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Volume expansion and pressure change

  1. Nov 9, 2003 #1
    A brass cube, 10 cm on a side, is heated with a temperature change of 200 deg. C. By what percentage does its volume change?

    Vo = 10cm = 1000cm^3 = .001m^3
    T = 200 C
    brass = 19x10^-6 coefficient of linear expansion

    V=Vo(1+3 x brass x T)
    change in V = Vo(3 x brass x T)
    = .001m^3(3 x 19x10^-6 x 200 C)
    = .0000114 = 1.14x10^-5

    % change = change in V/Vo
    = .0000114/.001
    = .0114 = 1.1%

    answer in the book is 1.1x10^-3 %

    What I am doing wrong here?

    A radial tire is inflated to a gauge pressure of 35 lb/in^2 at 60 deg. F.
    If the temperature increases to 100 deg F while the volume of the
    tire remains constant, what is the tire's new pressure?

    T1=60 F = 289 K
    T2=100 F = 311 K
    P1= 35 lb/in^2
    Find P2


    P2 = (35 lb/in^2 x 311 K)/289 K
    P2 = 37.7 lb/in^2

    Also tried this:

    P=pa + pg
    14.7 (atm) + 35 lb/in^2
    = 49.7 lb/in^2 = absolute pressure

    P2 = (49.7 lb/in^2 x 311 K)/289 K
    P2 = 53.5 lb/in^2

    answer in the book is 39 lb/in^2

    any help would be appreciated
  2. jcsd
  3. Nov 9, 2003 #2
    In the first one:

    You wrote it correctly here:
    but when you put in the numbers you lost the "1 +":
    Fix that & you'll get the correct answer.

    [Edited to delete erroneous comment about question 2.]
    Last edited: Nov 9, 2003
  4. Nov 9, 2003 #3


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    Brass? You're fine, the answer key is in error --- lotsa luck with your teacher on that detail.

    Tire pressure? The T conversion is fine, the answer key is correct, and you have omitted ONE step in one of the two calculations you did --- the other needs more than one correction.

    Need a hint beyond that?
  5. Nov 9, 2003 #4
    For the brass:

    No, the answer key is correct.

    this line:
    = .001m^3(3 x 19x10^-6 x 200 C)

    should be

    = .001m^3(1 + 3 x 19x10^-6 x 200 C)

    Fix that, & you will get the same answer that the book gives.

    For the tire:
    Sorry about my erroneous comment, which I'll now delete to avoid confusion. (As Bystander said, you've got it, if you just do one more step.)
    Last edited: Nov 9, 2003
  6. Nov 9, 2003 #5


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    A 19 ppm/K LINEAR expansion coefficient integrated over 200K produces a 10 ppm change in VOLUME? Not on this planet, not in this solar system, ....

    You are seeing a very common phenomenon occurring in answer keys --- student help working the problems and not being proofread by textbook authors.
  7. Nov 9, 2003 #6
    No, it isn't.

    Fine. Then you get V = 0.0010114 m3. If you calculate the percent change in V, that's

    ΔV/V0 x 100% = (V-V0)/V0 x 100% = 1.14%

    as fish obtained.

    (Actually, you can get it quicker by just noting that the coefficient of volume expansion β is defined by ΔV/V0 = βΔT; if you use the fact that β = 3α where α is the linear coefficient fish calls "brass", you immediately obtain ΔV/V = 3αΔT = 0.0114.)
  8. Nov 9, 2003 #7

    Sorry, I read it too quickly & didn't realized he was shortcutting to just the change in volume when he wrote ".001m^3(3 x 19x10^-6 x 200 C)"

    Beyond that, I'll just plead insanity, or something....(& get back to my own homework)
  9. Nov 9, 2003 #8
    I bet the solutions manual mistakenly listed the change in volume (nonsensically multiplied by 100), instead of the fractional change in volume.
  10. Nov 10, 2003 #9
    For the tire pressure,
    P2 = 53.5 lb/in^2 - 14.7 lb/in^2 = 38.8 = 39 lb/in^2

    In the preface of the book, it says each solution in the solutions manual has been checked for accuracy by a minimum of 5 instructors. I've found other errors in the answer keys and also errors in the example problems of the main textbook. What Bystander said was pretty interesting...not in this solar system

    thanks for the help guys
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