Physics Forums

Physics Forums (http://www.physicsforums.com/index.php)
-   Introductory Physics Homework (http://www.physicsforums.com/forumdisplay.php?f=153)
-   -   volume expansion problem, is my concept wrong? (http://www.physicsforums.com/showthread.php?t=46774)

nemzy Oct9-04 01:44 AM

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 9.6e-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?

Spectre5 Oct9-04 02:08 AM

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.

nemzy Oct9-04 02:19 AM

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

Spectre5 Oct9-04 02:22 AM

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


All times are GMT -5. The time now is 01:37 PM.

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums