Why is this not a violation to the 2nd law of thermodynamics?

1. Feb 14, 2007

LTP

An ideal gas is enclosed in a cylinder with a tight-shutting piston, which moves without friction. There is a bowl with fine grained sand on top of the piston. The cylinder is placed in a headed bath which keeps the gas a constant temperature, see attached file.

One sand grain is then removed. The piston moves an infinitesimal step out, and the pressure in the cylinder drops infinitesimally. If we place another sand grain in the bowl, the pressure rises by an infinitesimal step and work is converted to heat.
For an expansion of an ideal gas at constant temperature is
$$\Delta U = 0$$
which yields
$$Q= -W$$
So if we remove a sand grain from the piston, we are actually converting heat (from the heated bath) to work (lifting the piston) which is a violation to the 2nd law of thermodynamics.

Why is this possible? Is it because it's a reversible process? What if we remove all of the sand grains one by one, would the work then become non-infinitesimal?

Attached Files:

• 2HS.PNG
File size:
1 KB
Views:
131
2. Feb 14, 2007

HallsofIvy

What work is being done to remove and replace the grain of sand? The movement of the piston is directly dependent on the weight of the grain of sand and so is the work required to move it off and put it back on the plate.

3. Feb 14, 2007

Q_Goest

Hi LTP. That's a thought experiment used in virtually all texts on thermodynamics. The intent is to aid in understanding a reversible process. In the idealized case, the process is seen to be "reversible". Of course, it's impossible to actually create the ideal case in reality.

4. Feb 14, 2007

tim_lou

where in the second law says does it say you cannot convert heat to work? the process described here is infinitesimal and quasistatic. change in entropy is zero. What's wrong with it?

yeah, Q=-W, why not?

Last edited: Feb 14, 2007
5. Feb 14, 2007

Andrew Mason

There is only a violation of the second law if there is a reduction in the total entropy of the system and surroundings, which would occur only if heat flowed from colder to hotter reservoir without the addition of work.

But in this case the removal of a grain of sand decreases the pressure so the volume expands and in doing so cools slightly. The heat from the surroundings flows in to raise the temperature back to the original. The work done in lifting the sand is stored as gravitational potential energy of the sand.

When you add back the grain of sand (after having lifted it slightly from where your removed it), the pressure increases and the volume decreases causing a slight increase in temperature. Heat then flows out to the surroundings. The work done to the gas is the reduction of potential energy of the sand.

But for the work done in lifting that one grain of sand, this would be a reversible process with $\Delta S = 0$.

AM

6. Feb 17, 2007

haiha

I think deltaS is not zero when you remove one grain of sand. V increases (although very small) and consequently S increases, that's why the gas should be cooler if it is fully adiabatic.

7. Feb 19, 2007

jeroen

If this bath keeps the cylinder at constant temperature then it must do two things:
-Cool the cylinder when heat is added to the cylinder. (when you add a grain of sand)
-Warm the cylinder when heat is removed from the cylinder. (when you remove a grain of sand)

Assuming that this is not a magic bath, it would transfer heat between the system you described and the outside world.

Last edited: Feb 19, 2007