Problem Three Part A

Moderators: Chem_Mod, Chem_Admin

Ethan_Kato_1D
Posts: 25
Joined: Fri Jul 15, 2016 3:00 am

Problem Three Part A

Postby Ethan_Kato_1D » Thu Jan 19, 2017 10:15 am

8.3: Air in a bicycle pump is compressed by pushing in th ehandle. if the inner diameter of the pump is 3 cm and the pump is depressed 20 cm with a pressure of 2atm, (a) how much work is done in the compression?

Why when finding the amount of work, Area times Distance is equal to the change of volume? The units do not equal liters, unless you convert cm to liters. So how exactly does Area times Distance equal liters?

Michelle_Li_1H
Posts: 32
Joined: Wed Sep 21, 2016 3:00 pm

Re: Problem Three Part A

Postby Michelle_Li_1H » Thu Jan 19, 2017 10:34 am

Work is equal to -P multiplied by the change in volume. The equation for volume is area of the base times the height. In this case the distance that the pump has moved would equal to the height, therefore the work would equal to -P*A*d. However, because the units are given in cm, we would convert to m.

You would have -2 atm * (pi *(0.015 m)^2) * -0.2 m = 2.827*10^-4 atm*m^3.

To get our answer in Joules, we would use the given unit conversions, 1 L= 10^-3 m^3 and 101.325 J = 1 L*atm. The final answer should be 28.6 J.

Also, work would be positive in this case because work is being done on the system, since someone is pushing the handle of the pump down.

Hope this helps!


Return to “Heat Capacities, Calorimeters & Calorimetry Calculations”

Who is online

Users browsing this forum: No registered users and 2 guests