Homework Problem 4A.3

[Talia Dini - 3I]

Hi! I'm having some trouble figuring out this problem. Could someone please walk me through it?

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?

[Kelly Yun 2I]

I think the first way to approach this would be to use the formula w=-P∆V, and ∆V would be found by finding the volume of the cylinder using the given diameter (cm^3 = L), and since we have the pressure as well, we could plug the values in and get w. Hope this helps..!

[Arya Adibi 1K]

given the diameter and height, you can calculate the change in volume. The work is then equal to w = - P(v2 - v1).

[Asia Yamada 2B]

For part a, you would use the formula: w = -P•deltaV, since the external pressure is constant. To find the change in volume, you would use the formula for the volume of a cylinder which is πr^2•deltah. The work would be positive with respect to the air in the pump and since heat doesn’t affect this reaction, the change in internal energy of the system is equal to the value you found for work in part a.

[Sera Aintablian 2E]

Hi, I was also having trouble with this question at first, but I noticed that the only unfamiliar thing was having to calculate the change in volume, which I found we can find through the equation, (pi)r^2, given its diameter. Afterwards, the calculation is very straightforward, as we just use the equation w = -P(deltaV)

[Lesly Lopez 3A]

When I did this problem, I think the first way to approach this would be to use the formula w=-P∆V. In this case ∆V would be found by finding the volume of the cylinder using the given diameter. Then since we have the pressure as well, we could plug the values in and get w.