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Imagine a container that hold gas at atmospheric pressure at earth's surface. It is elevated by a space elevator to a point high enough that there is a significant pressure difference (but not in orbit). An opening is opened in the container and the gas runs through a turbine generating electricity while it leaves, then falling back to the atmosphere. Gas would return to the atmosphere with lower thermal energy, due to some of it's energy being converted to mechanical or electrical energy.

My question is that if this system converts thermal energy to mechanical or electrical energy, wouldn't this mean that the entropy of the universe would decrease violating the second law of thermodynamics? If the gravitational work that must be done is greater than energy generated by the turbine, what about an atmosphere with a gas of different density?

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    $\begingroup$ Hauling the container up will use more energy than you'd get from the turbine, so it is just a fairly inefficient way of shipping energy up the space elevator. $\endgroup$ – zeta-band Jan 18 '18 at 22:34
  • $\begingroup$ Converting thermal energy into mechanical or electrical energy doesn't mean there's entropy decrease. For example, if you have a container of hot gas and you let it expand, it'll do work on the environment ($W = p dV$) and cool down. You'll have converted heat energy into mechanical. But entropy wouldn't have decreased. $\endgroup$ – Allure Jan 18 '18 at 22:42
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The comments have gotten this one answered pretty well, actually. Remember that the 2nd law operates on the entire universe, not just the object itself. So while yes, your generator makes itself some energy, it loses more energy than it makes. Let's take a look at some equations:

Gravitational Potential Energy is equal to $a_g*m*h$. If our magic box starts at sea level and rises to 3 km in the atmosphere, and its total mass is 100kg, then the amount of energy needed to raise it there is $\approx9.81*100*3000$ or 2943000 J.

Now, let's assume that the box is 90% atmosphere and 10% mechanisms. Simply by returning the box to the ground, we get back just under 10% of the energy. Now, what about the pressure differential?

Our turbine is a wind turbine, basically, so we can use wind turbine equations to solve its energy generation. Basically, through a complex derivation process, you get $P = \frac{1}{2}\rho A v^3$, which describes the density of air ($\rho$), the velocity of the air (We'll use the maximum of 343m/s) and the cross sectional area of the turbine.

Now, how do we go about finding A? Since we know that the velocity is 343 M/s, we can use $\frac{flow\space rate}{area} = velocity$. Flow rate is expressed in volume per time. The density of air at sea level is 1.2$\frac{kg}{m^3}$, and we have 90 kg, so we have a volume of 108$m^3$. Our equation seems to look like

$$\frac{Joules}{second} = \frac{1}{2}*(1.2\frac{kg}{m^3})*area*\frac{\frac{volume}{second}}{area}$$

Simplifying, we get $J = .5*\rho * volume$, which yields... 64.8 J. 64.8 J is far less than the couple million it took to get the thing up there.

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  • $\begingroup$ Could the last line be simplified from .5*p*V=.5*(m/V)*V=.5*m? If so, what about a hypothetical scenario that acceleration due to gravity was less than 1/(2*h), making gravitational potential energy less than .5*m. $\endgroup$ – user182015 Jan 19 '18 at 1:29
  • $\begingroup$ Acceleration due to gravity is completely dependent on distance between two bodies. To use your hypothetical situation, we'd need to rewrite the base conditions of the freaking universe. Needless to say, thermodynamics is one of those conditions. Also, no, the last line could not be simplified as such, because ρ isn't 1$\frac{kg}{m^3}$, it's 1.2. Otherwise, the dimensional analysis works. $\endgroup$ – Jakob Lovern Jan 19 '18 at 2:39

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