0
$\begingroup$

Consider an ideal gas inside a cylinder-piston setting. Now, we know that the temperature of the system reduces under adiabatic expansion. For an ideal gas we know that the molecules move about separately possessing kinetic energy without having any potential energy between them.
One way I think to intuitively understand the reduction of temperature is that the total energy inside the system remains constant due to no exchange of heat but the volume increases due to expansion. Thus, the internal energy density decreases resulting in reduction of temperature.

My question is: whether there is any form of exchange of energy between the system and the surroundings in a form other than heat due to the movement of the piston as the gas expands or the reduction of temperature is purely due to increase in the volume?

$\endgroup$
2
$\begingroup$

One way I think to intuitively understand the reduction of temperature is that the total energy inside the system remains constant due to no exchange of heat but the volume increases due to expansion.

You're not thinking about it correctly. The total energy of the system, which in the case of an ideal gas is purely kinetic energy, decreases as a result of the expansion. This is due to the fact that the system does work and expends some of its internal energy in the process of doing so. From the first law,

$$\Delta U=Q-W$$

Where $Q$ is heat and is positive if heat transfers to the system, and W is work and is positive if done by the system. For the adiabatic expansion, $Q=0$ and therefore $\Delta U=-W$. In this case $W$ is positive when the system does work, which decreases internal energy. For an ideal gas, any process, $\Delta U=mC_{v}\Delta T$. So a decrease in internal energy results in a decrease in temperature.

My question is: whether there is any form of exchange of energy between the system and the surroundings in a form other than heat due to the movement of the piston as the gas expands or the reduction of temperature is purely due to increase in the volume?

Yes, the other form of energy exchange (transfer) besides heat is work. As explained above, that's why the internal energy and temperature decreases when the gas expands adiabatically. When the gas expands it transfers internal energy to its surroundings if no heat transfer to the system is possible to supply energy for doing work.

I am confused whether the U in this equation is the total internal energy or internal energy density?

$U$ in these equations is the total internal energy. The specific internal energy, or energy density as you call it, is the total internal energy divided by the mass. It is usually shown as a lower case $u$. Then in the first law equation you would also have lower case $q$ and lower case $w$, meaning heat and work transfer per unit mass of the system. In the equation $\Delta U=mC_{v}\Delta T$ if you divide by the mass, you get the specific internal energy, or $\Delta u=C_{v}\Delta T$.

Secondly, even if so energy is transferred out of the system in the form of work done by the gas, the increase in volume also must contribute to the decrease in temperature from a purely intuitive point of view since the temperature depends on the internal energy density rather than just the total internal energy. Is my reasoning correct?

No that is not correct because temperature does not depend on the internal energy density rather than the total internal energy. It's the same for both.

Temperature is what we call an intensive thermodynamic property, meaning it does not depend on the amount of mass or volume of the system. Pressure is another intensive thermodynamic property.

Let's say you have a room full of air and its temperature is 20 $^0$C. If you divide the room in half each half has half the volume and half the mass of air of the whole room, but the temperature of each half of the room is still 20 $^0$C.

If we take the equation

$$\Delta U=mC_{v}\Delta T$$

where $\Delta U$ is the change in total internal energy and divide it by the mass, we get

$$\Delta u=C_{v}\Delta T$$

Where now $\Delta u$ is the specific internal energy (energy density). But $\Delta T$ is the same in both equations.

Follow up question: doesn't halving the room also halves the total internal energy? But the temperature remains the same.

Yes each half has half the total internal energy because each half has half the mass of the total but the energy density of each half is the same and the temperature of each half is the same. That's because internal energy is an extensive property, meaning it depends on the mass. But temperature is an intensive property and does not depend on the mass. That's my point.

Here’s another perspective. The temperature of the room full of an ideal gas is a measure of the average kinetic energy per molecule of the gas molecules in the room. If you divide the room in half each half has half of the total kinetic energy (half the total number of molecules and thus half the internal energy), but the average kinetic energy per molecule in each half is the same so the temperature is the same.

Hope this helps.

$\endgroup$
7
  • $\begingroup$ Thanks for defining the positives of Q and W in your 1st law equation. People often skip that. $\endgroup$
    – Bill N
    Sep 11 '20 at 16:31
  • $\begingroup$ I am confused whether the U in this equation is the total internal energy or internal energy density? Secondly, even if so energy is transferred out of the system in the form of work done by the gas, the increase in volume also must contribute to the decrease in temperature from a purely intuitive point of view since the temperature depends on the internal energy density rather than just the total internal energy. Is my reasoning correct? $\endgroup$
    – D.Mason
    Sep 11 '20 at 17:11
  • $\begingroup$ @D.Mason I have updated my answer to respond to your follow up questions. Hope it helps. $\endgroup$
    – Bob D
    Sep 11 '20 at 17:41
  • $\begingroup$ Follow up question: doesn't halving the room also halves the total internal energy? But the temperature remains the same. And doesn't the fact that temperature is an intensive quantity mean that it only depends on the energy density? $\endgroup$
    – D.Mason
    Sep 11 '20 at 17:46
  • $\begingroup$ @D.Mason I have updated once again to answer your follow up question. $\endgroup$
    – Bob D
    Sep 11 '20 at 17:55
1
$\begingroup$
  1. The energy of an ideal gas independent of volume. This is due to lack of distance-dependent potentials between the molecules of the gas.
  2. Under the kinetic theory perspective, the molecules of the gas have net kinetic energy associated with it. The total internal energy is distributed among the gas molecules as a gaussian function dependent on temperature. (Refer maxwell -Boltzmann distribution curve)
  3. As we increase the temperature, the width of the gaussian increases but the peak speeds decrease . This means that the speeds more evenly distributed with fewer molecules having dramatically different speeds.

Edit: As correctly pointed out by D.mason, the statement that peak speed decrease is wrong. The correct statement is just that the distribution of speeds of molecules become more spread out as we increase the temperature.


Summary: The volume, for an ideal gas, does not have any control over actual internal energy. You could have rapidly moving molecules in a container of any arbitrary size. The energy changes associated with temperature changes should be interpreted using the kinetic picture of molecules.

Caveat: What I said above is only true for an ideal gas. If it's not an ideal gas, then the internal energy function may have a dependence on the volume and you'd have to involve the joule Thompson co-efficient in your calculations of internal energy


Refer ChemLibre text for further readings on maxwell boltz-man distribution.

Also, refer to this other chemlibre text chapter for more information about joule Thompson co-efficient


Further notes: The process by definition can not exchange heat with surroundings due to it's adiabatic nature.


If any part is not clear, feel free to comment and I'll revise my answer accordingly.

$\endgroup$
7
  • $\begingroup$ I understand the fault in my reasoning now. I thought temperature to be dependent on the internal energy per unit volume when in fact it depends on the internal energy per unit mass. $\endgroup$
    – D.Mason
    Sep 11 '20 at 18:12
  • $\begingroup$ But I don't understand why you said the peak speed decreases as temperature increases . Did you mean that the number of molecules having the most probable speed decreases? $\endgroup$
    – D.Mason
    Sep 11 '20 at 18:15
  • $\begingroup$ That is another question to discuss entirely xD but you can check that it's true from the links that I have provided. By the way, I still suggest that you use the kinetic theory picture to think of temperature, as in the temperature is an indirect measure of how fast molecules are moving here and there in the container. $\endgroup$
    – Buraian
    Sep 11 '20 at 18:19
  • $\begingroup$ From the link you gave I see that the probability of the most probable speed decreases. $\endgroup$
    – D.Mason
    Sep 11 '20 at 18:21
  • $\begingroup$ Ah, my mistake I'll fix that. I got confused between the speed which the maximum number of molecules have and the actual maximum speed. (It only increases with temperature as you have noted ) $\endgroup$
    – Buraian
    Sep 11 '20 at 18:23
1
$\begingroup$

If there was no energy exchange between the ideal gas and the surroundings, there would be no change in temperature, because the temperature is the gas total (kinetic) energy divided by the number of molecules (and a constant). Volume has no role to play in that.

The exchange happens at the piston. Any time a molecule hits the piston it transfers a part of its kinetic energy to it. You can see the molecules as responsible for moving the piston, loosing energy during this process.

$\endgroup$
0
$\begingroup$

For free expansion of an ideal gas there is no change in temperature. For a real gas, there is a slight change in temperature after free expansion, but this is little importance in most engineering applications. (See discussions of the Joule experiment of 1843.)

If the expansion is not free- specifically for your case expansion with an external pressure on the piston- the adiabatic expansion process reduces the internal energy of the gas due to the work the gas does in moving the piston against the external pressure, and assuming an ideal gas, since internal energy is a function of temperature only the gas cools. Adiabatic means no heat transfer, but work can be done during an adiabatic process. You can model this process using the first law of thermodynamics applied to the closed system defined as the mass of gas (no mass transfer in/out of a closed system); here, the boundary of the closed system (at the piston) moves. See any book on thermodynamics, such as any of the textbooks by Sonntag and van Wylen.

Bottom line: for adiabatic free expansion of the piston the gas slightly cools but this is usually a negligible change. For a non-free adiabatic expansion the gas cools noticeably due to work done by the gas on the surroundings.

This discussion assumes no presence of condensable vapors in the gas. For example, room air normally contains water vapor (amount dependent on the humidity) and if saturation conditions are reached during a process, the vapor can condense to a liquid. One of the supposed benefits of filing auto tires with nitrogen instead of air is to remove the effect of condensation of water vapor on lowering tire pressure.

$\endgroup$

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.