Perpetual Motion Type II:

Dunking for Dynamics

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In 1965, before I had learned enough physics to know better, I conceived a perpetual-motion machine of the second kind, a device that does not violate the first law of thermodynamics (conservation of energy), but rather violates the second law, the law of entropy. In concept, imperfect as it was, the device would have drawn heat out of a system at thermal equilibrium and converted it into useful work. Drawing the randomized energy of heat out of a system and converting it into the organized energy of work reduces the system’s entropy without producing the required compensating increase of entropy somewhere else, so the machine will not move. To be certain of that statement, let’s take a more detailed look at what happens in the device I designed.

A tour of the device begins with the center of our attention, a cylinder fitted with a piston and an air-tight seal that allows the piston to slide freely within the cylinder. To the piston I attach a rod that will transmit the piston’s motion and force to an electric generator (Gen-1) and I add any subsidiary mechanisms as might be required. To the cylinder I attach a rod that lets me move the cylinder up and down in a pool of some suitable liquid (water in this case) and I attach it to a second electric generator (Gen-2) that can also work as a motor. When I have the device completely assembled, I set it into a pool of water in such a way that the cylinder lies just under the water’s surface. In that state, because of the way in which I’ve made it, the cylinder has neutral buoyancy.

For the working fluid inside the cylinder I want to use an ideal gas, but that’s not available. However, at normal temperatures and pressures ordinary air gives us a good approximation to the ideal, so I will use that.

Now I can manipulate the device in such a way as to compress the air in the cylinder and then allow it to re-expand, doing so in such a way that, even though there is no temperature difference in and around it, the device will convert heat into useful work (it won’t, of course). I have available to me two different kinds of thermodynamic process – isothermal and adiabatic – and I figure that at least one of them will enable me to achieve my goal (and again, it won’t).

The Isothermal Process

For the convenience of calculation I refer locations of objects to a vertical y-coordinate for which y=0 at the surface of the water. In the device’s initial state the cylinder lies at an average location of y0 and the air inside it occupies a volume of V0. The water pressure at any given depth accords with the equation

(Eq’n 1)

in which rho represents the density of water and g=9.81 meters per second squared represents the acceleration due to gravity (measuring the stiffness of Earth’s gravitational field at or near sea level). The pressure inside the cylinder conforms to the average pressure that the water exerts on the piston, so we have

(Eq’n 2)

In an isothermal process any work done in compressing the gas turns into heat, which seeps out of the cylinder and into the water, which serves as a heat reservoir, thereby keeping the temperature of the system effectively unchanged. Likewise, any work done by the gas as it expands comes from the heat content of the gas, whose diminution is compensated by heat flowing into the gas from the surrounding heat reservoir, again keeping the temperature of the gas constant. To express those facts algebraically we write the first law of thermodynamics (conservation of energy, which we are not trying to violate, so I don’t expect a problem here) in the form

đQ=dE+đW.

(Eq’n 3)

In that equation đQ represents a minuscule change in the heat content of the gas (or of any system in general), dE represents a differential change in the gas’s internal energy, and đW represents the increment of work done on or by the gas. The barred dees represent inexact differentials, whose integrals take values that depend upon the specific path over which we carry out the integration. With appropriate substitution, Equation 3 becomes the fundamental equation of classical thermodynamics,

(Eq’n 4)

With the heat content of the gas not changing, we have TdS=0, which means that the entropy of the gas does not change. The internal energy of the gas also does not change, so we have

(Eq’n 5)

which integrates into Boyle’s law,

(Eq’n 6)

That equation, taken along with Equation 1, tells us that at any location in the water the volume of the gas in our cylinder conforms to

(Eq’n 7)

In the device’s initial state the air in the cylinder displaces a volume V0 of water and thus has a buoyancy equal to the weight of that water (I’m assuming that the weight of the air is negligible). Because I have assumed that the diving component of the device has neutral buoyancy at y0, we know that it has a weight of

(Eq’n 8)

Now let the cylinder descend to a location at y1. As the cylinder goes deeper into the pool, the pressure on it increases and the piston moves inward, compressing the air in accordance with Equation 7. The descent occurs slowly enough that the pressures on both sides of the piston remain equal (so the device does no work on Gen-1) and the heat produced by the work done upon the gas has time to seep out of the cylinder. At any given depth y the cylinder will have a net weight of

(Eq’n 9)

That force, acting through the vertical motion of the cylinder, does work upon Gen-2, doing so in the net amount

(Eq’n 10)

With the cylinder at y1 we lock the piston into place and use Gen-2 as a motor to pull the cylinder back up to y0. The work done against the net weight of the cylinder comes out to

(Eq’n 11)

Finally we unlock the piston and allow the air to expand isothermally against the pressure p0, thereby doing work on Gen-1. The work done conforms to the description

(Eq’n 12)

in which we integrate the pressure difference acting on the piston with respect to the accompanying change in volume. Using Equation 7 to express the pressure of the gas in terms of the gas volume, we get

(Eq’n 13)

In the last step in the transformation of that equation I used both Equation 2 and Equation 7 to replace the pressures and volumes with the appropriate equivalents in terms of the y-coordinate.

For the net work done by the device in that closed cycle we get

(Eq’n 14)

We must infer that no purely isothermal closed cycle will ever succeed in converting heat into work without a temperature difference to drive the conversion. But perhaps an adiabatic cycle, which does change the temperature of the gas, will succeed?

In order to understand the adiabatic compression or expansion of an ideal gas, we must return our attention to Equation 3. In this case we intend to change the internal energy of the gas by changing the gas’s temperature, so we have the fundamental equation of thermodynamics as

đQ=dE+đW

=ncVdT+pdV=0

(Eq’n 15)

Again the equation zeroes out because no net heat enters or leaves the system. In that equation n represents the number of moles of gas in the system (in dry air one mole ponders 28.97 grams, an average of the molar masses of, almost entirely, nitrogen and oxygen) and cV represents the molar specific heat of the gas measured with the gas occupying a constant volume.

In order to eliminate the change in temperature from Equation 15 (because we want to use only pressure and volume as our variables) we must refer to the equation of state of an ideal gas,

(Eq’n 16)

in which R (the universal gas constant)=8.3145 joules per mole per degree Kelvin and k (Boltzmann’s constant)=1.38065x10-23 joule per particle per degree Kelvin. Differentiating Equation 16 then gives us

(Eq’n 17)

Solving that equation for the differential temperature and making the appropriate substitution into Equation 15 gives us

(Eq’n 18)

Dividing that equation by cVpV/R gives us

(Eq’n 19)

in which

(Eq’n 20)

the ratio of the specific heats measured at constant pressure and at constant volume. For dry air, which consists of over 99% diatomic molecules, γ=1.4 as our ratio. Integrating Equation 19 gives us

(Eq’n 21)

whose antilogarithm comes out as

(Eq’n 22)

which is analogous to Equation 6. It also gives us the equivalent of Equation 7,

(Eq’n 23)

Now we return our attention to the apparatus described above. This time, though, we have so insulated the cylinder and its piston that no heat can flow between the trapped air and the surrounding water. Repeating the same steps as before, we must repeat, with the new equations, the calculations that go with them.

In this case the force equivalent to that in Equation 9 conforms to

(Eq’n 24)

From that equation we calculate the work done on Gen-2 in the first step as

(Eq’n 25)

In the second step Gen-2 does work on the device in pulling it back up to y0, the amount of work given by

(Eq’n 26)

In the final step the piston does work on Gen-1 as the gas expands adiabatically from V1 to V0. We start with Equation 12 and make the appropriate substitution for the pressure. Using Equation 22, we can describe the pressure in the gas as

(Eq’n 27)

The equivalent of Equation 13 then becomes

(Eq’n 28)

As in the isothermal case, we sum the contributions described in Equations 25, 26, and 28 to get

(Eq’n 29)

I have laid out the algebraic reduction of that sum in the appendix because it’s not immediately obvious that the terms do, if fact, cancel out completely.

In either the isothermal case or the adiabatic case we can’t get this device to convert uniform heat into work. In neither case can we devise a closed-cycle process that will violate the second law of thermodynamics. As with other perpetual-motion machines, this one looked good on first impression, but when subject to a proper mathematical-physics analysis it simply fell apart.

Appendix: W4+W5+W6=0

Combining the formulas from Equations 25, 26, and 28 and dividing out the common factor of ρgV0 gives us

The first and third terms in that expression cancel each other because they represent the work involved in moving the weight of the device up and down. We make the appropriate substitutions for the volumes in the fifth term and get

The first part of the second term and the second part of the fourth term cancel each other and we can combine the second part of the second term and the first part of the fourth term into a new term. We can also multiply the third term by (-1/-1) and get

Finally we multiply the third term by (1-γ)/(1-γ) and we get

as we require. Q.E.D.

eabf

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