## 8.3 6th Edition

Alexa Tabakian 1A
Posts: 38
Joined: Fri Sep 28, 2018 12:20 am

### 8.3 6th Edition

I have no idea how to approach this problem. If somebody could please explain how to solve how much work is done, I would really appreciate it.

Joon Chang 2F
Posts: 59
Joined: Fri Sep 28, 2018 12:25 am

### Re: 8.3 6th Edition

Work due to a change in volume is called expansion work. The equation is $w= -P_{ex}\Delta V$ where P is pressure, and Delta V is change in volume.
For part a) you are given the pressure and the dimensions to find the volume of the pump. Volume of cylinder is $\pi r^{2}h$. So you find the volume of the pump and negate the value because in the problem, the volume is being compressed. Insert the value for pressure that was given and multiply by the negated value of volume, giving you the work.

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