# Why is it more difficult to raise the temperature when the degrees of freedom are higher?

Some materials are hard to raise the temperature the reason for it are the degrees of freedom of the molecules. But why is the existence of many ‘ways to move’ of molecules so counter intuitive as heat is movement of molecules?

With other words if molecules has many freedoms that means (for a layman) that the molecules can vibrate/move more easily and by this has more kinetic energy and so heat. But the opposite is true, so what is the difference of movements in freedom degrees and movements which raise temperature?

• With more degrees of freedom, there are more places for the energy to go. And thus less energy in each "type of motion" which is a smaller temperature. Silly analogy: A bucket of water spread into a number of glasses. Water = energy, number of glasses = degrees of freedom, amount of water in one glass = temperature. – George Herold Dec 21 '16 at 20:12
• You could split this up into separate, more focused question on each of the three phases, imo you are trying to ask/learn too much with one question. – user139561 Dec 21 '16 at 22:45

Consider every "degree of freedom" of every molecule in the system as an independent place where you can store energy.

If I have $N$ molecules with $m$ degrees of freedom, that gives me $N\times m$ places to store energy. "Temperature" relates to "how much energy is there in one of these states", because by equipartition$^*$, every state will on average have the same amount of energy.

Now double the number of molecules to $2N$ - I now have $2\times N \times m$ places to store energy, and clearly I need twice as much energy to heat the system to the same temperature.

But if instead each of my $N$ molecules had $2m$ degrees of freedom, I would still have $2\times N \times m$ , and I would need the same amount of energy to heat the system that had twice as many molecules with half the number of degrees of freedom.

The situation is more complex when you have liquids - the energy is no longer stored simply in the motion of the molecules, but also in their interactions; in particular, intermolecular forces come into play, and depending on the structure of the molecules this can really change the heat capacity. A particular example of this is water, with a very high heat capacity due to the hydrogen bonding between its molecules: the energy needed to break these bonds is relatively large and dominates the heat capacity of water.

$^*$ When two systems are in thermal contact, they will tend to the same temperature; energy will transfer from the high temperature state to the lower temperature state. If you have two different ways that a system can contain energy (e.g. horizontal motion; vertical motion) then these will exchange energy until they are at the same temperature. Now if you add a third way (degree of freedom), you will exchange energy until all three are at the same temperature; and so, with more degrees of freedom, you can store more energy for the same temperature. This is one way to state the equipartition principle.

• You are ahead of me by 10 secs. From reading the answers, although the site is enhanced by a wide spread, I was just about to ask George above to expand his answer specifically for the OP, but you have done that. – user139561 Dec 21 '16 at 22:40
• Does this all actually mean that we can say that if we consider atoms instead of molecules, then temperature corresponds to their mean energy, irrespective of their interactions? Then it becomes much more obvious e.g. why we have heat capacity $\propto\frac32$ for monoatomic gas but $\propto\frac52$ for diatomic: in the latter case we have $2N$ atoms, but one "direction" of moving (vibration — for ideal gas) is "frozen", so heat capacity increases to $\propto\frac52$, less than $2\times$. – Ruslan Dec 22 '16 at 4:47
• @Ruslan as long as you recognize that some motions are "impossible" (so that the final number of allowed motions == number of degrees of freedom) then yes. – Floris Dec 22 '16 at 12:13

But why is the existence of many ‘ways to move’ of molecules so counter intuitive as heat is movement of molecules?

Here's your first mistake. Do not think of objects as containing "heat", any more than they contain "work". Instead think of heat and work as being things that are transferred between an object and its environment as the object's thermodynamical state changes.

The reason you should not think of objects as containing heat (or work) is that the amount of heat (and work) transferred between an object its environment as the object is taken from some initial state to some final state depends on the path between those initial and final states. If objects contained heat, a heat engine cycle that goes from an initial state to some intermediate states and then back to the initial state could produce no work. Heat engines work (pun intended) precisely because heat and work are path dependent rather than just state dependent.

With other words if molecules has many freedoms that means (for a layman) that the molecules can vibrate/move more easily and by this has more kinetic energy and so heat. But the opposite is true ...

The opposite is not true. Consider two objects call them object A and object B, both comprising the same number of molecules, both at the same temperature, but having different degrees of freedom, $D_A$ and $D_B$. The internal energies of the objects are proportional to the products of the number of degrees of freedom and the temperature: $U_A = \alpha\,D_A T$ and $U_B = \alpha\,D_B T$, where $\alpha$ is some constant of proportionality.

The object with the larger number of degrees of freedom does indeed have a a greater internal energy than does the object with the lesser number of degrees of freedom. A consequence of this is that it takes more energy input to increase the temperature of the object with the larger number of degrees of freedom than it does to increase by the same amount the temperature of the object with the lesser number of degrees of freedom.

Suppose you transfer the same amount of energy to the two objects, call it $\Delta U$. What is the change in temperature? For object A, the temperature change is $\Delta T_A = \frac {\Delta U}{\alpha D_A}$, but for object B it is $\Delta T_A = \frac {\Delta U}{\alpha D_B}$. So the object with the larger number of degrees of freedom undergoes a smaller temperature change than does the object with the lesser number of degrees of freedom.

• It cleared up a bit, so heat is defined as something transferred. But can you give an example of a motion which is not transferred as heat? Because it is still a bit unclear, as when you touch something wouldn't you be able to feel all movements in a way. It is hard to understand the line between the two? – Marijn Dec 22 '16 at 11:47
• @Marijn -- A thermally insulated cylinder and a piston form a good example. Plunge the piston into the cylinder and the internal energy, temperature, and pressure of the gas inside the cylinder increase, adiabatically. Adiabatic means "no heat transfer". Energy is being transferred to the gas, but via work rather than heat transfer. Slowly let the gas push the piston out and the gas's internal energy, temperature, and pressure will decrease, again adiabatically. Once again it is work rather than heat that is being transferred. – David Hammen Dec 22 '16 at 12:12
• But do you also know an example with a degree of freedom in molecules? – Marijn Dec 22 '16 at 20:40

This is an answer for gases. The image below features a monatomic (3 DOF), diatomic (5 DOF), and polyatomic (6 DOF) gas.

Notice for a monatomic gas we have only the 3 translational degrees of freedom, for the diatomic gas we have the 3 translational and 2 rotational, and lastly for the polyatomic gas we have all 3 translational and 3 rotational (could be more depending on the complexity of the molecule and vibrational modes). From Kinetic Theory, we have the generalized relationship for the specific heat at constant volume of gas, $$C_v = \frac{f}{2} R$$ where $f$ is the degrees of freedom of the gas molecule, and $R$ is the universal gas constant, $R = 8.314 \ \text{J/mol$\cdot$K}$. Hence, from this description it is obvious that the heat capacity is larger for molecules with higher degrees of freedom. In general, for idealized gases we have, $$\text{Monatomic}: \quad C_v = \frac{3}{2} R$$ $$\text{Diatomic}: \quad C_v = \frac{5}{2} R$$ $$\text{Polyatomic}: \quad C_v = \frac{6}{2} R$$

Now this would indicate that it requires more energy to raise 1 mole of the polyatomic gas 1 K than it would the diatomic and monatomic gases. Also, it would require more energy to raise 1 mole of the diatomic gas 1 K than it would the monatomic gas. The simplest explanation for this is that the energy becomes equally partitioned in such a way that all the translational and rotational modes are excited simultaneously with the same average energy provided the gas is in thermal equilibrium. This is the law of equipartition of energy. So rather than dumping all of the energy into translational degrees of freedom like the monatomic gas, the diatomic and polyatomic gases are partitioning some of that energy into the other modes like rotational and even vibrational at higher temperatures. An easy way to think of it is that assume you only have so much water "energy" in a bucket, and you have to partition that water "energy" over a given number of cups "degrees of freedom". If you have only 3 cups "degrees of freedom" you have more water "energy" going into each of those 3 cups "degrees of freedom", which in return puts more average kinetic energy or molecular motion into the gas. On the other hand, if you have 5 or 6 cups "degrees of freedom", then you have less water "energy" to partition into the first 3 cups "translational degrees of freedom" since all cups must have the same average energy due to the equipartition of energy law.

• Yes. Equipartition is the key (for a gas). You might want to elaborate on the reasons for the equipartition a little more; also - do you think it is equally valid for liquids? If not, what else comes into play? – Floris Dec 21 '16 at 20:08
• Unfortunately, I only commonly work with gases. I am not all to familiar with the thermophysical properties of liquids. – TRF Dec 21 '16 at 20:13
• Can I conclude by this that rotational and vibrational energy doesn't support to EM-waves. Or is with heatcapacity meant more than EM-waves, is that also including for example conduction? But if the latter is true than why aren't rotational movement not contributing to conduction? – Marijn Dec 21 '16 at 21:19
• You are grouping together many concepts with very vaige connections. I am not sure exactly what you are asking. – TRF Dec 21 '16 at 21:22
• Well there are, as far as I know, three different kind of heat production, infraredwaves, conduction and convection. Now if some freedoms can't produce if-waves perhaps they can produce conduction by there movement. But I'm certainly not an expert, so I could things mixe up. – Marijn Dec 21 '16 at 21:45

In terms of basic thermodynamics we have the fundamental relationship between entropy $S$, temperature $T$ and internal energy $U$:

$dS = \frac{1}{T} dU$

and the definition of specific heat $C$: $dU = C dT$, so that

$dS = C \frac{dT}{T}$

so materials with larger specific heats have larger entropy changes for a (logarithmic) change in temperature than ones with smaller specific heat.

Or if you like:

$\frac{dS}{dT} = \frac{C}{T}$

systems that exhibit large changes in entropy under temperature changes have to have large specific heats.