Homework chapter 8 #3 part a  [ENDORSED]

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Nicholas Fierro 1E
Posts: 9
Joined: Wed Nov 18, 2015 3:00 am

Homework chapter 8 #3 part a

Postby Nicholas Fierro 1E » Sat Jan 21, 2017 9:13 pm

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 20cm with a pressure of 2.00 atm (a) how much work is done in the compression?
I am not quite sure how to approach this problem because the height of the bicycle pump isn't given therefore I am not sure how to go about finding the total volume of the bicycle pump and cannot find the change in volume as a result. Any suggestions? Thank you!

Swati_Sharma_2D
Posts: 21
Joined: Wed Sep 21, 2016 2:59 pm

Re: Homework chapter 8 #3 part a  [ENDORSED]

Postby Swati_Sharma_2D » Sat Jan 21, 2017 10:35 pm

Hi,

In order to solve this problem, you can use the fact that F (force) = P (pressure) x A (area) and that W (work) = F (force) x D (distance). Combining these two equations, we get W = (A x P) x D because we substituted P x A for F.

The pressure is given in the problem and the area A can be calculated because we are given the diameter so we can do pi x r^2. Make sure to use correct units though and convert to m since the values are given in cm. We also have to convert the pressure into pascals (1 atm = 10^5 pascals). So, we get that P = 2 x 10^5 pascals and that A = pi x (0.015m)^2. We are also given the fact that the pump is depressed 20 cm which is our distance. We convert that value to m as well to get 0.2 m.

Now, we plug in our values into W = (A x P) x D and we get that W =28 J.

Noelle Min-1N
Posts: 10
Joined: Wed Sep 21, 2016 2:58 pm

Re: Homework chapter 8 #3 part a

Postby Noelle Min-1N » Sun Jan 22, 2017 9:42 pm

How do you know that pressure needs to be converted to Pa instead of kPa or another conversion? Is there a chart or other resource that tells me which units to use?


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