2
$\begingroup$

Let us say I have a liquid of mass $m$ at its boiling point and add heat to increase to cause turn it all into a vapour. Under what conditions will the heat I actually add be equal to: $$Q=ml$$ Where $l$ is the specific latent heat of vaporization of the liquid.

$\endgroup$
1
$\begingroup$

The equation $$Q=ml$$ Only holds in a system at constant pressure (isobaric) and constant temperature (isothermal) and generically it is not true.

This can be shown as follows:

If we look at the change in the internal energy (which is a state function) of the system: $$dU=dQ+dW$$ We see that for the above process we have: $$U=mL-p(V_2-V_1)$$ Where $V_1$ and $V_2$ are the initial and final volumes respectivly (with generally $V_2>V_1$)*. Now take a different process which has the same intial and final state but which does it in the following steps: 1. Free expansion to the final volume. 2. Isochoric (constant volume) vaporization of all of the liquid In this process no work is done but we still have the same change in internal energy (since it is a state function). In this case all that energy must come from the heat supplied and is thus simply not equal to $ml$ as one may naively expect. (i.e. for this process the heat added would be: $$\tilde Q=mL-p(V_2-V_1)$$ Which is actually less then that for the standard isobaric, isothermal phase change.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can you please elaborate on the "different process." I'm having trouble understanding how you can get to the same final state doing this. You start out with liquid in a closed container (either filling the container or with vapor above in equilibrium) and you add heat to vaporize all the liquid. Does this occur at constant temperature to arrive at the beginning of step 2? If so, how (since the pressure has to be increasing)? $\endgroup$ – Chet Miller Jan 7 '16 at 13:01
  • $\begingroup$ @ChesterMiller Thanks for your comment, I wasn't really thinking about the details when I wrote it :). I believe the points you raise go way if we let step 2 becomes step 1 and vice versa (I will change it in my answer). Starting with a liquid and vapour in equilibrium we can perform a Joule expansion on the gas so its volume is $V_2$ (for an ideal gas this will keep the temperature constant). We then add heat to evaporate more liquid until we once again reach the equilibrium pressure. $\endgroup$ – Quantum spaghettification Jan 7 '16 at 13:17
  • $\begingroup$ During this process there will be variations in temperature and pressure, but we can just stop when we get to that of the final state of the first process $\endgroup$ – Quantum spaghettification Jan 7 '16 at 13:17
  • 1
    $\begingroup$ That makes much more sense. So the heat for the new version of the "different process" is just equal to the change in internal energy in going between the initial and final equilibrium states. I still don't get that $=p(V_1-V_1)$. I think it should just be $\bar{Q}=\Delta U$ $\endgroup$ – Chet Miller Jan 7 '16 at 14:40
  • $\begingroup$ @ChesterMiller That was a typo, you are correct. $\endgroup$ – Quantum spaghettification Jan 7 '16 at 14:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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