# IPhO 2012 problem

1. Jul 23, 2013

### jaumzaum

I was attempting to solve a problem from IPhO 2012, but got in trouble with some concepts. I'd apreaciate if someone could help me

Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is r0 and the mass is m. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly T0. The gas may be assumed to be ideal. The average molar mass of the gas is μ and its adiabatic index is γ > 4/3. Assume that Gmμ/r0 ≫ RT0, where R is the gas constant and G is the gravitational constant.

a) "During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings (...)"

b) At some radius r3≪r0, the gas becomes dense enough to be opaque to the heat radiation. Calculate the amount of heat Q radiated away during the collapse from the radius r0 down to r3

I solve b) like this: Q = ΔU + W, where U = ΔEt + ΔEp (Et is the thermal energy and Ep is the potential energy of the system)
As T=constant, ΔEt = 0, so Q = ΔEp + W
As the gas is "sparse", P is very low, so W << ΔEp
That way Q = ΔEp = 3/5 G M² (1/r0 - 1/r3) ≈ -3/5GM²/r3

But the solution is the following:
As T=constant, ΔU=0, so Q=W=nRTln(V3/V0) = -3nRTln(r0/r3)

Why did they assume ΔU=0? As far as I know, the only quantity that depends only in the temperature (when n is constant) is Et, not U. For systems in which ΔEp=0, ΔEt = ΔU, but in our case ΔEp≠0, so why should we say ΔU=0?

Last edited: Jul 23, 2013
2. Jul 23, 2013

### jaumzaum

I think I've got a) but still don't understand b)

Any help would be appreciated

3. Jul 24, 2013

### haruspex

I don't see any reason why the answer should be unbounded as r0 tends to infinity. The PE remains finite, and nothing else in the initial set-up becomes nonsensical. So I don't see how the given answer can be right.

4. Jul 24, 2013

### Basic_Physics

Compressing an ideal gas would not alter its potential energy since we assume that there are no forces acting between the atoms when they are not colliding (one of the assumptions in constructing the ideal gas theory).

5. Jul 24, 2013

### collinsmark

6. Jul 24, 2013

### jaumzaum

So do you think I'm right?

I'd agree with that is most cases, but should in a supernova explosion the potential energy be negleted too? The gravitational field plays an important hole in the process, and, as the problem said, GM μ/r0 >> RT0, shouldn't we consider it?

I know, I've got the question from there
Thanks

7. Jul 24, 2013

### Basic_Physics

The internal energy of a confined ideal gas depends only on its temperature. It is the only indicator that a change in its internal energy took place.

Last edited: Jul 24, 2013
8. Jul 24, 2013

### voko

U is the internal energy. Temperature is another word for internal energy. When the heat of compression is radiated away, the internal energy and the temperature stay constant. The radiated heat is the work of isothermal compression.

9. Jul 24, 2013

### haruspex

I wasn't saying you were right, merely that I had an objection to the given answer. But maybe the ideal gas model would break down at sufficiently large r0. Since you're told to assume that model, it may be unreasonable to expect the answer to be valid as r0 tends to infinity.

10. Jul 24, 2013

### Basic_Physics

The 1st law of thermodynamics relates that the change in energy of a system is given by

ΔU = Q - W

It can be changed by either adding heat to the system or/and work being done by the system. Such processes alters the internal or thermal energy of the system. This change in energy is at the expense of the kinetic energy of the atoms of the ideal gas. Any change in this energy will thus affect the temperature of the gas.

11. Jul 24, 2013

### jaumzaum

Thanks for all the responses

Everyone is saying ΔU is zero when T does not change, that's what I want to undersand
Actually, in every exercise I've already done in my entire life I've assumed the same, when T is constant, ΔU = 0. But that's because I've never done a problem where I had a strong field going on inside the gas. At my first thermodinamic classes, I was tauch there is a quantity called thermal energy, that measures the kinetic energy (translational, rotational and vibrational) of all atoms in the system, and (according to wikipedia), it's the only quantity (different from ΔU) that depends directly on the temperature, an increase on T would be an increase on Et, i.e when T is constant, ΔEt=0.
ΔU is defined by them as the sum of the thermal energy to the potential energy of the system (caused by an electric, magnetic or gravitational field). They give an example when ΔU is zero and T is not constant: Consider a gas into a adiabatic rigid recipient, a combustion occurs without difference in the number of moles between reagents and products, and so on the degrees of freedom of them, so by Q=ΔU + W, ΔU=0. But the released energy (that was released breaking the molecular bonds, i.e reducing Ep) will be used to heat the system (increasing Et and T). I know in this example we could not assume the model of an ideal gas because the particles are directing interacting one to another in the break of the molecular bonds, but in the previous exercise, they are doing the same with the gravitational field. They also say GMμ/r0 >> RT0, so the expression given for the gravitational work is much larger than that of the pressure work. But you are saying -3/5GM²/r3 are not considered, ok, so where is goes in? We know in fact that the gravitational potential energy of the system deacreases by -3/5GM²/r3, so what other quantity will increase for it to deacrease? Another doubt, the particles will start acceleratig in direction to the center of the cloud, so their velocity will increase, and so their kinetic energies. So we can say that when the macroscopic velocity is high enough (i.e higher than √3RT/M), the temperature will increase beacause of that?

12. Jul 24, 2013

### voko

Obviously, you can define any symbol any way you like, but ΔU in thermodynamics and kinetic theory typically stands for "internal energy", which is the kinetic energy of the molecules in the gas. Potential energy is not covered by it.

The kinetic energy of molecules in the gas increases, but, as stated in the formulation, this increase in the energy is immediately radiated away.

13. Jul 24, 2013

### haruspex

That's not what it says here: http://en.wikipedia.org/wiki/Internal_energy
But regardless of terminology, the work done by gravity in the collapse is clearly (3/5) G M² (1/r0 - 1/r3), as stated in the OP, while the work needed to compress it, viewed as an ideal gas, is 3nRTln(r0/r3). The work done in compressing the gas raises temperature, and that's what has been radiated away, so that is the answer to the question. The only remaining issue is why the difference from the work done by gravity.
It seems to me that for the collapse to occur we must have (3/5) G M² (1/r0 - 1/r3) ≥ 3nRTln(r0/r3), and the difference will take the form of non-thermal kinetic energy - i.e. the bulk inrushing of the collapsing ball. Since the acceleration will be proportional to radius, the density remains uniform.

14. Jul 24, 2013

### jaumzaum

What would be an example non-thermal kinetic energy? I thought thermal energy englobed all types of kinetic energy.

15. Jul 24, 2013

### voko

Note that I equated internal energy with kinetic energy when speaking specifically of a gas. I do not think there is any disagreement.

I am not sure we should really be considering this here. The problem is clearly intended to be solved by ignoring the true effects of gravity, so theorizing where (and what) gravitational energy is spent seems rather pointless to me.

16. Jul 24, 2013

### haruspex

It is entirely relevant to the question of why jaumzaum's original analysis produced the wrong answer. jaumzaum presumed that all the work done by gravitation went into the radiated energy. That overlooked the non-thermal kinetic energy.
Thermal kinetic energy is disordered. A bulk flow within a system is ordered, to some extent, so non-thermal. Strictly speaking there is no hard boundary between the two - it's all a question of entropy and enthalpy. The closer you have to look to see bulk flows, the higher the entropy. In the case of a concentric implosion or explosion, you don't have to look very closely, so the entropy is low.

17. Jul 25, 2013

### Basic_Physics

The work done by gravitation in compressing the gas cannot be stored as potential energy since we assume that there are no forces between the molecules of an ideal gas. The potential energy mentioned in this context refers to chemical energy which is a result of the electrical forces between the atoms in the molecules of the gas. Vibrations of the atoms in such molecules alternate between kinetic and potential energy. So in theory such vibrations should pick up on the increase in potential energy. How does this come into play in the theory? Such vibrations will also result in an increase in the temperature.

18. Jul 25, 2013

### haruspex

I don't think anyone was suggesting that. The question is whether the gravitational potential energy of the gas cloud counts as internal energy. As I read the wikipedia article, it does. I'm not taking that as the final arbiter, and it is not crucial to answering the OP, but it would be nice to have a clear definition.

19. Jul 25, 2013

### jaumzaum

Thanks everyone. I think I have a clear understanding now. The work done by gravity is much larger than the work done by the pressure, but only the second one could increase in the internal energy (i.e the temperature). As the temperature is constant, all the work done by pressure will be radiated (as stated in the IPho original solution). The difference in the works, so, will be added to the non-thermal kinetic energy of the system (that is the macroscopic/ordered kinetic energy), and the particles will start accelerating in direction to the center. The second part of the problem isto determinate the orbit a particle in the boundary of the cloud would describe, and the answer is a elipse, that they find applying Kepler laws. But Kepler laws only are true when the mechanical energy is constant. But in our problem there is a variation in the mechanical energy (that is the work done by pressure). However, as they told in the begginning of the problem GMμ/r0 >> RT0, so this work is insignificant compared to the potential/kinetic energy, this way mechanical energy is conserved. However there comes a time when the work done by pressure is not insignificant and part of the potential energy is being used to increase the temperature and also there is a time when the gravitational field is not enough to compense the work done by pressure, the non-thermal energy will do that. This is all covered by the problem in the last part. But as they said the temperature is not yet high enough to induce fusion, although the particles still have on-therma kinetic energy l, and they will use it (i.e starts desaccelerating) to increase the internal energy and the temperature, and to feed the residual work done by pressure. Although there could be a time before this energy ends (and there certainly will for a supernova) that the temperature is high enough to ignite nuclear fusion, and so you start getting other elements and blablabla

Is everything I said now correct?

Last edited: Jul 25, 2013
20. Jul 25, 2013

### Basic_Physics

The gravitational attraction does positive work on the gas during compression. So in this way it is incorporated into the internal energy.

Last edited: Jul 25, 2013