Calculating Work

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elizabethrojas1G
Posts: 44
Joined: Fri Sep 25, 2015 3:00 am

Calculating Work

Postby elizabethrojas1G » Thu Jan 14, 2016 11:27 am

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. How much work is done in the compression?

I know how to get to the answer because my TA briefly went over this problem, but can someone thoroughly explain each step please?

Nathan Danielsen 1G
Posts: 17
Joined: Fri Sep 25, 2015 3:00 am

Re: Calculating Work

Postby Nathan Danielsen 1G » Thu Jan 14, 2016 3:12 pm

Work is equal to -P*deltaV. The problem tells you that the pressure being applied to the system is 2 atm, so now you just need to find the change in volume. The volume of a cylinder is equal to the area of the base multiplied by the height. The diameter of the pump is 3, which means that the area of the base is about 7.07 cm^3. You know that the pump is depressed by 20 cm, which means that the volume of the pump has decreased. You multiply the area of the base by the amount the pump has been depressed to get the change in volume: 7.07 x 20 = -141.4 cm^3. One liter is equal to 1000 cm^3 so the volume change in liters is -.1414 L. Then you multiply that by the pressure to get -.2828 L*atm, which then becomes 0.2828 because the negatives cancel out. Finally, to convert the answer to J, you multiply by 101.325 J/L*atm, to 28.7 J.

hannaluong4E
Posts: 22
Joined: Fri Sep 25, 2015 3:00 am

Re: Calculating Work

Postby hannaluong4E » Thu Jan 14, 2016 3:28 pm

First step would be to consider the work formula, w=-P∆V. And we know to use this formula because there's a constant ext. pressure of 2 atm.

Then to find the unknown ∆V, we use the formula for the volume of the cylinder V=πr2h, where ∆V=πr2∆h because the pi and r are constants. ∆V=π(1.5 cm)2(-20 cm)= -141.37 cm3. The volume change is negative because the pump is getting DEPRESSED 20 cm.

To plug back into the work formula you would convert the ∆V to L using .001 L=1 cm3. So that would be -.14137 L. And w=-(2 atm)(-.14137 L)=.283 L x atm = 28 J.

The work is positive with respect to the air because the air isn't doing any work; the work is being done on it, compressing it.

The change in internal energy change is ∆U=q+w but in the absence of heat, ∆U=w=28 J.

AKatukota
Posts: 100
Joined: Thu Jul 25, 2019 12:18 am

Re: Calculating Work

Postby AKatukota » Thu Jan 30, 2020 9:52 pm

Thank you!! This really helped because I was having trouble figuring out the internal energy.

Madelyn Romberg 1H
Posts: 102
Joined: Tue Oct 02, 2018 12:16 am

Re: Calculating Work

Postby Madelyn Romberg 1H » Sat Feb 01, 2020 6:45 pm

Use w=-P∆V
-w=work
-P= pressure
-∆V= change in volume

you are given P and the change in volume can be calculated using the given diameters and some geometric formulas.


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