Thermodynamics: $T\mathrm dS$ equations and second law of thermodynamics

Question

Let's assume that an ideal gas undergoes an irreversible process of heating at constant volume. Using the fundamental relation $$ T\,\mathrm dS = \mathrm dU + p\,\mathrm dV, $$ and $\mathrm dV = 0$, so $$ \mathrm dU = T\, \mathrm dS. $$ But simultaneously, $\mathrm dU = \mathrm dQ$ (as no work is done on the system), and from the second law of thermodynamics, $T\, \mathrm dS > \mathrm dQ$, leading to $T\, \mathrm dS > \mathrm dU$, and contradicting the fundamental $T\, \mathrm dS$ relations! What wrong assumption did I make?

Answer 1

What wrong assumption did I make?

The assumption that the second law requires $TdS\gt \delta Q$ for the system.

In the case of an isochoric (constant volume) heat addition, the heat added to the system between the same two equilibrium states is the same for an irreversible heat transfer as a reversible heat transfer process. Moreover, for any process, reversible or not, the change in entropy of the system between the same two equilibrium states is the same and is calculated on the basis of a reversible path between the states. Consequently, for the system, $TdS=\delta Q$ for both a reversible and irreversible isochoric process.

However, for the irreversible process, the decrease in entropy of the surroundings is less than the increase in entropy of the system, so that $\Delta S_{sys}+\Delta S_{sur}\gt0$ in compliance with the second law. Essentially, the reason the decrease in entropy is less is that the heat out of the surroundings occurs at a higher temperature at the boundary for the irreversible heat transfer, as shown below:

1. For calculating the change in entropy of the system for the irreversible process you assume reversible heat transfer with an infinite set of thermal reservoirs. You can always assume a reversible path since entropy is a system property, independent of the path. Thus the entropy change of the system for both the reversible and irreversible isochoric heat addition process is

$$\Delta S_{sys}=C_{v}\ln\frac {T_f}{T_i}\tag {1}$$

2. For the reversible process, where heat leaves the surroundings

$$\Delta S_{sur}=-C_{v}\ln\frac {T_f}{T_i}\tag {2}$$

3. Thus, for the reversible process

$$\Delta S_{tot}=\Delta S_{sys}+\Delta S_{sur}=0\tag{3}$$

4. But for the irreversible process, where heat leaves the surroundings at a single reservoir temperature equal to the final system temperature

$$\Delta S_{sur}=-\frac{C_{v}(T_{f}-T_{i})}{T_f}\tag{4}$$

5. It can be shown that the right side of eq (4) will always be less than the right side of eq (1). Thus, for the irreversible process

$$\Delta S_{tot}=\Delta S_{sys}+\Delta S_{sur}\gt 0\tag{5}$$

In compliance with the second law.

Hope this helps

Answer 2

You have to set a control volume with a fixed boundary whose outside is a thermostat at some temperature, say $T_0$. If you assume that the system is at temperature $T \ne T_0$ and is subject to only thermal interaction then $T_0 >T.$ If $dS_0$ is the entropy supplied by the thermostat, that is $T_0 dS_0 = \delta Q,$ then because the interaction is not reversible $dS \ne dS_0,$ in fact since $TdS=\delta Q$ (energy conservation) and $dS > d S_0$ we must have $T <T_0$ as required for natural heat conduction.