# How is the $2^\mathrm{nd}$ law of thermodynamics valid?

For example, if we have an infinite cylinder fitted with a piston having adiabatic curved walls and a diathermal base. Now, bring the base in contact with some higher temperature body and let the cylinder absorb some heat. After that, we cover the base with an adiabatic material. Then, we let the gas to expand forever, thereby converting almost all of its energy into work by the mechanical motion of the piston. I don't understand why we have to give some heat to a lower temperature body. We can have as much efficiency as we want by letting the gas expand forever.

I wanted to ask one more thing. In my book, while explaining the efficiency of Carnot engine, there is a graph between Temperature and Entropy. The curve is cyclic and rectangular in shape. It says that the engine goes through four states $a,\ b,\ c,\ d$ corresponding to points $(T_1,S_1)$, $(T_1,S_2)$, $(T_2,S_2)$ and $(T_2,S_1)$. So, the entropy increases while going from $a$ to $b$ but decreases from $c$ to $d$. But in the same book, it was written that once entropy has been created, the universe has to carry the burden of it forever or in other words it can't be destroyed. I think that means entropy can't be decreased. Then, what did it decrease from $c$ to $d$? And why entropy can't be decreased. For example, entropy of electrons moving randomly is large but when we apply an electric field, electrons begin systematic motion and hence randomness is decreased. Then, isn't entropy destroyed"

• You proposed process isn't a cycle, so using forms of the 2nd law that refer to heat engines (Carnot's version, perhaps) doesn't make sense. Construct a cycle or use a different form of the second law. Jan 20, 2017 at 20:56
• "For example, if we have an infinite cylinder...". Spherical Cow interrupts and says "Nooooo... we do not deal in infinite objects". Well it would have... but it is in a vacuum. Interesting question: if you attempt to speak, but there is no air to carry the sound, have you then spoken? (Version of "If a tree falls...") Oct 17, 2017 at 10:48

You are doing a common misconception of the second law.

The Kelvin statement of the Second Law says it impossible to have a process whose only effect is convert heat into work. The term "only effect" means that the system must return into its initial state, that is the process must be cyclic. That is not the case in your example.

The entropy does not decrease for isolated systems. When you plot those diagrams for a heat engine you are dealing with a non isolated system. The engine (the system) is in thermal contact with the sources (the surroundings). If you consider engine plus sources as your system, then the entropy never decreases.

• Does that mean that we can draw heat from a high temeperature body and convert it all to work as long as we don't have to make the system return to its original state? Then why can't ships be driven by drawing heat from the ocean and converting it all to work done on the ship eithout using fuel?
– Dove
Jan 20, 2017 at 15:41
• @Dove, Yes, it is possible integrally convert heat into work, however the engines of the ship work in cycles. Therefore the second law prohibits them to transform all heat into work. Jan 20, 2017 at 15:47
• Dove, there is a large amount of "low grade" heat, both in the sea and in areas such as deserts and even from highways. One big problem is that we would normally expend far more energy gathering together all this free heat compared to the useful work it would do. So we concentrate on building dams say, as the rivers and lakes have done the collection work for us.
– user140606
Jan 20, 2017 at 16:40
• Do engines always have to work in cycles? If we, somehow, had an infinite cylinder in the ocean fitted with a piston. And, if we used that piston to transport the ship from one place to another by drawing heat from the sea, then would the second law won't bring any limitation to such a device?
– Dove
Jan 22, 2017 at 3:31
• @Dove You definitely could have a ship with the non cyclic engine you propose. It would have so big though that it would be feasible. Jan 23, 2017 at 18:02

Then, we let the gas to expand forever, thereby converting almost all of its energy into work by the mechanical motion of the piston.

The gas could not expand forever, this does not make sense, I am sorry to say. (Even without the Carnot Engine example). The return stroke of the piston will compress it, driving it towards the cold reservoir.

once entropy has been created, the universe has to carry the burden of it forever or in other words it can't be destroyed. I think that means entropy can't be decreased. Then, what did it decrease from cc to dd?

On the path from c to d, entropy and energy were transferred to the cold reservoir, keeping the entropy of the working fluid as low as possible, at the cost of losing some energy, which is then restored to working fluid from the hot reservoir.

Eventually, the machine will stop working, as both reservoirs will be in thermal equilibrium.

• If there is no external pressure on the piston then the gas must expand forever because there is always a non zero pressure given by the equation P=$frac{nRT,V}$and hence a non zero upward force on the piston.
– Dove
Jan 20, 2017 at 15:50
• But forever is a bit extreme, both in time and space. Anyway, I know what you mean. I think you asked a good question, as other people probably want this sorted out. +1
– user140606
Jan 20, 2017 at 16:09
• There is no problem with the process he proposes beyond it being states in PhysicLand terms. He just can't use a version of the second law that is predicated on a cycle without having a cycle. Jan 20, 2017 at 20:57
• @dmckee point taken, thank you, it diverted from the question. Sorry Dove.
– user140606
Jan 20, 2017 at 21:26

For example, entropy of electrons moving randomly is large but when we apply an electric field, electrons begin systematic motion and hence randomness is decreased. Then, isn't entropy destroyed"

I think you have a wrong picture here.

If electrons don't interact, their initial speed randomly distributed $\bf v_0$ will evolve as ${\bf v_0}+q{\bf E}t$, which as the same statistical properties as the initial distribution because $\bf v_0$ is somehow preserved. The disorder is not changed, the entropy is constant, which corresponds to the field only providing work (and no heat) to the distribution.

If you want to get rid of $\bf v_0$, the electrons need to interact together or with a surrounding lattice. In this case, the energy provided by the field will be redistributed, leading to an increase of the disorder and of the entropy - typically, particles will reach a thermal state with higher temperature.