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Calculating percentage change in volume at constant pressure.

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data

    The coefficient of the thermal expansion of ethanol is given by:

    [itex]\alpha[/itex] (per degree Celsius) = 1.0414 x 10^(-3) + 1.5672 x 10^(-6) t + 5.148 X 10^(-8) t^2

    Calculate the percentage change in volume when ethanol is heated from 0 to 100 (celsius) at constant pressure.

    2. Relevant equations

    alpha = (1/V)(partial(V)/(partial(T))(at constant P)

    3. The attempt at a solution

    I simply rearranged the above equation and integrated with limits T1 and T2:

    delta(V) = (alpha/2)(100)^2 = 5.123..something but the answer is 13.8 %

    Can someone also comment on the units of alpha? Its confusing because the units t and t^2 are included. From a mathematical point of view, we can't just add them..( or thats what i think at least).
     
  2. jcsd
  3. Oct 12, 2012 #2

    rude man

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    You're wise to question units (dimensions). Too few people do that, including a former PhD boss of mine! Yet it's a very powerful way to check your computations as you go along.

    The answer of course is that the t coefficient has units of K-1and the t2 coefficient has units of K-2. All terms must be dimensionless as is α.

    Just out of curiosity, why did you use t instead of T in your expression for α?

    You need to integrate to get ΔV/V over 0 to 100 deg C.
     
  4. Oct 12, 2012 #3

    Hey, thanks for you help! I got ΔV/V = (alpha)(T dt), which I integrated from 0 to 100. The answer I get now is 20.86..I'm not sure what I'm doing wrong..

    As for the units of alpha..I dont know why its t and not T. This is a question from Chemical Thermodynamics by Peter A. Rock.
     
  5. Oct 12, 2012 #4

    rude man

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    I don't see your integral. What is it?

    I assume t is in Kelvin, not centigrade. Double-check this!
     
  6. Oct 12, 2012 #5
    I integrated both (ΔV/V)= (alpha) dt [from 0 to 100] = alpha (100) = wrong answer
    and (ΔV/V) = (alpha)(T) dt [ from 0 to 100) = alpha x 2 x ( 100^2 - 0^2 ) = 20.86

    the answer in the back of my textbook says the answer is 13.8 %
     
  7. Oct 12, 2012 #6

    rude man

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    V/V. What was your expression for Δ
    OK, I got the same answer (13.8%) if I assume t is in deg C.

    You did right by setting ΔV/V = αΔt or, let's be sophisticated physicists, dV/V = αdt. I need to see your work step-by-step to see what the error was you made, because you took the right tack in integrating this equation.
     
  8. Oct 13, 2012 #7
    Here's what I did:

    dv/V = [itex]\alpha[/itex] dt = [itex]\alpha[/itex] T (100-0) [ after integration ] = [itex]\alpha[/itex] (100-0) = [itex]\alpha[/itex] (100) = 10.43 for me..

    I added the three quantities of alpha and then multiplied by 100. I also multiplied each individual quantity to 100 and then added them together. the answer is still wrong..
     
  9. Oct 14, 2012 #8

    rude man

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    dV/V is a differential fractional increase in volume. You did not compute the total fractional change in volume.

    And you did not integrate α(t)dt from 0 to 100C either.

    Integrate both sides of this equation correctly & what do you get?
     
    Last edited: Oct 14, 2012
  10. Oct 17, 2012 #9
    Got it. Thank you so much!
     
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