homework question 4A.3

[Jade Hinds 2B]

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 negative with respect to the air in the pump? (c) what is the change in internal energy of the system?

[Benjamin Feng 1B]

Work is equal to negative pressure times change in volume. You can use the equation to calculate all the numbers.

[005321227]

how do you go about the conversions for this problem? the answer is given using L times atm.

[McKenna_4A]

Wouldn't using -P mean the system lost energy? Lavelle said this in lecture today but the solutions manual says that the system has work done on it.

[bellaha4F]

^^to answer the conversion question, you would use 1 L*atm = 101.325 J conversion. it's on lavelle's constant and equations sheet!

[Cavalli_1H]

Wouldn't using -P mean the system lost energy? Lavelle said this in lecture today but the solutions manual says that the system has work done on it.

Yes! I thought it should be positive too because the question said there is compression, and today Lavelle said expansion should be negative. What am I missing?

[Cavalli_1H]

Wouldn't using -P mean the system lost energy? Lavelle said this in lecture today but the solutions manual says that the system has work done on it.

okay, after looking more closely at the solutions manual, it says that you should calculate the change in V first, which is negative because of compression. The formula is still w= -PdeltaV, but deltaV is -0.14L so w=-P(-deltaV) which makes the work on the system positive because the two negatives cancel.