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This question already has an answer here:

I have read in Zeemansky's physics

$dQ=dU+pdV$ for first law of thermodynamics

But when I came across another book of thermal physics,it says

$δQ= dU +pdV$. So what us the difference ? enter image description here

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marked as duplicate by joshphysics, Jon Custer, John Rennie energy Nov 30 '17 at 6:48

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The notation $\delta Q$ is sometimes used to denote the fact that there may not be a well defined function $Q$. It simply means "small increment". It is used to underlie the distinction between say the energy $U=U(p,V)$, which is a function of state and stuff like $Q$ and $W$, which are not functions of state. So for example I can write: $$dU = \frac{\partial U}{\partial p}dp+\frac{\partial U}{\partial V}dV$$ You can't do this with $Q$, say, because it's not a function of state.

If one wants to be more mathematically precise, both are mathematical objects called differential forms. Only $dU$ is an exact form, $\delta Q$ is not.

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  • $\begingroup$ +1: Probably more clear to most readers than my answer. $\endgroup$ – joshphysics Nov 29 '17 at 20:47
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For heat, the difference is purely notational. In both cases, infinitesimal heat transfer is being modeled as a differential form -- the authors are simply using different notations for that same differential form.

Often authors rightly prefer to use "$\delta$" to indicate that heat is an inexact differential form so the "$d$" can be reserved for exact forms which, by definition, can be written as the differential (aka exterior derivative) of another form. See, for example, closed and exact differential forms

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  • $\begingroup$ See the edit I made. You could use $\delta V$ if you really wanted to -- after all you can use whatever notation you want as long as you know what you're talking about, but that would probably not be the most informative notation because by using the "$d$" in that case, you are making explicit the fact that $dV$ is an exact differential form. $\endgroup$ – joshphysics Nov 29 '17 at 20:40
  • $\begingroup$ Yes that's basically it. I would be careful, however, not to call $Q$ or $W$ themselves "exact" or "inexact." It's the corresponding differential forms to which that terminology applies. And yes, your last equation is standard. In fact, it's more correct/informative than something like $dQ = dU + dW$. $\endgroup$ – joshphysics Nov 29 '17 at 20:45