Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Does the volume of a thermodynamic system always have to change for it to do work? If yes,could you explain why? And if no, could you provide the example of a system, where it is not neccesary.

share|improve this question
add comment

3 Answers

up vote 0 down vote accepted

Yes, because work is force times distance moved. It's not immediately obvious that this means work can only be done if the volume changes, but have a look at How much work is needed to compress a certain volume of gas? where I went into this in some detail. If any parts of this aren't clear comment here and I can go into them in more detail.

share|improve this answer
    
Nathanial and Juanrga make good points. I must admit that I was only thinking about mechanical work. –  John Rennie Nov 4 '12 at 14:06
    
John Rennie then can I say, "If a thermodynamic system consists of charges, and has an electric/magnetic field, then it is not necessary for the volume to change. But in the absence of charge, by the very definition of work (PdV), then yes, the volume MUST change." –  Adwait Kumar Nov 4 '12 at 14:52
    
@AdwaitKumar: possibly. The trouble is that if you accept a broad definition of work then it's hard to be sure you've excluded all the possibilities except for volume change. –  John Rennie Nov 4 '12 at 15:06
    
@JohnRennie: Thank you! –  juanrga Nov 6 '12 at 12:04
add comment

It depends on how you define work. Work is sometimes defined as pressure times a change in volume ($p\Delta V$, or $\int p dV$ if the pressure is not constant), which is equivalent to force times distance, as John Rennie defines it in his answer. In this case it's necessary for the volume to change, purely from the definition.

However, another way to define work is something along the lines of "a change in energy that is not due to the transfer of heat or thermal radiation." In this case it is not necessary for the volume to change. A particular example where the word "work" is used in this way is in electronics, when one calculates the work involved in charging a capacitor as $\int V dQ$, where $V$ here is the voltage rather than the volume, and Q is the charge. In this case neither the volume of the capacitor nor the battery charging it changes, but work is said to be done because the internal energy of the capacitor has changed reversibly, without a transfer of heat.

To put it more mathematically, consider the fundamental equation of thermodynamics: $$ d U = T dS - pdV + \sum_i \mu_i dN_i + \dots $$ (where the "$\dots$" can include many other optional terms, including a $V dQ$ one for a capacitor). The right-hand side of this equation represents all the ways in which a system's energy can change. The first term ($T dS$) represents a transfer of heat. Some people define "work" as just the second term ($-p dV$), whereas others define it as the sum of all the other terms apart from the first one. Thus, with the second definition you can have work that's associated with a change in volume, but you can also have work that's associated with a change in the chemical composition, charge, or any other conserved quantity.

share|improve this answer
    
The additional terms require either interchange of molecules with some other system (or non equilibrium for the to be not zero. –  Eduardo Guerras Valera Nov 4 '12 at 10:42
    
(sorry) for them to be not zero. So adding/subtracting species while in thermal equlibrium (or not) is another way of doing work on the system. –  Eduardo Guerras Valera Nov 4 '12 at 10:52
    
@Eduardo this is true. Note that this can in principle be done without a change in volume, however. For example, you could remove molecules of some species $A$ while adding molecules of some other species $B$ that happen to occupy the same volume as $A$ molecules, but have a different chemical potential. Then you've done $\mu d N$ work without a change in volume. –  Nathaniel Nov 4 '12 at 11:22
    
Yes, I only wanted to add some to your correct answer (I gave the point). Note too, that electric currents, macrowave oven and any other energy transfer with no volume change or species interchange, enters the equation in the Tds term and thus, is considered heat transfer. –  Eduardo Guerras Valera Nov 4 '12 at 11:49
    
Kelvin would have loved a microwave oven as a present... –  Eduardo Guerras Valera Nov 4 '12 at 11:51
add comment

Mechanical work $dW_\mathrm{mech} = -pdV$ is due to a volume change for a pressure $p$. But other kind of thermodynamic works exist:

  • Chemical work $dW_\mathrm{chem} = \mu dN$ involves change in composition $N$ for a chemical potential $\mu$.

  • Electrical work $dW_\mathrm{elect} = -E dP$ involves change in electric dipole moment $P$ in presence of an electric field $E$.

  • Magnetic work $dW_\mathrm{mag} = -B dM$ involves change in magnetic dipole moment $M$ in presence of a magnetic field $B$.

  • Etcetera.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.