# 1st law- thermodynamics

1. Feb 1, 2009

### v_pino

This is an example of the application of first law of thermodynamics to closed systems.

-There is a bursting of a diaphragm separating high pressure gas compartment and vacuum compartment in a controlled volume. Consider the gas as the system. Assume that initial and final velocity are zero.

I know that there is no work done and there is no heat transferred to the system. Therefore, there is no change in internal energy.

What I don't understand is how we can use specific heat capacity at constant volume to explain that temperature (and hence internal energy) is unchanged for this ideal gas.

My teacher wrote (capacity at constant volume)*(final temperature) = (capacity at constant volume)*(initial temperature)

Would we say that this system is at constant volume?

2. Feb 1, 2009

### stewartcs

We're not really saying the system is at a constant volume or pressure - we're just using that relation. For an ideal gas, Cv is a function of temperature alone (mostly). The internal energy of an ideal gas is determined by using the Cv and temperature: du = Cv(T)dt. In other words you can use the relation since it is known. For a free expansion like you described, the internal energy remains constant (proven empirically). Hence the temperature must have remained constant. This of course assumes that the Cv is constant which is a safe assumption for an ideal gas.

Another way to think of it is that since the expansion doesn't require any work (due to the vacuum chamber) there is no extra energy required (unlike with Cp). Which is also why the Cv is normally lower than the Cp. In other words, a system with constant pressure is allowed to expand and the energy for this expansion work must also be supplied to the system.

Hope this helps.

CS

3. Feb 1, 2009

### Mapes

This is a quite common question. It seems so wrong to apply cV to processes in which the volume changes. After all, cV is defined as

$$c_V=T\left(\frac{\partial S}{\partial T}\right)_V$$

where we're considering a constant-volume process (and S is the molar or specific entropy so that things are normalized). But we've also found that for an ideal gas,

$$U=nc_VT+U_0$$

The key here is that we're no longer considering a process; cV is now just a material property that we're using in an equation to calculate internal energy. We might also write it as

$$U=n(c_P-R)T+U_0$$

and take the V out of it. It would be wrong to spot the P in this equation and conclude that now the equation requires not constant volume but constant pressure. It's the same equation!

That's why you can apply

$$\Delta U=nc_V\Delta T=n(c_P-R)\Delta T$$

to all ideal gas processes.

4. Feb 2, 2009

### Chuck St. Lou

If the diaphragm had a small hole pricked in it would there be any temporary temperature change in a closed system such as this while the system is out of equilibrium?

5. Feb 2, 2009

### Mapes

This isn't something I have any references for, but intuitively, wouldn't the first atoms/molecules to enter the evacuated side have a relatively high speed, and thus a relatively high temperature? This would leave the side that was originally filled with a relatively low temperature. The imbalance would equalize in time, leaving a gas with the same temperature as it originally had.

6. Feb 2, 2009

The ideal gas equations enable us to get approximate answers and as such are very useful.If the gas were real then work is done for example the bulk potential energy of the gas changes during the expansion.

7. Feb 2, 2009

### GT1

Why Cp and Cv should have different values ?

8. Feb 2, 2009

### Mapes

I disagree with the non-zero work assertion. For the case of a real gas expanding into an evacuated area inside an isolated box, the heat transfer to the gas is (still) zero and the work done on/by the gas is (still) zero. What's different is that the internal energy is a function of pressure and temperature instead of temperature alone; thus, we can no longer conclude that unchanged internal energy implies unchanged temperature. Does this make sense?

9. Feb 2, 2009

But the ideal gas equations are just that ideal-they are derived theoretically by making simplifying assumptions and they break down more appreciably under those conditions where the assumptions break down.The experimentally derived equation are also limited in their scope for example Boyles law breaks down under high pressures or for gases approaching their critical points.

10. Feb 2, 2009

### stewartcs

During the expansion there may be a differential. The typically free-expansion problem deals with initial and final states (i.e. after equilibrium).

CS

11. Feb 2, 2009

### stewartcs

But that doesn't change the fact that the gas is still not doing any boundary work (i.e. there is still a vacuum).

CS

12. Feb 2, 2009

### stewartcs

Using the definition of enthalpy and writing the differential of enthalpy, the relationship between the specific heats for ideal gases is:

$$h = u + Pv$$

$$dh = du + d(RT)$$

$$C_pdT = C_vdT + RdT$$

$$C_p = C_v + R$$

Hence Cp is always greater than Cv.

A system with constant pressure is allowed to expand. The energy for this expansion work must also be supplied to the system.

CS

13. Feb 2, 2009

Hello stewartcs.Work is done in increasing the separation of the molecules and this done at the expense of the internal energy of the gas which cools slightly(assuming no energy inflow from the surroundings)

14. Feb 2, 2009

### stewartcs

The ideal gas assumption ignores inter-molecular forces. During free expansions, the temperature is constant.

In real gases, the temperature may decrease or increase depending on the inversion point.

In either case (ideal or real), there is no boundary work being done in a free expansion. The temperature changes comes from converting kinetic energy to potential energy so as to keep the total energy the same (i.e. conservation of energy law).

CS

15. Feb 2, 2009

### Chuck St. Lou

The part I am having a problem with is that in this closed system i.e. no heat in or out same number of moles and ideal gas, if I move a piston in a cylinder so it can occupy V2 or I open the diaphragm so that the gas flows & is sucked in and occupies V2 in each case the net result is volume equals V2. After settling to equilibrium the kinetic energy should be the same because the boundary is the same. Maybe I'm missing something.

16. Feb 3, 2009

### Mapes

The difference between extracting work and free expansion (I believe these are the two processes you're comparing) is that when you extract work via a piston, the gas molecules transfer momentum to the piston and lose energy. In the free expansion they don't. So in the first case the gas ends up cooler after expanding to V2; in the second case it ends up at the same temperature. Make sense?

17. Feb 3, 2009

### Chuck St. Lou

With a closed system no heat in or out, V1 with piston=V1 with diaphragm. Like wise V2 with piston = V2 with diaphragm. Because no heat goes in or out of the system the final state is the same for each system. The pressure on the redefined boundary is the same. Does it really matter how it became that way? Because it is a closed system no work/heat comes out of the system energy is conserved. My original question is weather or not there would there be any temporary temperature change in a closed system such as this while the system is out of equilibrium?

In stewartcs earlier post says;In either case (ideal or real), there is no boundary work being done in a free expansion. The temperature changes comes from converting kinetic energy to potential energy so as to keep the total energy the same (i.e. conservation of energy law). I think I am satisfied with that and it answers my question.

Mapes said;thus, we can no longer conclude that unchanged internal energy implies unchanged temperature. Does this make sense?

The heat or energy stays the same but the temperature does change. Yes this makes sense to me thank you Mapes.

18. Feb 3, 2009

### vanesch

Staff Emeritus
Eh, maybe I'm missing something, but you do not end up in the same situation. After all, in the case with the piston, *work* has been extracted from the system, and hence the internal energy (first law) diminished. In the second case, no work has been extracted from the system. During a certain time, part of the internal energy of the gas has been converted into kinetic energy of the gas, but that kinetic energy has, through turbulence and viscosity, again been converted into internal energy after the macroscopic gas flow came to rest. So the internal energy in the case of the free expansion is larger than in the case of the piston moved because in the first case (first law) it wasn't diminished with external work, while in the second, it did.

Now, for an ideal gas, the internal energy is only function of the temperature, and not of the pressure (this is not the case anymore for real gasses), so if the internal energy didn't change, then the temperature didn't change. The free expansion keeps the temperature constant. The adiabatic expansion with the piston made the gas cooler.

You can still see this differently. In the expansion with the piston, you can do the expansion very slowly, and have a reversible process. The entropy doesn't change in this case. (adiabatic).

You cannot have the free expansion "reversible". It must be irreversible (hence, gain in kinetic energy in the gas jet, followed by irreversible dissipation in turbulence and viscosity). The entropy increases.

Or still differently: instead of a free, irreversible expansion, you could have an "internal" piston, against which the gas expands adiabatically, but now, the work done by the piston (say, connected to an electrical generator) is transformed in heat again (with a resistor) which heats the gas after it expanded adiabatically (and cooled). The heat that is given back to the gas is in the case of an ideal gas, exactly equal to the heat needed to bring it back to its original temperature.

No, it won't be, because the pressure will be higher in the case of the free expansion (temperature higher, same volume and amount of gas, ideal gas law).

Last edited: Feb 3, 2009
19. Feb 4, 2009

### Chuck St. Lou

If I understand this correctly then, in free expansion into a vacuum the molecules pass through the diaphragm and don't run into other molecules which would slow it down. No collisions, no loss of kinetic energy I.E. no change in temperature. Do I have this much right?

20. Feb 4, 2009

### Mapes

I don't believe any problems arise from elastic collisions in an ideal gas.