What is the difference between temperature and internal energy of a gas?

I came across this doubt multiple times while studying thermodynamics. Aren't both temperature and internal energy fundamentally due to the kinetic energy of the molecules? An even crazier doubt of mine is that pressure is also in some sense due to the kinetic energy of the molecules (which on collision with walls of the container transfer momentum and in turn cause pressure). But why are pressure, temperature and internal energy not the same? Why are we treating them as different thermodynamic state variables? This doubt prevents me understanding other concepts in thermodynamics.

For example, in an adiabatic change of an ideal gas from the state ($$P_1$$, $$V_1$$, $$T_1$$) to the state ($$P_2$$ ,$$V_2$$ ,$$T_2$$),if work is done on the gas, $$T_2 > T_1$$, i.e., the temperature of the gas rises. We know that in an adiabatic process work done on the gas goes only in increasing the internal energy of the gas. Now, immediately a doubt arose in me: "What is the difference between increasing internal energy of a system and increasing the temperature of the system?"

I don't know how mature this question is or how accurately this question conveys my doubt. Please bear with this question if it has some inaccuracies while dealing with technical terms.

You happen to have picked a model for a state of matter—gas, modeled as an ideal gas—where temperature, internal energy, and pressure are coupled.

Now generally, the internal energy $$U$$ is

$$U\equiv TS-PV+\mu N,$$

with temperature $$T$$, entropy $$S$$, pressure $$P$$, volume $$V$$, chemical potential $$\mu$$, and amount $$N$$. (That is, temperature differences drive shifts in entropy, pressure differences drive volume shifts, and concentration/chemical potential differences drive shifts in matter. These compose heat transfer, work, and mass transfer, respectively. If we wish to consider additional types of work, we supplement the $$PV$$ term.)

The temperature is defined as

$$T\equiv\left(\frac{\partial U}{\partial S}\right)_{V,N},$$

or the internal energy increase per entropy increase for a closed system at constant volume.

The pressure is defined as

$$P\equiv-\left(\frac{\partial U}{\partial V}\right)_{S,N},$$

or the internal energy increase per volume decrease for a closed system at constant entropy.

But the ideal gas is a very simple model: Its two equations of state are

$$U=CT+U_0;$$

$$P=\frac{NR}{V}T,$$

with heat capacity $$C$$ and gas constant $$R$$, and the internal energy is entirely kinetic.

So a temperature change in the ideal gas really is exactly an internal/kinetic energy change and a pressure change, mediated by those respective coefficients.

This is not generally the case. Condensed matter, for example, stores potential energy in the form of molecules spaced differently than their equilibrium spacing. Condensed matter also sits in an energy well provided by intermolecular bonding.

Here, one can have, for example, (1) a system that's very hot but at low pressure—or even in a vacuum—or (2) a system that's very cold—toward 0 K—but very strongly compressed.

In (1), the system tends to interact with its surroundings in a way that randomly excites their molecular movement. In (2), the system tends to interact with its surroundings in a way that energizes molecules in concert, through strain energy storage or momentum kicks from an expanding boundary. The first exchange is what defines heating; the second is what defines work (specifically, expansion work).

So temperature, kinetic energy, and pressure aren't necessarily coupled; it may seem that way only because you've considered only the ideal gas, where pressure is a solely a manifestation of impacts associated with the bulk temperature and the individual kinetic energy of freely moving molecules.

• Id just consider van der Waals gas as an example whereby the difference from ideal gas behaviour is explicitly broken, so that the 3 quantities are distinguished concretely. After all, your answer is already so complete... Commented Apr 18 at 6:05
• That would make for a great answer. I don't have a good sense of the nuances of how the van der Waals gas decouples the direct temperature–pressure–kinetic energy coupling, so I'd love to hear about it. Commented Apr 18 at 6:28
• Oh no, I was not meaning much with that. It has potential energy, and thus it wont be having the direct relationship between E and T (except Virial theorem insisting, lol), whereas the equation of state of van der Waals gas explicitly breaks the linear relationship between T and p by the $a/V^2$ term Commented Apr 18 at 6:50

Aren't both temperature and internal energy fundamentally due to the kinetic energy of the molecules?

Internal energy is the sum of the kinetic and potential energy at the atomic and molecular level. It is not "due" to kinetic energy. Molecular potential energy is due to intermolecular forces. Temperature is a measure of the average translational kinetic energy of the molecules.

A more crazier doubt of mine is that pressure also in some sense is due to the kinetic enrgy of the molecules (which on collision with walls of the container transfers momentum and in turn causing pressure).

Pressure is due to the rate of change in momentum of the molecules when colliding with the walls of the container.

But why aren't pressure, temperature and internal energy are not the same?

They are independent system properties. They are not the same, but are related to one another at equilibrium. For an ideal gas, they are interrelated by the equations

$$PV=nRT$$ and $$\Delta U=nC_{v}\Delta T$$

For example, In an adiabatic change of an ideal gas from the state (P1, V1, T1) to the state (P2 ,V2 ,T2),if Work is done on the gas, then $$T2 \gt T1$$ i.e., the temperature of the gas rises.

Yes. That follows from the first law of thermodynamics for a closed system, i.e.,

$$\Delta U=Q-W$$

for an adiabatic process $$Q=0$$. For an ideal gas, $$\Delta U=nC_{v}\Delta T$$ and therefore

$$nC_{v}(T_{2}-T_{1})=-W$$

For work done on the gas (compression work), $$W$$ is negative, and therefore $$T_{2}\gt T_1$$.

Now, immediately a doubt arose in me, "what is the difference between increasing internal energy of a system and increasing the temperature of the system?"

For an ideal gas only, there is no difference. That's because the internal energy of an ideal gas is considered to be kinetic energy only.

Hope this helps