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In my textbook it says for an ideal gas, as volume decreases at constant pressure, there is a decrease in temperature. At first, I fully agreed because as volume decreases the temperature decreases by the gas laws, but then when I tried to apply $P\times{\Delta{V}}$ to the process, it appears that $\Delta{U}$ increases as the volume decreases, which does make sense since decreases volume at constant pressure should intuitively increase some type of energy; in this case internal energy.

Obviously, this is my faulty reasoning but the contradiction arises when one takes not that if $\Delta{U}$ is positive, then $\Delta{T}$ is positive as well. This tells us that T increases.

Where did my reasoning go wrong and what is the correct way of view this isobaric process?

EDIT:

So, when looking at Greiner Neise thermodynamics and statistical mechanics, the general equation is reported as:

$\ dH|_p = \delta Q|_p + \delta W^{rev}_{other}|_p$

Where the $\ W^{rev}_{other}$ is utilizable work, not simply volume work against the constant external pressure.

I apologize, I wanted to edit my own but edited this instead. Please remove if not suitable.

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  • $\begingroup$ You're overthinking this. Look at the $pV$ diagram for an ideal gas: en.wikipedia.org/wiki/… . The curves are isotherms for ever increasing temperatures. Watch what happens to $T$ for isochoric ($V=$constant) or isobaric processes ($p=$constant). $\endgroup$ – Gert Aug 4 '16 at 0:11
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Far be it from me to challenge pro physicists who would provide you with a much more formally correct answer, but I, being as an applied metallurgical scientist, will give you my take on it: From my experience, sometimes even pretty seasoned physio-chemists and engineers can loose their way in the 'forest of just three trees' - classic simple mono-atomic ideal gas processes. To avoid mistake for me it's always helped to imagine a concrete experiment with concrete conditions to avoid mistake when talking about isobaric, isothermal etc process. Let's imagine a very simple isobaric process that you can perform. You have a syringe half empty (piston half way inside), and inside the syringe chamber is just air - or suppose helium (as monotonic gas). The piston is oiled very well so there is almost no friction with the walls. You plugged the needle orifice, then you took a usual lighter (the one we use for cigarettes), lit it, and slowly drew the flame to your syringe (not too close, so that the heat could be transferred approx. in a steady manner through the wall of the syringe chamber to the inside helium). What you will see, is that the piston will start slowly crawling out of the syringe and the volume of the part of the syringe with helium will correspondingly start increasing - the volume of helium growing. What's going on? Typical isobaric process, for the piston (hence helium) produces work against the outside pressure, which is 1 atm (~1 bar). The system needs to produce work AGAINST the environment (which already has 1 atm pressure, just like helium inside by the way) just to increase its own volume. You transferred from your lighter an amount of heat Q, which went into increase of ΔU of the helium and the work, which equals PΔV (conveniently P is constant by definition and equals in our case just 1 atm or 100000 Pa). You can see easily that NOT all of your heat went into work (that's possible only for isothermal process), so that internal energy U increased, which automatically means that temperature INCREASED too for an ideal gas (the quantity of atoms not changing). At the beginning of your experiment the volume of helium was Vo, and its energy according to ideal monoatomic gas laws was Uo = 3/2nRT (n- number of moles of helium, R is universal gas constant); because PVo = nRT for ideal gases you can see that Uo = 3/2PVo - that's at the start before setting the lighter. Hence the volume during the experiment went up from Vo to V, that automatically means that U went up from 3/2PVo to 3/2PV (P being constant). Which means that the new temperature of your helium will be (2/3)*(3/2PV)/(nR) = PV/(nR). Increase in temperature will be accordingly ΔT = P(V-Vo)/(nR) or simply PΔV/(nR).

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  • $\begingroup$ @IanLimarta In your example imagine the reverse process of what I described w lighter and syringe. Now, after th first experiment your syringe is warm (or even hot), you started to cool down that syringe ever so slowly - by itself in the ambient air. And it will start contracting again slowly back to Vo: now the environment performs work on your syringe, BUT for that to happen th syringe should give off MORE heat that the energy it gets from that work, so at the end you end up with negative change in U. That work is NOT ENOUGH to compensate for th loss of heat. $\endgroup$ – Philipp Aug 4 '16 at 1:52
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If the compression process is adiabatic and reversible, then the pressure will increase as the volume decreases, and the temperature will also increase.

If the adiabatic compression is carried out irreversibly, then viscous stresses contribute to the force per unit area on the piston, and the force depends not only on the volume but also on the rate of change of volume. We call an irreverisble adiabatic compression process isobaric if the external force per unit area applied the piston is constant throughout the process. If the piston is massless and frictionless, then, at the piston face at least, the force per unit area of the gas acting on the piston matches the force per unit area applied externally. But neither is determined by the ideal gas law applied to the overall volume of gas. This is because not only are viscous stresses present within the gas (to contribute to the force per unit area), but also the pressure within the gas is not even uniform spatially. So, in an irreversible compression, the force per unit area on the piston can only be controlled externally.

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The ideal gas law gives you the answer for this problem. For $PV = nRT$ , with constant number of moles, with R also being constant, the problem statement said that the system is held at constant pressure. The only way to do this is to decrease temperature as volume is decreased. In other words, P, n, and R are constant, which makes this a linear equation in V and T. As V decreases, T decreases. Another way to look at it: as V decreases, heat must be transferred out of the system to hold the pressure constant. This is NOT an isothermal or adiabatic process.

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  • $\begingroup$ Then how do you explain how a constant pressure yields a work done of $P\times{\Delta{V}}$?(or area I suppose) I thought this was still applicable even for an isobaric process? Is there a greater heat loss still that outweighs the work done on the system? $\endgroup$ – Ian Limarta Aug 4 '16 at 0:47
  • $\begingroup$ @IanLimarta Yes, that's exactly it. $\endgroup$ – knzhou Aug 4 '16 at 0:50
  • $\begingroup$ @IanLimarta , delta-U equals heat or work entering the system, minus heat or work exiting the system. P is constant, delta-V is not, so there was work done on the system in decreasing the volume, but there was ALSO heat transfer out of the system. The details would tell you if delta-U increases, decreases, or remains the same. $\endgroup$ – David White Aug 4 '16 at 2:10

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