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Volume expansion problem, is my concept wrong?

  1. Oct 9, 2004 #1
    here is the question:

    The density of gasoline is 730 kg/m^3 at 0 degree celsius. Its average coefficient of volume expansion is 9.6e-4. If 1.00 gallon of gasoline occupies .00380 m^3, how many extra kilograms of gasoline would you get if you bought 15.0 gal of gasoline at 0 degrees celsuis rather than at 20 degrees celsius from a pump that is not temperature-compensated?

    Here is what i did:

    since i know the avg coefficient of volume expansion, initial volume, and change in temp, i was able to find the change in volume using the formula:

    change in v= (avg. coeff. of volume)(initial volume)(change in temp)

    then i found the final volume. Since i know the density of gasoline at 0 degree celsius, i found the density of gasoline at 20 degree celsius. After that, i came up with these numbers:

    Final volume: .00387296 m^3
    Density of gasoline at 20degree celsius: 716.248 kg

    now, i multiplied the initial volume by 15, then multiplied it again by the density to find the # of kg's that the initial volume had. here is the work:

    (.0038 m^3)(15)(730 kg/m^3) = answer in kg (at 0 degree celsius)

    then i did the same with final volume:

    (.00387296)(15)(716.248 kg/m^3) = answer in kg (at 20 degree celsius)

    then i took the diff bw the two to find the extra kilograms of gasoline...however my answer is wrong...anyone know where i went wrong?
     
  2. jcsd
  3. Oct 9, 2004 #2
    The way you did it you should have gotten the same answer for the two temperatures becuase you are counting for the temperature-compensation. To not account for it, the machine would assume that it dispenses the same volume for each temperatue.
     
  4. Oct 9, 2004 #3
    i dont think i understand what you are trying to say? sorry i've been up all night procrastinating my physics hw, and im down to my last one and i cant seen to solve it
     
  5. Oct 9, 2004 #4
    I believe you want to do this:

    (0 degrees) = (.0038 m^3)(15)(730 kg/m^3)

    (20 degrees) = (.0038 m^3)(15)(716.248 kg/m^3)

    Thus the pump would not account for the change in volume due to the temperature difference
     
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