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If a system with const. pressure can change it's volume free, a heat transfer into the System results in a partial work done, which gives back the inner energy addet partial back to the environment. (That was roughly the introduction of a book on enthalpy.)

ΔU < q

However, I now wonder why the following does not apply:

+q = |-w|

(suppose i have a gas in a container with a piston, there is mechanical equilibrium, the internal pressure is equal to the external pressure. Shouldn't +dq (energy added to the system in the form of heat) not directly convert to -dw (that the piston moves?))

So shouldn't all the energy that was added in the form of heat end up completely and not just partly in volume work?

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  • $\begingroup$ What book are you referring to, and what is the exact text? $\endgroup$ May 13, 2022 at 23:37
  • $\begingroup$ @Chemomechanics It's from Atkins Physical Chemistry 11th Ed. Here is a screenshot of the page: imgur.com/a/hJrs9PT $\endgroup$
    – iwab
    May 13, 2022 at 23:48
  • $\begingroup$ Atkins allows the possibility that the system heats up (or changes phase, for instance, or otherwise exhibits an increase in internal energy). This means that not all the energy entering through heating can necessarily be removed through work; some energy may stay in the system. Does this resolve the issue? $\endgroup$ May 14, 2022 at 0:05
  • $\begingroup$ @Chemomechanics Not quite :/, if you say that the system has the possibility to heat up, this heating up must lead directly to a volume change, because if p=const. pV=nRT the volume change depends only on T. Do you have an example where increasing the internal energy does not lead to a volume change (if p=const.)? Actually, that can't happen, because according to the law of equal distribution, it only depends on T. $\endgroup$
    – iwab
    May 14, 2022 at 0:33
  • $\begingroup$ Why are you applying the ideal gas law? I don't see that assumption made anywhere. In any case, even assuming an ideal gas, if the volume increases and the temperature rises, then the gas has both done work and increased in internal energy. The heat input must equal the work done plus the increase in internal energy, and so the increase in internal energy must be less than the heat input, as Atkins writes, and the heat input does not exactly equal the work output ($|q|\neq |w|$). $\endgroup$ May 14, 2022 at 0:47

2 Answers 2

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When we heat a system at constant pressure, its volume tends to change. (For many materials and many temperature ranges, the volume tends to increase, but thermal contraction can occur, as with liquid water over the temperature range of 0–4°C.) As a result, the system does work (or has work done on it). The first law reads

$$\Delta U=Q+W$$

(or, in differential form, $$dU=q+w=q-P\,dV=q-P\alpha V\,dT,$$

where $U$ is internal energy, $q$ and $w$ are infinitesimal heat and work inputs, $P$ and $V$ are pressure and volume, and $\alpha$ and $T$ are the thermal expansion coefficient and temperature, respectively).

Thus, we can't write $Q=-W$ because we're controlling the pressure—not the internal energy—to be constant. (For that matter, we can't write $\Delta U<Q$ unless $\alpha>0$, which Atkins implicitly assumes when transitioning from his mention of expansion and contraction to expansion work only.)

(The content in bold represents important content for recalibrating your intuition of constant-pressure heating.)

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  • $\begingroup$ Ok, i've always thought (under the given conditions) that internal energy have to be proportional to the internal pressure. Can you maybe give me a molecular interpretation, how an increasment of internal energy doesn't need to cause more momentum exchange between molecules and piston in a gas? $\endgroup$
    – iwab
    May 14, 2022 at 11:29
  • $\begingroup$ Is it maybe because only translation cause the piston to move and others like rotation, oscillation doesn't? $\endgroup$
    – iwab
    May 14, 2022 at 11:41
  • $\begingroup$ And in condensed matter, you get additional energy stored in springs between molecules. $\endgroup$ May 14, 2022 at 17:27
  • $\begingroup$ But the whole thing also applies in an ideal gas, where there are only the translational degrees of freedom. You say "because we're controlling the pressure—not the internal energy—to be constant", that's right, but my question is, why isn't the internal energy constant. I imagine, that +dq accelearates the molecules, so U rises, therefore moving the piston to keep pressure constant. While moving the piston they lose the same amount of previously gained extra speed (momentum), U falls again, and therefore U is the same as before. $\endgroup$
    – iwab
    May 15, 2022 at 14:09
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However, I now wonder why the following does not apply:

+q = |-w|

It applies if the change in internal energy $\Delta U$ is zero. Then, from the following version of the first law for a closed system (no mass transfer)

$$\Delta U=Q-W=0$$ $$Q=W$$

So shouldn't all the energy that was added in the form of heat end up completely and not just partly in volume work?

Again, it can if $\Delta U=0$. For example, for a isothermal (constant temperature) expansion of an ideal gas, any process, $\Delta U=nC_{v}\Delta T$. For an isothermal process $\Delta T=0$ and $Q=W$.

Hope this helps.

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