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April 1, 2015

April 1, 2015

Posted by **Xi** on Friday, November 28, 2008 at 8:55pm.

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!

- Chemistry -
**GK**, Saturday, November 29, 2008 at 1:41amReferring 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

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