Chemistry
posted by Lauren .
Consider an ideal gas encloesd in a 1.00 L container at an internal pressure of 10.0 atm. Calculate the work, w, if the gas expands against a constant external pressure of 1.00 atm to a final volume of 20.0 L.
w = ____ J
now calculate the work done if this process is carried out in two steps.
1] first let, the gas expand against a constant external pressure of 5.00 atm to a volume of 4.00 L.
2] From there, let the gas expand to 20.0 L against a constant external pressure of 1.00 atm.
w = _____ J
i know that the formula to solve this is w=(p)(Delta V) but when they give you 2 pressures does that mean the formula turns to w=(delta p)(delta v)?

1)
Pdelta V = 1 atm x (v2v1) = 1(201) = 19 L*atm work.
2)
The first one is 5 x (41) = 15 L*atm
Second stage is 1 x (204) = 16 L*atm
Total = 31 L*atm
Here is a good site you can read; in fact there is an example problem that is ALMOST the same as this one.
(Broken Link Removed) 
Thank you Dr. Bob! That was very helpful

No prob. I always help sexy ladies ~

2=2
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