## homework question 4A.3

Posts: 75
Joined: Tue Sep 10, 2019 12:16 am

### homework question 4A.3

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
Posts: 102
Joined: Sat Sep 07, 2019 12:19 am

### Re: homework question 4A.3

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

005321227
Posts: 90
Joined: Sat Sep 07, 2019 12:15 am

### Re: homework question 4A.3

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

McKenna_4A
Posts: 115
Joined: Sat Aug 24, 2019 12:18 am

### Re: homework question 4A.3

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
Posts: 104
Joined: Wed Sep 18, 2019 12:20 am

### Re: homework question 4A.3

^^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
Posts: 92
Joined: Thu Sep 19, 2019 12:15 am

### Re: homework question 4A.3

McKenna_4A wrote: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
Posts: 92
Joined: Thu Sep 19, 2019 12:15 am

### Re: homework question 4A.3

McKenna_4A wrote: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.