One statement of the Clausius inequality is
$$dS \geq \frac{\delta q}{T}$$
where $\delta q$ is the exchanged heat. My understanding is that $T$ refers to the temperature of the surroundings rather than the temperature of the system itself. This is because when deriving the Clausius inequality, we use
$$dS = dS_\text{proc} + dS_\text{exch}$$
where the total entropy change of the system is the change in the entropy due to a spontaneous process plus the change in entropy due to heat exchange with surroundings. Since the surroungings are kept in thermal equilibrium, it follows that $dS_\text{exch} = -dS_\text{surr} =\frac{\delta q}{T}$ where $T$ is the tempreature of the surroundings. Since $dS_\text{proc}\geq 0$ (with equality only in the case of a reversible process), we obtain the Clausius inequality.
However, when considering thermodynamic potentials, it seems that $T$ is used to refer to temperature of the system rather than the temperature of the surroundings. For example, to derive the spontaneity criterion for Helmholtz free energy, we use
$$dA = dU – TdS – SdT = \delta q + PdV – TdS – SdT.$$
At constant $T$ and $V$, this becomes
$$dA = \delta q – TdS$$
and we use the Clausius inequality to state $dA \leq 0$. In this derivation, however, it seems that $T$ refers to the temperature of the system rather than the surroundings, so I am not sure how the Clausius inequality can be correctly applied here.
Best Answer
Your interpretation of the Clauius inequality is correct, and T is the temperature of the surroundings. However, your interpretation of how the Helmholtz free energy is applied is incorrect. In the case of applying the Helmholtz free energy. we assume that for both reversible and irreversible processes on a closed system, the surroundings are constantly maintained at the same temperature as the initial temperature of the system T throughout the process for both reversible and irreversible processes. So from the first law of thermodynamics, we have $$\Delta U=Q-W$$ and, from the 2nd law of thermodynamics we have $$\Delta S=\frac{Q}{T}+\sigma$$, where $\sigma$ is the generated entropy. If we combine these two equations, we obtain: $$\Delta U=T\Delta S-T\sigma-W$$or under these constant external temperature conditions, $$\Delta A=-W-T\sigma$$or$$W=-\Delta A-T\sigma$$So, for a given pair of end states, the maximum work that the system can do is for a reversible path, and is equal to $-\Delta A$. For irreversible paths between the same two end states, the irreversible work is less than for the reversible path. Again, all this applies only to cases where the surroundings are maintained at the same temperature as the initial temperature of the system.