if gas heats up when condenses why does it become a liquid?

[GASPER20]

I am assuming if a gas warms when it is compressed, it also gains energy? (energy~temp).

If it gains energy then why do gases at high pressure and constant temperature lose energy and become liquid? Wouldn't they gain energy and stay in the gas phase?

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[Morsetlis]

Depends on the process (adiabatic, isovolumetric, etc.) and where it is on the phase diagram. Increasing pressure while keeping temperature constant means squeezing the molecules closer together (decreasing volume), which drives it more toward cohesion -> precipitation.

I can't think of how you can increase pressure, while keeping temperature constant, without also decreasing volume, so there you go.

 

[GASPER20]

thanks for the response. looking at the phase diagram helps a lot. but i don't understand why it would lose energy when it compresses? when a gas compresses it warms, so doesn't that mean it gains energy?

 

[RogueUnicorn]

it's a phase change, not gas compression

 

[KD1655]

Remember, there is some

H (enthalpy/heat/energy) associated with condensation. This enthalpy is numerically equal to the heat of vaporization of the material. Grpahically, these enthalpy values correlate to the plateus on the energy vs. temperature graph. The energy can be calculated:

Q= n*

H​

Based on your equation, it seems that you are assuming that a transition from gaseous form to liquidius form requires some decrease in temperature. However, a true phase transition is isothermic and isobaric; if you are familiar with partial differential equations, it is essentially a boundary layer.

If you are just posing this question for MCAT purposes, then disregard the information below because it is probably a bit advanced for the MCAT.

Since you are interested in the pressure, temperature, enthalpy/energy, and a phase transition, you can utilize the Clausius-Clapeyron, which is a partial differential that describes dP/dT. This equation can be derived by knowing that two phases of matter at the same temperature and pressure, ie the plateua described above, have the same Gibb's free energy (G1=G2). The change in Gibb's free energy of a system is described: dG= VdP-Sdt. Since dG is zero because dP and dt are zero, the relationship G1=G2 is confirmed. We can assume that dG1=dG2. Therefore, we can apply the equation for dG to this equality:

V1dP-S1dt=V2dP-S2dt​

Following some algebraic manipulation: dP/dT= dS/dV. Another thermodynamic relationship that you should be familiar with is dS= dQ/T. During a phase transition, dQ= n*dH (as described above by the delta function). For simplicity sake, lets assume that the molar quantity in this case is n=1. Therefore, we have the dS=dH/T for the transition. Therefore, we can rearrange to yield teh Clapeyron equation:
​

dP/dT= 1/dV(dH/T)= dH/TdV​

 

[GASPER20]

im still confused. so decrease in volume doesn't mean gas compression?

if you're increasing the pressure and keeping temperature constant, then volume has to decrease. if volume decreases, aren't you compressing the gas?

here's the answer from the book, "as pressure of a gas is raised at constant temperature, there comes a point at which many molecules no longer have sufficient kinetic energy to remain in the gas phase. at this point, the gas begins to condense into liquid."

my question is why would they lose energy? i guess the increase in temperature when a gas compresses is a result of the exothermic process of condensation and not that their kinetic energy is increasing?

 

[Flapjacks]

The energy is stored as potential energy in chemical bonds and intermolecular forces, which is why there is no temperature change.

 

[KD1655]

There is also kinetic energy which is a function of the of molecular vibrations. These vibrations "slow" as the temperature decreases.

 

[fizzgig]

my understanding... years old but there you go.. I feel like looking at these forum questions is a great motivation to get off my butt and study… I had no idea how much I’d forgotten…

1) decrease in volume does not necessarily = 'compression', no. if you cool a gas in a box it can decrease in vol with constant pressure, so you’re not compressing (pushing harder to hold the box shut) but the density does increase.

2)if you do increase pressure by decreasing volume, gas tends to heat up (volume change is work on the system, and Uenergy = Qheatloss/gain + Work, so if heat loss isn’t allowed, the work done by volume change goes to internal energy, thus Temp).

my understanding of the book statement is that heat is allowed to leave the box. if you compress a box of gas, you do work on it, and isothermally the energy has to be lost as heat. if you let that heat escape you'll end up with an isothermal but more dense gas.

Since you’re right, and energy ~temp, as you keep smushing the box and releasing Q, that’s the 'loss' you’re thinking about. The pressure goes up bc you have the same number of items bouncing around the box, but the volume is smaller, even though their energy is not increased. Eventually, still having the same energy (T) as they had earlier in the process, but running into each other many times more often because of the increased gas density, molecules are likely to stick together and become liquid. at this phase they have about the same energy as liquid or as gas, so it's just a probability game as to whether they'll escape the water surface and if they do when they’ll condense back. this is at the beginning of condensation.

I hope that was semi understandable. I would have just PMed but I think I need it to be out here so if there’s stuff I got wrong others can correct it for you (and me).

I fear the mcat.

 

[KD1655]

Does it mention whether the system is isolated, closed, or open? That certainly makes a difference in fizzgig's response. If you are unaware of the difference, an isolated system does not allow the transfer of anything to its surroundings; a closed system allows the transfer of heat and energy to its surroundings but not matter/mass; in an open system, anything goes. I don't know what the "assumption" for thermodynamics on the MCAT is; it wouldn't make sense for it to be an isolated system because that is much less prevalent than open or closed systems but I have stopped trying to figure out the rationality of the MCAT by now. In reality, neither a closed nor an isolated system are actually possible.

 

[docelh]

I am assuming if a gas warms when it is compressed, it also gains energy? (energy~temp).

If it gains energy then why do gases at high pressure and constant temperature lose energy and become liquid? Wouldn't they gain energy and stay in the gas phase?

Two factors determine the physical state of a compound: temperature and pressure. Looking at only temperature is viewing this problem through one half of the lens. The simple answer is that it depends on whether the effects of the pressure or temperature is greater.

Increasing pressure via compression is transferring energy into the system via work. This reduces entropy. In simpler terms, the molecules are forced closer together, and their increased intermolecular attraction counteracts the kinetic energy from the added temperature. Assuming this is a closed system, the compound must establish equilibrium with its vapor pressure. The compression raises that vapor pressure, and thus, the bar is higher --> the ones that don't meet the bar stay in the liquid state.

 

[BerkReviewTeach]

I am assuming if a gas warms when it is compressed, it also gains energy? (energy~temp).

I'll toss my hat into this ring too. Meaning no disrespect to the above replies, I think there is a strong tendency to overcomplicate things. While they all explain the phenomenon, the goal on the MCAT is to keep things simple and answer the question quickly.

The scenario you describe is based on conversion of energy from one form into another form. In this case from chemical potential into heat energy. As a gas is compressed (open, closed, isolated or whatever), the particles get closer together. While no bond is formed per say, it is similar to the bond-forming process because intermolecular attraction will increase. This is assuming the gas is made up of molecules that can condense if the pressure is high enough (a highly probable assumption). Bond formation is an exothermic process, so as the particles get closer together, they release heat energy. In essense, chemical potential energy has been converted into heat energy. The molecules in the system absorb this thermal energy and thus experience an increase in their temperature. If the system is open, then this heat energy readily escapes. If the system is isolated, then the system experiences an increase in temperature due to this extra heat energy.

A phase change of a gas into a liquid driven by compression is just an extreme case of this example.

If you are having trouble with this example, you can always consider the reverse process. If you expand a gas, then it should become colder. You have no doubt felt this first hand (pun intended) when holding an aerosol can and releasing the contents. It gets cold to the touch, because as the gas particles remaining in the can move away from one another, it is in essense a bond-breaking scenario, and therefore is endothermic. The molecules in the system absorb the surrounding heat energy so they can expand, making it feel cold to the touch (the system is stealing your heat). Hopefully this helps with the visualization process.

If it gains energy then why do gases at high pressure and constant temperature lose energy and become liquid? Wouldn't they gain energy and stay in the gas phase?

The key constraint you added to your question is constant temperature. As the gas is compressed, heat will be released into the environment. That would raise the temperature. The only way to maintain the temperature would be to remove that heat (cool the system by allowing heat to escape to the environment). Because you have allowed for the loss of heat to the environment, you have in fact taken the system to a lower energy state, which explains the phase change.

 

[GASPER20]

thanks for the responses. so my understanding is that the increased (excessed) in temp is loss to the surroundings while the original heat~kinetic energy in the system is converted to stronger bonds and thus the molecules loses kinetic energy.

thinking in terms of entropy also helps because as you increase pressure, less entropy is favored. thus, gas will become liquid, which is more stable.

 

[Flapjacks]

You have the second part right. However, the increased temperature is due to the surroundings doing work on the system ( you stated that the volume decreased E = q - P * delta V).