## Question 8.3 (Sixth Edition)

Steve Magana 2I
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

### Question 8.3 (Sixth Edition)

Question: Air in a bicycle pump is compressed by pushing in the handle. If 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?

I am lost on how to start this problem, can someone start me off please? Thank you!

Jesse Kuehn 1B
Posts: 40
Joined: Wed Nov 14, 2018 12:23 am

### Re: Question 8.3 (Sixth Edition)

One equation for work is w=-P(change in Volume)(the constant for Lxatm=J) the change in volume can be calculated using the area of the bike pump cylinder times how far it moved.

Posts: 37
Joined: Mon Apr 23, 2018 3:00 am

### Re: Question 8.3 (Sixth Edition)

For part a, you want to find the work so you would use w=-PV. V=π(r^2)(h), so you would use the values given for the diameter of the pump and the distance the pump is depressed to find the volume.

For part b, if work is being done on the system (the air in the pump), work is positive. If work the system is doing work, work is negative.

For part c, use deltaU=q+w