Strictly speaking, a gas whose behavior follows the ideal gas law. The concept of an ideal gas is an approximation often used in physics and chemistry in order to simplify calculations. In general at moderate conditions a gas can be assumed to behave as an ideal gas. Under high pressures and low temperatures (as determined by their relation to the gas in question's critical temperature and critical pressure) certain assumptions made by the ideal gas model no longer hold true, and you will have a nonideal gas on your hands.

*That's great, but what really makes an ideal gas ideal?*

In order to understand the nature of an ideal gas it's important to discuss the Kinetic Theory of Gases. Kinetic Theory in a nutshell:

- Gases are composed of discrete particles (either molecules or atoms) and these particles are in a constant
state of motion (That's the kinetic bit).
- When these particles collide with each other it results in a perfectly elastic collision, and the force of these
collisions is the only force between particles of the gas.
- The sum total of the volume of these particles is much smaller than the volume being taken up by the gas itself,
therefore it can be assumed these particles have (effectively) no volume.

Using these base assumptions and Newtonian Mechanics, one can derive your fundamental gas laws:

By combining all of these we can arrive at our Ideal Gas Equation: `P*V` = `n*R*T`, where
`R` is the ideal gas constant.

This means that the state of an ideal gas has three degrees of freedom. That is to say, you can define the state of the gas by assigning a value to any three of the terms, then the fourth can be calculated. Often times this equation is reduced further by assuming just one mole of gas, leaving `P*V` = `R*T`, and only two degrees of freedom. Since for most processes gas pressure and temperature are much easier to measure than volume, often the system of a gas is simply described in terms of its temperature and pressure.

*So we have all these little particles bouncing around, now what?*

Well, let's think about what we know, and how we can expand on it:

*(For the sake of those of the non-scientific persuasions, I will try to avoid pure mathematical derivations, and use conceptual ones where I can)*

**The Speed of Particles in an Ideal Gas**
We know that our ideal gas has a pressure. Pressure is a force divided over an area and we know the only force our little gas particles have comes from their collisions. Given that the kinetic energy of a body is `1/2*mass*velocity`^{2}, it is possible to find a relation between the force these particles are exerting and the speed they must be traveling. Thanks to the work of some of math and science's heavy hitters (Bernoulli, Maxwell, and Boltzmann) a probability function was found to show the distribution of velocities that models the behavior of our gas.

**Ideal Gases and Thermodynamics**

While the ability to have a simple equation of state is an attractive idea in and of itself, from a thermodynamic perspective, applying the assumptions of ideal gases can greatly simplify things.

**Internal Energy of an Ideal gas**

Internal energy is the amount of energy that is stored at the atomic level of a molecule. This may be in any of a variety of forms, including but not limited to: vibrating bonds, shaking molecules, or energy from intermolecular forces. Internal energy is a state function, which means it is solely dependant on the current conditions of the system. Given that, our internal energy (we'll go ahead and call it `U`) must be some function of temperature, pressure and volume (since we can use those to define our system state), giving us `U(P,T,V)`. As stated earlier, if we assume the amount of gas is constant we can remove one of these variables, so now let's call it `U(T,V)`. Since one of our initial assumptions was that the only forces acting on our particles of gas are those from physical collisions, then it doesn't matter from an energy standpoint how close these particles are to each other. Therefore the total internal energy of a given ideal gas would be unaffected by any state change that only affected the volume. This means that the internal energy of an ideal gas is dependant upon only one variable, temperature, giving us `U(T)`.

Before we move on, let's take a moment to reflect on the ideas of energy contained by an ideal gas and our ideal gas equation. On one side we have `P*V`, which is pressure (a force spread across two dimensions) multiplied by volume (an extension across three dimensions). Their product would therefore be equivalent to a force applied in one dimension, which is the same thing as Work, which is a form of energy. Now according to our ideal gas equation, this energy is directly related by a constant to the amount of gas we have multiplied by the temperature of that gas. Thus it makes sense that the energy of this gas is dictated solely by the temperature. The other interesting thing to note from all this is that though we have found that internal energy is a function of only temperature, we have no means of calculating an absolute internal energy, instead only the difference in energy between a given state and some reference. In the end this is only a small limitation because most calculations are based around the changing of state, so the difference in energy is what really matters.

*But wait, you said that internal energy was a reflection of energy contained by the molecules. Whereas Pressure is a force being*

done by the collisions of molecules, a macroscopic force we can measure. Wouldn't that mean they are different energies?

Ah, you are an observant soul, and yes you are correct. Fear not, I'm not trying to pull a fast one on you. Internal energy and the energy being denoted by the `P*V` term are different, and in fact there is another important property that is assigned to their sum.

**Enthalpy of an Ideal Gas**

So we have found that the ideal gas not only has this internal energy (`U(T)`), but there also must be some energy in the motion of the particles themselves (`P*V`), so it would perhaps be a good idea to have a new energy term that combines the two. Let this term be called enthalpy, and we can denote it `H`=`U`+`P*V`. Since both `U` and `P*V` functions of temperature, it would be more accurate to refer to this new term as `H(T)`=`U(T)`+`P*V`. Since internal energy, pressure and volume are all state functions, enthalpy too is a state function of the gas.

*While this discussion of state is all very interesting, I thought you said something about thermodynamics, doesn't that imply it will be a little more...dynamic?*

Ah, yes, worry not, that's exactly the direction we're headed toward. So far we have explored how temperature is the determining factor for energy in the state of an ideal gas, but we have a problem applying: We have no way to directly put temperature into a system. Instead let's take a look at the reverse of the relationship we've just demonstrated. Instead of using temperature to define the energy of a system, what if we used the change in energy to define the corresponding change in temperature?

*A slight disclaimer from this point on: For the sake of keeping it simple we are going to assume all processes described are done in a reversible manner. For more detailed discussion of irreversible processes check out entropy.*

**Heat Capacity of an Ideal Gas**

Since the energy in the gas is directly related to temperature, we can assume that should one add energy to the gas system, the temperature will correspondingly rise. The quantitative measure of this fact is known as heat capacity. According to the first law of thermodynamics, there are two ways we can add energy to a closed system. The first is through heat energy(Heat, which we will call `Q`), and the second is through mechanical energy (Work, which we will call `W`). Assuming that in adding energy we are not also adding mass, the net result on the system is one of three effects:

- The internal energy of our system is raised.
- The system has been accelerated and now contains more kinetic energy.
- The system has been physically raised and now contains more potential energy.

We are going to assume that our system of gas has no net movement, so the second and third terms will be zero. This means that all the energy we add will go directly into internal energy, which in turn raises the temperature of the gas. Mathematically speaking: `Q`-`W`=`ΔU` or if we are looking at differential changes, `dQ`-`dW`= `dU` (It's important to note the sign of `W` varies depending on how it is defined. In this case it is work being done by the gas on the surroundings. Many places will define it as work on the gas by the surroundings and say `Q`+`W`=`ΔU`). If you recall from earlier, for this gas we've expressed work in terms of `P*V`. Throw in a little algebriac rearrangement to get: `Q` = `U` + `P*V`, or differentially `dQ` = `dU` + `P*dV`.

*Wait, the right side of that looks awfully familiar...*

As it should, it is the expression we found for enthalpy! This means that you can equate heat added to an ideal gas directly with the enthalpy. However, since there were two terms in enthalpy that are important to us, in order to carry this further, we're going to define two separate means of heating: heating at constant pressure, and heating at constant volume. The reason for this is because according to our equation from earlier (`dQ` = `dU` + `P*dV`), if volume is constant, all the heat we add will go directly into increasing the temperature (remember `U` is a function of only `T`, so as one changes so will the other), whereas heating at constant pressure will have some energy used in increasing the temperature, and also some of it will be used in the work of physically expanding the volume of the gas.

**Heating at Constant Volume**

As long as volume is constant, the term `PdV` = 0 (since `dV` is change in volume), meaning that `dQ` = `dU`. This means that the total heat added is equivalent to the difference in our final internal energy and initial internal energy. This means that not only is there a function relating the amount of heat added and the temperature, but that it will only vary in terms of temperature. This function (we will call it `C`_{v}(T) or just `C`_{v} is approximated by actually measuring the amounts of heat added and noting the
temperature change. In the end we find that `Q` = `ΔU` = `∫C`_{v}*dT, for constant volume heating.

**Heating at Constant Pressure:**

Now we tackle the same issue, only this time we can't ignore `P*dV` since the volume is changing. We have our equation: `Q` = `ΔH` = ∫`C`_{v}*dT + ∫`P*dV`. In order to help us simplify this a bit, recall our original relation we derived `P*V` = `R*T`. If we still assume constant pressure, and we derive it we get `PdV` = `R*dT`. We can plug this back in to get `Q` = `ΔH` = ∫`C`_{v}*dT + ∫`R*dT` = ∫(`C`_{v} + `R`)`dT`. The term `C`_{v}+`R` is also very important in dealing with ideal gases, and it is often abbreviated `C`_{p}. Our final equation for constant pressure heating is then just: `Q` = `ΔH` = ∫`C`_{p}*dT.

*These equations are great, but what about in the real world were we can't assume constant pressure or constant volume?*

Ah, yet another good question. If we take a look at the terms that have been appearing on the right side of all of these equations, even before we started making assumptions about out process: pressure, volume, and internal energy. These are all state functions. Since the entire process can be described in terms of state functions, and state functions are path independent, our process is path independent. All we need to know is the state we start at, and the state we will end at, and we can determine how much energy is required. Thus any process involving the expansion, compression, heating or cooling of an ideal gas can be described by adding a heating (or cooling) at constant pressure, and a heating (or cooling) at constant volume.

So far we have described two important processes in ideal gas behavior, the constant volume process (a.k.a.: isochoric, `dV`=0), and the constant pressure process (a.k.a.: isobaric, `dP`=0). In addition to these, there are two more processes of note we will consider. The first is a process where instead of volume or pressure, temperature is held constant (a.k.a.: isothermal, `dT`=0). The last process is one where no heat is exchanged between the gas and the surroundings (a.k.a.: adiabatic, `dQ` = 0). These last two are of particular importance when discussing heat engines (such as the kind that drive your automobiles) where gases undergo compression/expansion cycles.

**Expanding at Constant Temperature**

If we assume that the temperature of the gas does not change, this means that the internal energy of the gas will also not change (`ΔU`=0), so that means all heat put into the gas will be recovered through work the gas does on the surroundings (`Q` = `W`). Since work is pressure multiplied by the change in volume (`dW` = `P*dV`), and pressure is inversely related to volume (`P` = `R*T/V`), we know that the final amount of work (and likewise the amount of heat) will be proportional to the natural log of the ratio of the final volume to the initial volume (`dQ` = `dW` = `R*T`*`dV`/`V` integration yields: `Q`=`W`= `R*T`*Ln(`V`_{2}/`V`_{1})). Once the volume has been calculated it is a simple matter to find the pressure using the ideal gas equation, in fact, since they are inversely proportional we know that `V`_{2}/`V`_{1} = `P`_{1}/`P`_{2}, which you can plug into the previous equation to get: `Q`=`W`= `R*T`*Ln(`P`_{1}/`P`_{2}) .

**Expanding With No Heat Exchange**

The last process we will explore in the world of the ideal gas is the adiabatic process. Since there is no energy transferred in the form of heat, this means that all work done on the gas will be converted directly to temperature. For a real world example, the compression of gas in an internal combustion engine can be reasonably approximated this way, as the compression is considered to happen fast enough that heat is lost during the compression process is negligible. To quantitatively measure this we should once again fall back on our first law equation: `dQ` = `C`_{v}*dT + `P*dV`. By definition `dQ` = 0, leaving us with `C`_{v}*dT = -`P*dV`. From here we can either substitute for `P` or `dV` giving us: `C`_{v}*dT = -`R*T*dV/V` = -`R*dT` - `R*T*dP/P`.
Using a little bit of algebra and a little bit of calculus we arrive at this: `T`_{2}/`T`_{1} = (`V`_{1}/`V`_{2})^{R/Cv} = (`P`_{2}/`P`_{1})^{R/Cp}.

The use of ideal gases doesn't stop here. This write-up was only meant to scratch the surface of what ideal gases are all about. Even though here in the real world there is no such thing as an ideal gas, all throughout chemistry and physics they are the gold standard that all real gases are compared to, and even just the ability to use them as a reference point allows for some pretty nifty calculations. If you find you like ideal gases be sure to check out their cousin: ideal solutions.

Sources: Too many years of chemistry with a little bit of double checking in *Introduction to Chemical Engineering Thermodynamics 7*^{th} ed. By: J.M. Smith, H.C. Van Ness, and M.M. Abbott.