone check my working and let me know if it is correct?

     PV =RT

     P = \frac{RT}{V}

     \int^{v_{2}}_v_{1} \frac{RT}{V} dv

    since R and T are constants it can be left out of the integration so;

     RT \int^{v_{2}}_v_{1}\frac{1}{V} dv

    =  ln(V) evaluating that I get  ln(V_{2})- ln(V_{1})

    is this correct?

    Also need help with part 'b' and 'c' please.

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398

    Thermo- Reversible Work

    Quote Originally Posted by Tweety View Post
    I think I am able to do part 'a', can someone check my working and let me know if it is correct?

     PV =RT

     P = \frac{RT}{V}

     \int^{v_{2}}_{v_{1}} \frac{RT}{V} dV

    since R and T are constants it can be left out of the integration so;

     RT \int^{v_{2}}_{v_{1}}\frac{1}{V} dv

    =  ln(V) evaluating that I get  ln(V_{2})- ln(V_{1})

    is this correct?

    Also need help with part 'b' and 'c' please.

    Thank you.

    You dropped the negative sign, then you dropped the RT.

    Multiply your answer by -RT.

    Part (b);
    Simply substitute KV^{-\gamma} in for P.

    W_{rev}=-\int_{V_1}^{V_2} KV^{-\gamma}  d V, K is a constant, so it comes out of the integral. That's a fairly basic integral.

    Part (c)
    For a van der Vaal's gas:

    \displaystyle W_{rev}=-\int_{V_1}^{V_2} \left(\frac{RT}{(V-b)} - \frac{a}{V^{2}} \right)  d V =-RT \int_{V_1}^{V_2} \frac{1}{(V-b)}\ dV +a \int_{V_1}^{V_2}  \frac{1}{V^{2}}\  d V

    I expect that you can do these integrations also.
    Last edited by SammyS; November 29th 2010 at 07:27 PM. Reason: not finished.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integration Word Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 5th 2010, 08:06 PM
  2. Integration word problem please help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 18th 2009, 01:05 PM
  3. word problem with possible integration? a rate
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 5th 2009, 03:11 PM
  4. Integration word problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 18th 2009, 05:21 PM
  5. Word Problem Integration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 1st 2008, 07:28 AM

Search Tags


/mathhelpforum @mathhelpforum