Referring to Part (b)
••Assume [PV] is a state function:
PV = (5 L)(2 atm) = 10 L.atm
PV = (10L)(1 atm) = 10 L.atm
∆(PV) = [PV]2 - [PV]1 = 10L.atm - 10L.atm = 0 work (not true)
••Actual work = -P∆V which could not be equal to 0 since P is not 0, and ∆V = is not 0. P varies from 2 atm to 1 atm., and ∆V = 5L.
For more in depth information on this check:
http://www.chem.arizona.edu/~salzmanr/480a/480ants/pvwork/pvwork.html
I really don't get this. The calculations shown don't make sense.
Demonstrate that work is not a state function by calculating the work involved in expanding a gas from an initial state of 1.00: and 10.0atm of pressure to (a) 10,0 L and 1.0 atm pressure. (b) 5.00L and 2.00 atm and then to 10,0L and 1.00 atm pressure.
Solution
(a) w=-PV = -(1.00atm(10.0L-5.0L)
= -9.0L atm
(b) w=-PV = -(2.00atm)(5.0L-1.00L)-(0.0L- 5.0L)
=-13.0L atm
I don't see how the calculations correlate with what was given in the question.
Can someone please explain to me? Thanks!
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