Textbook question 4A.3

[Paige Lee 1A]

Could someone please explain why part c is 8J and not 28J? Also how do you get 8J?

Air in a bicycle pump is compressed by pushing in the handle. The inner diameter of the pump is 3.0 cm and the pump is depressed 20. cm with a pressure of 2.00 atm. (a) How much work is done in the compression? (b) Is the work positive or nega- tive with respect to the air in the pump? (c) What is the change in internal energy of the system?

[MAC 4G]

I believe that this is an error in the textbook, but it is 28J, as it is correctly written in the solutions manual.

[Chem_Mod]

Part c should be 28 J. The change in internal energy is equal to work, since there is no heat exchange. We calculate the work using the -PdeltaV equation where deltaV is the change in volume for the cylindrical pump. The volume for a cylinder is pi r^2 h. We also know since the volume is decreasing that this value should be negative. We multiply the deltaV by -P and then convert from l atm to J in order to get our final answer. Make sure to keep units in mind. If you do your volume calculation in cm then you should end up with cm^3 and 1 cm^3 =0.001 l