Textbook Question 4A.3

[Mia_Renna_1L]

Hi, I am confused on how to do this problem. I do not remember learning to calculate deltaV like this in class, so can someone explain this / do we need to know how to do this? I am also confused on the (101.325 J/L x atm) in the work equation.

I also am confused on b and c. If someone can help, that would be great. Thanks!

[Matthew Bolton 3L]

I can help at least with part a. You can calculate the change in volume using the diameter and length depressed of the pump. These values are 3.0 cm and 20. cm respectively. The equation for the volume of this cylinder is V=π(r^2)h. Now that you know the change in volume, you can multiply it by the pressure to find the work done.

Edit: I realized I left out an important step. After you have made this calculation, you must multiply it by 101.325, the amount of kilopascals in an atmosphere of pressure. This is simply a unit conversion.

[Shubhreet Bhullar 3C]

Hi, I can help you with part b. So the work done is positive, because the work is done on the system. The work is done against the pressure in the pump, so the Pex is going to be negative, so Pex delta V will be positive and so work will be positive.
As for part c, I thought that it would be the same as the work done on the system since there is no q. But the textbook says that the answer is 8J not 28J. I don't know if this is a typo or not.

[Nils Newman 1F]

There is a typo in the textbook, part c should also equal 28J, because there is no heat being transferred so deltaU = work. You can see it in the Solution Manual Errors at the course website.