Entropy of reversible and irreversible processes

As an exercise, let’s consider isothermal expansion of 2 moles of ideal gas from 0.5 m3 to 500 m3 at 300 K. We can do this process either reversibly or not reversibly, for example adiabatic expansion against vacuum. Let’s first calculate entropy changes in the reversible process. The work done on the system is - resisting pressure times volume changes dV. For the reversible isothermal process, the resisting pressure is the same with the gas pressure, the system pressure, so it is nRT over V. The work is then the integration of nRT over V for the volume changes and it gives -1.15 X 10 to 4 Joule.

For the isothermal process of ideal gas, the internal energy change is zero, since the internal energy of ideal gases is a function of temperature only. Thus from the first law, the heat delta Q is - delta W, and it is 1.15 x 10 to 4 Joule. Now, we can easily calculate the entropy change of the system . Since the process is reversible, the heat involved is the reversible heat. Inserting the heat just calculated into the definition of entropy gives 38.3 J/K For surrounding, the actual heat transferred is the reversible heat in this case, since the process is reversible.

By the sign convention, the heat transferred “to the surrounding” is -delta Q reversible, so the entropy change of the surrounding is -38.3 J/K The total entropy change of the universe is the summation of delta S of the system and delta S of the surrounding, thus is zero. Therefore, we see here that for a reversible process, the total entropy change of the universe is zero, and it is the 2nd law of thermodynamics. Now let’s consider irreversible case. The gas adiabatically expands against the vacuum. Since the resisting pressure is zero, the vacuum here, the work done by the system is zero. The heat is also zero since the process is adiabatic.

Then by the 1st law of thermodynamics, the internal energy change is zero. Since internal energy of an ideal gas only depends on temperature, the temperature is invariant in this case, at 300K. So, we can see that this irreversible process and the case one, the reversible process have the same initial and final states. Here, we intuitively know that this expansion occur spontaneously since its against vacuum. The entropy change of the system does not depends on whether the process is reversible or not. It only depends on the initial and final states since entropy is a state function. If different processes have the same initial and final states, the entropy changes of those processes are all the same.

It is the entropy change for the reversible process. So here the entropy change of the system is the same with that of the case I, the reversible process. The entropy change of the surrounding is now different from the case I. Since the entropy change of the surrounding is not a state function, and it depends on the actual heat. The actual heat here is zero, since the process is adiabatic. Now, the total entropy change of the universe is + 38.3 J/K, the positive value. So, for the spontaneous, irreversible process, the total entropy change of the universe is positive and it is the second law. Here we summarize the equations describing the 2nd law. The equations are entropy of the system.

For the close system, the entropy of the system is Q reversible divided by temperature, and also the reversible heat is the same with Q actual - lost work. For the entropy change of the open system, we need to consider the entropy comes in and out accompanied with the materials in transit. So the entropy change of the open system is the entropy change for the closed system + entropy goes in with the material - entropy goes out with the material.