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Enthalpy as a state function

by flamcsd
Tags: enthalpy, function, state
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flamcsd
#1
Jun8-12, 05:27 AM
P: 4
Consider gas here as idea gas.

The gas expands from state 1: P1, V1 and T1 to state 2: P2, V2, and T1 using two different paths:

Path A: reversible expansion at constant T
Path B: irreversible expansion by releasing the gas to a vacuum to achieve V2 at adiabatic condition.

Thing I confuse: consider that enthalpy as a state function: H1 to H2 from state 1 to state 2.

for reversible path A: dU = q + w. dU = 0. so q=-w. The system needs some q from surrounding to perform w. and dH = q. H2 = H1 + dH = H1 + q.

but for Path B: w=0, q=0, dU=0. dH=0. So, H2 = H1.

Why? my question is enthalpy can't be same for path B, because enthalpy is a state function which is independent from its path.
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Studiot
#2
Jun8-12, 07:16 AM
P: 5,462
How did you arrive at w = 0 for the second expansion?

Don't forget that full expansion into a vacuum implies that P2 = 0 & V2 = ∞

The work done is pv work which is the area in the pv indicator diagram shown.

Between some point C at P3, V3 (say 1 atmosphere) and vacuum (unattainable) the work is shown by the hatched diagram.

If you take it all the way to infinity as shown the work is the area under the whole graph.

You need to integrate an equation of state for the gas to obtain this figure.
Attached Thumbnails
pvwork1.jpg  
flamcsd
#3
Jun8-12, 07:20 AM
P: 4
Thanks for your reply

the V2 in path B is assumed equal to V2 in path A. therefore, Path B and Path A could both reach a same state, P2, V2, and T1.

the vacuum part is V2-V1, it's like lift a barrier and let the gas go to extra V2-V1 part and reach V2. I don't want V2 go infinite.

Studiot
#4
Jun8-12, 09:12 AM
P: 5,462
Enthalpy as a state function

The point of state variables is that at any state there is one and only one value available to each state variable.

So to compare enthapies, start and end points have to be the same for both processes.
You started well by noting this.

At every point on each path there will be a well defined P and V.

You didn't answer the question about why you think w = 0 for the adiabatic irreversible path.

Path B and Path A could both reach a same state, P2, V2, and T1.
You are trying to specify P, V and T. You only specifiy two of these the third is not independent it is determined by the other two already specified
flamcsd
#5
Jun8-12, 09:47 AM
P: 4
Hi Studiot,

Thanks for the discussion. To answer your quesion:

"why you think w = 0 for the adiabatic irreversible path"

my path B is a free expansion path. a free expansion path does zero work.
see Free_expansion in wikipedia.

And actually, I only try to specify two variables P and V, where T is kept constant as T1. sorry for not clear.
Studiot
#6
Jun8-12, 10:12 AM
P: 5,462
Wiki also says

.................................which implies that one cannot define thermodynamic parameters as values of the gas as a whole. For example, the pressure changes locally from point to point, and the volume occupied by the gas (which is formed of particles) is not a well defined quantity.
flamcsd
#7
Jun8-12, 10:21 AM
P: 4
Quote Quote by Studiot View Post
Wiki also says
yes, I am not picking any state variable in the middle of the free expansion. what i want to do is comparing the state function H enthalpy at the start and the end of the free expansion.

apparently, the enthalpy change for a free expansion is zero.

However, by taking another reversible Path A at constant T. the enthalpy change is not zero.

The thing is the start state (P1, V1, T1), end state (P2, V2, T1) of Path A and B are exactly the same. I want to expect the same enthalpy change for path A and B.

however, this is not the case. This confuses me a lot if we take enthalpy as a state function, who is independent of the paths.
Studiot
#8
Jun8-12, 11:46 AM
P: 5,462
OK what you are asking about is known as Joules experiment.

The second part of this might help your question (and save me a lot of writing).

http://wikieducator.org/The1stLawofT...ynamicsLesson2


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