[georg gill]

for a system with p=0.8 at sea level p=1 in surroundings andsurroundings pushes back system until p system is 1 and one calculate work by

[tex]w=-p_{surr}dV[/tex]


if p system is p=1.2 and it still is at sea level so p surr is 1  one still uses p surr to calculate work even though this time it is system that is pushing on surroundings until p system is 1. How come one cnat use p system. I mean it is unpractical but if one could one would get a work witrh pressure from 1.2 to 1 at same dV meaning a more negative w. How is that?

From definition of w

m=mass a=acceleration l=length that piston moves  A=area V=volume

Fl=w   mal=W   malA/A=w    plA=w  pdV=w

if one uses nonconstant p one have to integrate

  int(mdadl)=w    i am not sure how to write this how could one integrate it get work?

[wolfzhchk]

Why we choose the pressure of surronding is because that makes the calculation more easier.



The equation is actually a negative value of work that the surronding does to system. First of all, it should be clear that work is not state functionthe value of work depends on the what path to go. Consider a expansion process, the gas itself is not in equilibrium, which means there is no certain T, V or p which can describe a certain state of the system. Three values of them may probably change all the time. Therefore, strictly speaking, we can only discuss 'partial pressure' of a specific part in the system. If you choose the 'partial pressure' around the piston to calculate the work, that requires complicated hydrodynamics calculation.

Furthermore, if you get insight into the path of the expansion in more details, you will find the piston is vibrating affected by the gas, not 'smoothly' moving from oneside to another. The gas is like a oscillator, continuously doing positive and negative work for the system until equilibrium is obtained, in which the pressure of gas ceaselessly changes.

Wherefore, the main reason why we choose external pressure of the system is because the pressure of surronding can be treated as unchangeable value. Although the expansion work is still not the state function,  we can easily attain the expansion work by using initial and final state of the system which is the volume change of gas.

[juanrga]

for a system with p=0.8 at sea level p=1 in surroundings andsurroundings pushes back system until p system is 1 and one calculate work by

[tex]w=-p_{surr}dV[/tex]


if p system is p=1.2 and it still is at sea level so p surr is 1  one still uses p surr to calculate work even though this time it is system that is pushing on surroundings until p system is 1. How come one cnat use p system. I mean it is unpractical but if one could one would get a work witrh pressure from 1.2 to 1 at same dV meaning a more negative w. How is that?

From definition of w

m=mass a=acceleration l=length that piston moves  A=area V=volume

Fl=w   mal=W   malA/A=w    plA=w  pdV=w

if one uses nonconstant p one have to integrate

  int(mdadl)=w    i am not sure how to write this how could one integrate it get work?

I am not sure that one has to use external p to calculate mechanical work. I find this disturbing. For me p would be the pressure of the system. Although know the difficulties when applying this to vacuum expansion, I believe that could be solved.

[juanrga]

for a system with p=0.8 at sea level p=1 in surroundings andsurroundings pushes back system until p system is 1 and one calculate work by

[tex]w=-p_{surr}dV[/tex]


if p system is p=1.2 and it still is at sea level so p surr is 1  one still uses p surr to calculate work even though this time it is system that is pushing on surroundings until p system is 1. How come one cnat use p system. I mean it is unpractical but if one could one would get a work witrh pressure from 1.2 to 1 at same dV meaning a more negative w. How is that?

From definition of w

m=mass a=acceleration l=length that piston moves  A=area V=volume

Fl=w   mal=W   malA/A=w    plA=w  pdV=w

if one uses nonconstant p one have to integrate

  int(mdadl)=w    i am not sure how to write this how could one integrate it get work?

I am not sure that one has to use external p to calculate mechanical work. I find this disturbing. For me p would be the pressure of the system. Although know the difficulties when applying this to vacuum expansion, I believe that could be solved.

Gary L. Bertrand gives four different protocols for calculating mechanical work in «Thermodynamic Calculation of Work for some irreversible processes» [Journal of Chemical Education 2005: 82(6), 874-877]

Protocol 1 is ## W=- \int P_{opp} dV ##

Protocol 2 depends if the process is stopped by external factor or not

Protocol 3 is ## W=- \int P_{ext} dV ##

Protocol 4 is ## W=- \int P_{gas} dV ##

He states that 3 is used in most physical chemistry texts. He suggests an experiment where protocol 3 is inconsistent. He repeats the common claim that 4 is incompatible with expansion of a perfect gas into vacuum. He proposes a new protocol ## W=- \int f_{opperating} dl ##. I find all this completely misleading, but the article is interesting.

[georg gill]

I am not sure that one has to use external p to calculate mechanical work. I find this disturbing. For me p would be the pressure of the system. Although know the difficulties when applying this to vacuum expansion, I believe that could be solved.

what if p internal was zero then would not

\( W=- \int P_{ext} dV \)

be defined as zero while it is not if one uses formula w=-pdV with p being p external

[juanrga]

I am not sure that one has to use external p to calculate mechanical work. I find this disturbing. For me p would be the pressure of the system. Although know the difficulties when applying this to vacuum expansion, I believe that could be solved.

what if p internal was zero then would not

## W=- \int P_{ext} dV ##

be defined as zero while it is not if one uses formula w=-pdV with p being p external

I do not understand your question.

After studying Gary L. Bertrand JCE paper cited above it is now clear to me that ## W=- \int P_{ext} dV ## is wrong [1].

He gives two (not only one as I said above, but two) experiments with gases [2] where application of that expression gives wrong results, whereas the expression ## W=- \int P_{gas} dV ## gives right results in agreement with experimental results.

For vacuum expansion the process is not quasistatic and the quasistatic work has to be corrected to ## W=- \int P_{gas}(1 -\frac{u}{v}) dV ## with u being piston velicity and u an average molecular velocity. For vaccuum expansion u is of order of v and W is zero although pressure is not zero.

For quasistatic processes u << v and we recover the quasistatic o ## W=- \int P_{gas} dV ##. He states that only a chemical engineering textbook gives the correct expression for work, whereas most physical chemistry books give the incorrect expression using ## P_{ext} ##.

[1] Moreover it would be also incompatible with kinetic theory of gases.

[2] One experiment is about expansion and the other about compression.

[georg gill]

are u saying my textbook zumdahl is wrong

p.364

http://books.google.no/books?id=2OxrDtDaSqIC&pg=PR18&lpg=PR18&dq=zumdahl+chemical+principles+online+version&source=bl&ots=Br0UJmkYSz&sig=qCfZm5ODF6ZHNsd0hW-RjPBWRXg&hl=no&sa=X&ei=hWAtT5acBafE4gSQ-oGrDg&ved=0CEQQ6AEwBDgK#v=onepage&q&f=false

would then be misleading i believe.

[juanrga]

are u saying my textbook zumdahl is wrong

p.364

http://books.google.no/books?id=2OxrDtDaSqIC&pg=PR18&lpg=PR18&dq=zumdahl+chemical+principles+online+version&source=bl&ots=Br0UJmkYSz&sig=qCfZm5ODF6ZHNsd0hW-RjPBWRXg&hl=no&sa=X&ei=hWAtT5acBafE4gSQ-oGrDg&ved=0CEQQ6AEwBDgK#v=onepage&q&f=false

would then be misleading i believe.

Yes, that textbook is wrong, as many others. As Gary L. Bertrand explains in «Thermodynamic Calculation of Work for some irreversible processes» [Journal of Chemical Education 2005: 82(6), 874-877]

##W = - \int P_{ext} dV ##

gives wrong answers for two experiments where ##W = - \int P_{gas} dV ## gives the correct results. He reports that only a textbook in chemical engineering gives the correct expression for the work  ##W = - \int P_{gas} dV ##, whereas many physical chemistry and engineering textbooks give the wrong formula.

Of course, for quasistatic processes  ## P_{gas} = P_{ext} dV ##, and you can still apply the formula in your book, but you cannot apply the formula in your book to arbitrary examples. I repeat for the experiments reported in the above JCE article, the formula in your book disagrees with experimental results.