October 31, 2024, 09:03:42 PM
Forum Rules: Read This Before Posting


Topic: Maximum work done ??  (Read 2452 times)

0 Members and 1 Guest are viewing this topic.

Offline AdiDex

  • Full Member
  • ****
  • Posts: 305
  • Mole Snacks: +16/-12
  • Gender: Male
Maximum work done ??
« on: April 07, 2016, 04:03:35 AM »
I was reading University Chemistry- Bruce H. Mahan 4rth edition . In Which it was mentioned that Work done for Irreversible process is maximum . I have googled , I found that work done for reversible process is maximum . But when I looked closely I saw in Bruce H mahan He considered dw = -PdV . It all about sign convention . In chemistry we define dw= -PdV whereas in Physics we define dw= PdV .
So in there must be difference in Maximum work done in both of the cases (since there is difference is sign convention ) .
Eg. In physics Work done for a reversible process is 5 and for irrevesible process is 3 (Maximum work done -> Reversible), Then in Chemistry the work done for irreversible process must -3 and for reversible process -5 (Maximum work done = Irreversible ).

Same question was asked in my exam , So gave the explanation and drew the graph .  But my professor has written in his notes that Maximum work done in for Reversible .

I need some references to disprove him . I am not able to find out . Or I am wrong ??

Offline Arkcon

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7367
  • Mole Snacks: +533/-147
Re: Maximum work done ??
« Reply #1 on: April 07, 2016, 05:24:46 AM »
We're actually coverin a very similar topic in this thread:  http://www.chemicalforums.com/index.php?topic=85364.0;topicseen
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline AdiDex

  • Full Member
  • ****
  • Posts: 305
  • Mole Snacks: +16/-12
  • Gender: Male
Re: Maximum work done ??
« Reply #2 on: April 07, 2016, 06:45:23 AM »
Ofcourse in the case of expansion the magnitude of work done done by reversible process will be maximum (It will be more negative than that of Irreversible work done) .

As the discussion is being done in that thread is about expansion but I am talking about process (not just only expansion but Expansion and Compression both )  , In case of compression , Magnitude of Irreversible process will be maximum . 


Offline idest

  • Regular Member
  • ***
  • Posts: 11
  • Mole Snacks: +0/-0
Re: Maximum work done ??
« Reply #3 on: April 07, 2016, 08:01:05 AM »
As I know, using different sign convention in chemistry and physics(especially engineering) arises from difference of interest. Chemists focus on the system itself where chemical reactions occur, but engineers pay attention to things the system makes.

If we choose the formula, [itex]dW=-P_{\mathrm{ex}}dV[/itex], then we regard energy transfers(in the form of work or heat to change the internal energy) coming in the system as postive(negative, otherwise). Thus, we should consider [itex]-W[/itex] if we want to talk about effects the system makes on the surrounding.

I guess Mahan's book explains two processes, nearly the isothermal expansion and compression. Let's check the expansion case first.

It is obvious that [itex]P_{\mathrm{int}} \geq P_{\mathrm{ex}}[/itex] (otherwise expansion won't occur) ,where the equality is satisfied in a reversible process. Integrating from  [itex]V_{1}[/itex] to [itex]V_{2}[/itex] makes

[tex] W_{\mathrm{rev}}=-\int_{V_{1}}^{V_{2}} P_{\mathrm{int}} \mathrm{d}V \leq -\int_{V_{1}}^{V_{2}} P_{\mathrm{ext}} \mathrm{d}V = W_{\mathrm{irrev}} [/tex]

However, we should note the sign of those quantities, [itex]W_{\mathrm{rev}} \leq W_{\mathrm{irrev}} <0 [/itex]. The system pushes the surrounding making it shrinked in the expansion process. That makes the system use its internal energy to the surrounding in the form of work, which is expressed as (-) sign.

The statement that work done for a reversible process is maximum is right when considering the work done by the system on the surrounding (energy transfer out of the system). Therefore, work done in a reversible process is larger than one in a irreversible process, i.e. [itex] -W_{\mathrm{rev}} \geq -W_{\mathrm{irrev}} >0 [/itex].

Then move on the case of compression. Obviously, [itex]P_{\mathrm{int}} \leq P_{\mathrm{ex}}[/itex] ,where the equality is satisfied in a reversible process. Then

[tex] W_{\mathrm{rev}}=-\int_{V_{2}}^{V_{1}} P_{\mathrm{int}} \mathrm{d}V \leq -\int_{V_{2}}^{V_{1}} P_{\mathrm{ext}} \mathrm{d}V = W_{\mathrm{irrev}} [/tex]

Now we get [itex]0<W_{\mathrm{rev}} \leq W_{\mathrm{irrev}} [/itex]. We can explain it reversely like: The surrounding pushes the system making it shrinked in the compression process. That makes the system given energy from the surrounding in the form of work, which is expressed as (+) sign.

In the view of the surrounding, the system steals its energy in the form of work, and the amount stolen is also expressed as [itex]-W [/itex]. Consequently, we get the conclusion, [itex] 0>-W_{\mathrm{rev}} \geq -W_{\mathrm{irrev}}  [/itex].

I edited some mistakes.

« Last Edit: April 07, 2016, 08:46:04 AM by idest »
To love and win is the best thing. To love and lose, the next best.

- William Makepeace Thackeray

Sponsored Links