volume expansion problem, is my concept wrong?
here is the question:
The density of gasoline is 730 kg/m^3 at 0 degree celsius. Its average coefficient of volume expansion is 9.6e4. 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 temperaturecompensated? 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? 
The way you did it you should have gotten the same answer for the two temperatures becuase you are counting for the temperaturecompensation. To not account for it, the machine would assume that it dispenses the same volume for each temperatue.

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

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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