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)?
-Pdelta V = -1 atm x (v2-v1) = -1(20-1) = -19 L*atm work.
The first one is -5 x (4-1) = -15 L*atm
Second stage is -1 x (20-4) = -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 ~