# Adiabatic process and 2nd law violation

Consider a gas which expands adiabatically and reversibly. Since there is no heat in the process, the gas transform some of it's internal energy into work. As no entropy is exchanged or created, the entropy of the gas is constant.

Now, suppose we use the work given in the expansion to power a refrigerator. Assuming the refrigerator works in a cycle it creates no entropy. However, since refrigerator transfers heat from a cold reservoir to a hot one, this would mean a reduction in entropy.

In total, we obtain a reduction in entropy. Doesn't this violate the 2nd law of thermodynamics? If it does what is wrong with this scenario? I know that this cannot be done forever since our original gas cylinder ends up in a different state.

• As you pointed out yourself, it's possible to reduce entropy locally. Every refrigerator does that. However, it's impossible to reduce the entropy of closed system. Dec 25 '19 at 16:21
• I might be misunderstanding the scenario, but it seems to me like you are just analyzing one part of a cycle and assuming it describes the entire cycle? Dec 25 '19 at 16:31

If the refrigeration cycle is reversible, then the reduction in entropy of the low temperature reservoir from which heat is extracted equals the increase in entropy of the high temperature reservoir to which heat is transferred. For the reversible cycle the total entropy change is zero.

The change in entropy for the cold reservoir is

$$\Delta S_{cold}=$$.

The change in entropy of the hot reservoir is

$$\Delta S_{hot}=+\frac{Q_H}{T_H}$$.

For a reversible refrigeration cycle (Carnot):

$$\frac{Q_C}{T_C}=\frac{Q_H}{T_H}$$

Thus,

$$\Delta S_{tot}=+\frac{Q_H}{T_H}-\frac{Q_C}{T_C}$$

Finally, $$Q_{H}=Q_{C}+W$$, where $$W$$ is the work input required to transfer heat to the hot reservoir.

Hope this helps.

• You and Ofek are right, i forgot that even in the best case scenario when the fridge works as a Carnot cycle it cannot reduce the entropy since the heat taken from the cold reservoir is less that the heat given to the hot one. Dec 25 '19 at 17:01
• @FranciscoLoyolaReyes That's correct. $\frac{Q_C}{T_C}$ = $\frac{Q_H}{T_H}$ and $Q_{H}=Q_{L}+W$. Dec 25 '19 at 17:12

You forgot that refrigerators deliver more heat into the heat reservoir than is taken out of the cool reservoir. Meaning the work you did by expanding the gas will go to heat up the heat reservoir. (And thus the net entropy gain will be 0 or positive, never negative)