# What happens to the pressure and the temperature of a gas compressed adiabatically?

The set-up is basic. There is a piston that goes down, reducing the volume. The process is adiabatic, so the acquired energy is kept inside the gas. The ideal gas law states that as the volume decreases the fraction $$\frac{T}{P}$$ has to decrease as well. Pressure has to increase since we compressed the gas. My question is what will happen to the temperature ?

Thank you!

• If the system is perfectly insulated, then the tempo does not matter, as long as the compression speed is below the Mach number. – Drew Aug 5 at 11:58
• 1. the pressure goes up not down. 2. the question "what part goes in [increasing] the pressure" does not seem to me to make any physical sense; please try to frame your question again, more clearly. – Andrew Steane Aug 7 at 12:20
• Section 1–2 in the Feynman Lectures on Physics has an interesting discussion of what is happening at the microscopic level when compressing a cylinder of gas in this way, and why both temperature and pressure are increasing. – Erlend Magnus Viggen Aug 20 at 13:38

## 4 Answers

Well, in every ideal gas model I'm familiar with internal energy is directly proportional to temperature. And, since there is no heat exchange with the outside world in an adiabatic process ($$\text{đ}Q=0$$), the first law of thermodynamics $$\mathrm{d}U= \text{đ}Q-\text{đ} W$$ implies that all of the work goes into the internal energy of the gas, so it all goes to increasing the temperature.

Long story short, the pressure cares about two things: how fast the molecules in the gas are moving, and how often the molecules hit the side of the container. Heuristically, the higher the temperature, the faster the molecules move (not proportional), the more molecules there are the more often they hit the container walls, and the farther apart the container walls are the less often molecules hit them.

Throw in that pressure is force divided by area, and a few more assumptions about what makes a gas 'ideal' and you can derive from kinetic theory that $$P = k \frac{N T}{V}.$$

Looking at that process, you could say that whatever fraction of the work went in to making the molecules move faster went into making the pressure higher, and the rest of the energy didn't, but all of the energy increases the temperature.

In a little more detail, the equipartition theorem states that the internal energy of a body is given by $$U = \frac{N_{\mathrm{df}}}{2} kT$$ where $$N_{\mathrm{df}}$$ is the number of "quadratic degress of freedom", which is a fancy way of saying terms in the microscopic description of the energy that are the square of some coordinate (e.g. $$\frac{1}{2}m v_x^2$$ and $$\frac{1}{2}k x^2$$). Because there are three components to velocity, kinetic energy of movement always contributes $$3N$$ to $$N_{\mathrm{df}}$$. Other ways to increase $$N_{\mathrm{df}}$$ include energy bonds between atoms (like springs) and the ability of molecules to rotate (rotational kinetic energy).

Since the pressure only cares about the kinetic energy of translation for gases, you could say that $$\frac{N_{\mathrm{df}}}{3N}$$ is the fraction of the energy that increases pressure, but this is a little misleading. That is because part of equipartition is that all of the quadratic degrees of freedom get an equal share of the energy. If you could somehow pour energy into just, say, the kinetic energy degrees of freedom the system wouldn't be in equilibrium anymore, and random/chaotic processes would work to equalize the energy per degree of freedom again.

In other words, even though the energy that's not in the kinetic degrees of freedom doesn't directly increase the pressure, its presence is necessary to keep the energy that does at the level it is at.

My question is what part of the obtained energy goes into the temperature build-up, and what part goes in reducing the pressure?

If the process is adiabatic and the piston goes down the volume decreases but the pressure increases. The pressure does not decrease. This is called an adiabatic compression. If the process is carried out very slowly and without friction it can be considered reversible. The equation for the process is then

$$Pv^{ϒ}=C$$

where $$C$$ is a constant and ϒ is the ratio $$\frac{C_p}{C_v}$$. The initial and final states are then related by

$$P_{f}v_{f}^{ϒ}=P_{i}v_{i}^{ϒ}$$

So you see if the volume goes down the pressure goes up.

As has already been pointed out, since the process is adiabatic we have $$Q=0$$ and therefore

$$\Delta U=-W$$

In this version of the first law $$W$$ is negative when work is done on the gas (compression), to the change in internal energy is positive.

And for an ideal gas, any process, the change in internal energy depends only on temperature according to

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

All the energy transferred to the gas due to external work goes into increasing its temperature. But the temperature of an ideal gas is a measure of the average kinetic energy of the molecules so the average kinetic energy increases. This causes the molecules to collide with the walls of the container at higher average velocities. That in turn causes the pressure on the walls of the container to increase as well.

Bottom line: The pressure increase is a consequence of the energy that caused the temperature increase. It is not a separate part of that energy.

Update:

This will respond to your follow up question.

How a gas will respond to a volume reduction, that is a compression process will depend on the type of gas and how the compression process is carried out. The process in your example is an ideal gas adiabatic compression process and I gave you the equation for that process. The actual increase in temperature and pressure depends on the actual gas, since different gases have different specific heats that are in the equations I gave you.

A different ideal gas compression process is an isothermal process in which there is a volume reduction and pressure increase but no temperature change. The equation for that process is

$$pv=Constant$$

Which is different than the adiabatic process.

In summary, in order to know how a volume reduction affects temperature and pressure you need to know the specific gas properties and the specific equation(s) that governs the compression process.

Hope this helps.

• Thank you! That indeed helped! – CHILLQQ Aug 5 at 17:21
• I by no means tried to say that pressure is a part of energy. I wanted to say that it increases (not decreases of course: that was another really stupid mistake of mine) as a consequence of the acquired energy. What I couldn't understand is how to estimate how a gas will respond to the volume reduction: some gases may heat up more than others etc. – CHILLQQ Aug 5 at 17:29
• @CHLLQQ I have updated my answer to respond to your follow up question. Hope it helps – Bob D Aug 5 at 21:23

Odd question really, how can energy go into pressure? Pressure has different units, force/area not forcexdistance. The area in question is orthogonal to the distance. Work goes into volume or temp changes that then affect pressure. Temperature however, is a measure of kinetic energy.

• You are totally right of course. All I wanted to say that the pressure increases (not decreases of course: that was another really stupid mistake of mine) as a consequence of the acquired energy. I should be more careful with my questions. – CHILLQQ Aug 5 at 17:31
• how can energy go into pressure ahem. – Kyle Kanos Aug 6 at 11:35
• Well you give some energy to the particles; they speed up (temperature rises). But since they have more energy, they collide with the walls with higher momentum which means more pressure. Energy does not literally go into pressure, but the pressure goes up as a consequence of acquired energy. Does that make sense ? or do I not get something? – CHILLQQ Aug 7 at 16:54

Expanding Sean Lake's explanation,

Say the gas is argon. Argon atoms bounce off each other, and maybe it makes sense to say they spin.

Say the gas is oxygen, O2. Oxygeon molecules definitely spin, and they also vibrate compressionally -- they get close to each other and farther apart.

Say the gas is hydrogen cyanide, HCN. It spins, it has two kinds of compression, and it bends.

The more complicated the molecule, the more different ways it can absorb energy in addition to increasing the molecule's linear velocity. So it can increase its temperature more without increasing its volume as much.

We can't make one simple rule that works for all gases. Because they are not in fact ideal gases. But the ideal gas assumption works pretty well usually except in high pressure etc. The idiosyncracies for each gas tend to come out to roughly a constant fudge factor. Usually.