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Setting a general background for my question:

We know that the equation of state for an ideal gas is given by:

$$pv=RT$$

where:

  • $p$ is the absolute pressure of the gas.
  • $v$ is the volume per unit mass of the gas.
  • $R$ is the gas constant.
  • $T$ is the absolute temperature.

We also know that, following Maxwell equations, the internal energy solely depends on the temperature: $$e=e(T)$$

Introducing enthalpy ($h\equiv e+pv$) and using the equation of state for an ideal gas, we can conclude from Maxwell equations that the enthalpy also depends solely on the temperature: $$h=h(T)$$

Question:

The book I'm using to study Thermodynamics of propulsion, gives a word of warning:

Both the internal energy and the enthalpy can change in the complete absence of heat.

We are still considering ideal gases. However, heat is closely related to change in temperature, and both $e$ and $h$ depend solely on $T$.

My question might be considered a bit basic (I'm starting with Thermodynamics), but I can't quite understand the word of warning the authors gave. How is it possible that $h$ and $e$ change in complete absence of heat, if both depend on $T$? Can you give an example?

Trying to solve my own question:

One example I can think of where $e$ would change in absence of heat is when compressing a gas (I don't know if this is true at all). But I don't have a way to visualize the enthalpy.

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  • $\begingroup$ If you're looking for a physical interpretation of enthalpy, don't spend too much time looking. Enthalpy is just a convenient function to work with in solving many kinds of thermodynamics problems. In my judgment, it is best just to think of h as a function defined by the equation you gave. If e changes, then, for an ideal gas T must have changed, so h must have changed. h has changed as a result of e changing and also as a result of pv changing. $\endgroup$
    – Chet Miller
    Commented Mar 6, 2016 at 14:23
  • $\begingroup$ Thanks for the clarification @ChesterMiller. I'm also looking for an answer to my question above. $\endgroup$
    – Jose Lopez Garcia
    Commented Mar 6, 2016 at 14:39
  • $\begingroup$ As you yourself noted, if the gas is compressed (or expanded) adiabatically (i.e., "without heat"), it will do work on its surroundings (or vice versa) and its internal energy, enthalpy, and temperature will change. $\endgroup$
    – Chet Miller
    Commented Mar 6, 2016 at 16:00

2 Answers 2

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Don't associate heat too closely with temperature. That's a connection we make in everyday language, but can be dangerous is physics.

Better: don't make that association at all. Connections between the two exist, for example in the equation for specific heat, but relationships such as that only relate two different concepts.

Your own answer to your question is correct.

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  • $\begingroup$ Does this mean that a gas can exchange heat, $Q$, without changing its own temperature $T$? How? Thanks. (Can't upvote because I don't have enough reputation). $\endgroup$
    – Jose Lopez Garcia
    Commented Mar 6, 2016 at 14:59
  • $\begingroup$ Yes, if at the same time work is done. (For ideal gases only. The situation can be different for real gases. If a phase transition is possible, then $Q$ can be transferred and the temperature remains constant even in the absence of work.) $\endgroup$
    – garyp
    Commented Mar 6, 2016 at 15:01
  • $\begingroup$ @garyp is absolutely right. Even better, do not use the word "heat" as a noun, but only as a verb, that is in the sense of "to heat" and then no mistake will be made. And on occasions when you see it used as a noun substitute in your mind immediately the word "heat" with the words "internal energy exchanged because of temperature gradient". $\endgroup$
    – hyportnex
    Commented Mar 6, 2016 at 15:42
  • $\begingroup$ Thank you both. Excuse my insistence, but I have yet another question: does that mean that (for an ideal gas) Work implies Heat transfer? (assuming a non-adiabatic boundary and a "non-slow" process) $\endgroup$
    – Jose Lopez Garcia
    Commented Mar 6, 2016 at 15:57
  • $\begingroup$ I suppose it does. If the boundary can conduct thermal energy, then if you do work the thermal energy will increase, increasing the temperature, after which transfer of thermal energy because of temperature gradient (heat) is inevitable. $\endgroup$
    – garyp
    Commented Mar 6, 2016 at 16:15
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A change in pressure at a constant temperature causes a change in enthalpy H.

A sudden change in pressure, like popping a balloon does cause a drop in temperature, to occur at constant temperature the change must happen slowly, like a slowly leaking tyre - the miniscule drop in temperature per second is balanced by other heat from the surrounding air flowing in.

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  • $\begingroup$ Hey @Arif, thanks for your analogy with the balloon, made things simpler to visualize. $\endgroup$
    – Jose Lopez Garcia
    Commented Mar 6, 2016 at 15:58

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