## 6th edition problem

Tatum Keichline 2B
Posts: 64
Joined: Fri Sep 28, 2018 12:26 am

### 6th edition problem

I'm not understanding how to approach problem 8.3 in the 6th edition textbook. I was sick during this lecture, so I am behind on this section and I'm not understanding how 24J is the answer.

sharonvivianv
Posts: 63
Joined: Fri Apr 06, 2018 11:05 am

### Re: 6th edition problem

We haven't covered work in lecture yet.

The answer is actually 28J for a.

First, they based the calculations off of w=-P*deltaV

To find the volume they used deltaV=-pi(r^2)(d)=-pi(1.5^2cm)(20cm)(1L/10^3cm^3)= -0.14L
Volume is negative because it was compressed to a smaller volume.

They plug in the deltaV: w=-(2.00atm)(-0.14L)(101.325J/L*atm) = 28J

This should all make more sense by next week, hopefully.

Kevin Tang 4L
Posts: 83
Joined: Fri Sep 28, 2018 12:28 am

### Re: 6th edition problem

Looking through the textbook may help you a lot while doing this problem. Look at example 1 on page 264 of edition 6.
At the bottom of the example, it says Related Exercises: 8.3

To solve this problem, we first need to get the area which is pi(r)^2. They give the diameter which is 3cm. 3/2=1.5cm which is the radius. Plug into area equation to get 2.25pi which you mulitply by the length which is 20cm in order to get 45pi. Then multiply by the pressure, 2 atm, to get 90pi which is equal to 282.74/10(used cm instead of m) to get 28joules.

I did this out of order so I hope you don't get confused by this answer. Hope this helps