Work

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Victoria Liang 3L
Posts: 11
Joined: Fri Sep 26, 2014 2:02 pm

Work

Postby Victoria Liang 3L » Fri Jan 23, 2015 8:06 pm

Question #3 in workbook

Consider a gas at an initial pressure of 5.0 atm and an initial volume of 1.0 L that undergoes a change to 2.0 atm and 4.0 L by two different pathways. The gas is at 298 K.

Pathway 1:
P1 = 5.0 atm P2 = 5.0 atm P3 = 2.0 atm
V1 = 1.0 L V2 = 4.0 L V3 = 4.0 L

Pathway 2:
P1 = 5.0 atm P2 = 2.0 atm P3 = 2.0 atm
V1 = 1.0 L V2 = 1.0 L V3 = 4.0 L

Calculate the work for these two pathways..


The solution says
work = -p (delta)V
Pathway 1: -5 atm (3L)
Pathway 2: -2 atm (3L)

I don't understand why they are different pressures...

Kayla Denton 1A
Posts: 106
Joined: Fri Sep 26, 2014 2:02 pm

Re: Work

Postby Kayla Denton 1A » Fri Jan 23, 2015 10:06 pm

It's easy to visualize on the graph! If you plot the two pathways with pressure on the y axis and volume on the x axis, the work is simply the area under the graph.
If you're having trouble with that I'll try to explain it in words too. So w = -P x delta V. So you want to find a volume CHANGE and the constant pressure at which this change occurs.

Path 1 has the volume change at P = 5.0 atm so this is the pressure used for P x delta V.
Path 2 has the volume change at P = 2.0 atm so this is the pressure used for P x delta V.
Delta V is the same in both cases.

Hope this helps!

Justin Le 2I
Posts: 142
Joined: Fri Sep 26, 2014 2:02 pm

Re: Work

Postby Justin Le 2I » Sat Jan 24, 2015 7:59 pm

Also, for this problem there needs to be two separate pathways because doing both at the same time is too difficult. It is easier if we split them up into two parts, one where the pressure changes and one where the volume changes. This is similar to when we have to separate irreversible changes into multiple reversible changes.


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