Conceptual understand of why gases cool when expanding?

[sillyjoe]

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Can anyone please explain? @gettheleadout you are usually AWESOME at explaining things like this!

 

[pepocho]

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[whoadude]

Can anyone please explain? @gettheleadout you are usually AWESOME at explaining things like this!

Is this always true? What about isothermal expansions of ideal gases?

 

[pepocho]

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[Gauss44]

5:20


I think # of collisions with container determine pressure for an ideal gas.

 

[sillyjoe]

I believe: When gas expands, bonds are typically broken. Energy is required to break bonds. When energy from the environment is absorbed into the system (to break bonds), the surroundings become cooler. W=-PΔV, E=W+Q

Bonds are broken? What bonds?

 

[el_duderino]

I believe the way I thought about this is in terms of work. A gas expanding does work on its environment, and the energy for that work comes from the gas itself.

 

[Gauss44]

Bonds are broken? What bonds?

Scratched that. Replaced with video.

 

[pepocho]

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[gettheleadout]

Can anyone please explain? @gettheleadout you are usually AWESOME at explaining things like this!

For whatever reason I didn't receive a notification that I was tagged here (I just double checked too), sorry!

I've read through the thread so far and I just have a little bit to add and I'll do so later today, I haven't slept tonight so right now isn't a great time for me haha.


Sent from my neural implant using SDN Mobile

 

[gettheleadout]

So @jonnythan is on the right track here in my opinion. The best way to approach the concept of temperature and gas expansion is from a thermodynamic perspective. First I want to establish that for ideal gases temperature is a measure of the internal energy of the system (since ideal gases have no intermolecular potential energy.) If the temperature of the system changes, then energy has flowed into or out of the system. The two ways for energy to be transferred? Heat and work.

Not him, BUT I'll try my best! What is temperature? Temperature is the average kinetic energy in atoms. Some people like to define temp as "How fast atoms are moving and bumping into each other." As you increase temp, the average KE increases (and thus atoms collide more often) and vice versa. Turns out that by increasing pressure or decreasing volume, you can increase the temperature of a gas. Why is that? Well in the case of decreasing volume, the atoms themselves have less space in order to avoid collisions. So now they begin colliding more, and temp increases.

This is somewhat misleading. The rate of collision is only useful as a proxy for the rate of movement of the particles (which itself represents their kinetic energy). Simply observing two different same-gas systems with different rates of intermolecular collision is not sufficient to say the systems have different temperatures (the systems could have different numbers of moles, different volumes, etc.) When volume is decreased the system is being compressed, meaning work is being done on the system by the surroundings. This represents a transfer of energy into the system, and so the system's temperature increases. I will present an alternative conceptualization of this later on.

On the other end, when you increase volume, the space between atoms increases, collisions are less likely to occur! I hope this helps. This also works for pressure AND temperature. Pressure is the defined as the sum of collisions against the container(Force/area) in which the gas itself is in. So as temperature increases, KE increases, number of collisions against the container increases, and pressure increases. And vice versa

Again the key here is that the frequency of collisions is not directly relevant. An increase in the volume of the system can be thought of either as work done by the system on the surroundings (e.g. a balloon expands when I pull it out from underwater into the air) or negative work done on the system by the surroundings (e.g. I pull up on a gas-filled piston and the gas inside experiences a volume increase). Either way, you observe energy flow outward from the system via work.

Well in that case, no. However, in isothermal(which means same temperature for those who don't know) environments, other conditions(pressure, volume, and other factors) are changing. For isothermal conditions to work under changing conditions for example lets say, if one condition is changed (let's volume was increased), another condition must also change to counteract that change to avoid temperature change(perhaps the scientist will increase pressure to negate the effects of the increased space). So, the scientist will begin tinkering with pressure, volume, to avoid the spike or drop in temperature. My example would not apply. Usually, I am sure, the question or the information in the passage will provide sufficient information to let you know what kind of conditions the gas is in.

The missing factor here is heat flow. For isothermal processes, heat flow occurs in parallel to work performed in order to keep the internal energy of the system constant (∆U = 0 and Q = – W). Take your example of the increased system volume and pressure being increased to maintain the same temperature. How would one increase the pressure while increasing volume? Heat the system. So, it's not so much that the scientist is tinkering with the physical parameters so much as it is that he is controlling heat flow.

So I mentioned an alternative conceptualization of the temperature change. Let's approach it from a kinetic molecular perspective rather than a macroscale thermodynamic one. Recall the two examples of gas expansion I gave above: a system expanding against the surroundings due to pressure differential (balloon expanding as external pressure drops) and a system where the surroundings effectively contract (a piston rising by an applied force; the system itself is not causing the volume of its container to increase.) If you're comfortable with the overall flow of energy via work (∆U = – W) then let's look at it on the level of the gas particles. For ideal gases, the concept of kinetic temperature relates the temperature of a gas system to the average kinetic energy (translational only) of the particles. Also recall that one of the ideal gas postulates is elastic collision between particles and with the walls of the container. Let's consider the balloon first. The pressure differential (a compressed gas in the balloon has pressure P_g > P_atm) drives the gas system to expand until the pressures equalize. The pressure is also defined as the force per unit area upon the inside of the container walls. Thus, as the container expands (and its volume increases) the internal surface area increases, making the denominator of the pressure expression larger and lowering the pressure. Thus the pressure drops as it expands until P_g = P_atm. But why does the temperature of the system change? The key here is that temperature change isn't always related to heat flow. Recall that elastic collisions are characterized by conservation of not only momentum but energy as well. As each gas particle hits the inside of the container wall and pushes it out just a little further, that incoming gas particle is transferring some of its momentum and thus some of its kinetic energy to the container wall. As the particle bounces off, it flies away with slightly less kinetic energy than it had before the collision. As this happens often enough the energy distribution of the population of gas particles changes and the average shifts downward. This is the decrease in temperature.

Now lets hit the second example, the piston. In this case the walls of the container are literally pulling back away from the incoming gas particles flying at them. It's not that the gas particles are transferring enough momentum to the container walls to expand the container (though they are transferring some), it's that the container walls are moving away of their own accord. Consider an incoming gas particle. In a container of static volume the relative velocity between the particle and the wall is equal to the particle's velocity; the wall isn't moving relative to the particle. Recall that for elastic collisions the coefficient of restitution (e = |v_rel,f| / |v_rel,i|) is 1. This means the relative velocity will stay constant, and so, with the container volume constant, the wall will remain at rest and the particle will bounce away with an equal magnitude velocity as it had when it approached. The particle loses no speed and thus loses no kinetic energy, and this holds for every particle, so the average KE of the system doesn't change and temperature is constant. Now consider the forcibly expanding system. The relative velocity between the particle and the wall is no longer equal to the particle velocity, because the wall is retreating from the incoming particle. The magnitude of the relative velocity is now lower than the magnitude of the particle's velocity alone. When the collision occurs, because it is elastic and e = 1, the particle must bounce away with a speed such that the magnitude of the relative velocity between itself and the wall is the same as before the collision. Since the wall is still moving with the same velocity as before the collision, this means the particle flies away with a lower magnitude velocity than before the collision. The particle has lost KE and thus as this occurs the average KE of the system drops, and temperature goes down.

Consider a third case, where the system is forcibly compressed. Can you see how the magnitude of the relative velocity between the wall and incoming particle before the collision is now greater than the magnitude of the particle velocity? Thus, post-collision the particle flies away at a greater speed than it approached, in order to satisfy e = 1. The particle's KE has increased and as this happens, the system average KE increases and temperature goes up from compressing the gas. No heat flow required. 😀

The app that Bozeman uses in the video Gauss posted is pretty fun to play around with too: http://phet.colorado.edu/en/simulation/gas-properties

 

[pepocho]

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