# How can we prove the inequality TdS > dU + pdV?

I know that for reversible process

$$T\,\mathrm dS = \mathrm dU + p\,\mathrm dV,$$

but since all the terms in the given equation are state variables so the above equation should also be true for irreversible processes but in my textbook for irreversible process the inequality is used for irreversible process.

• Lowercase letter $s$ is used to denote specific entropy. Are you absolutely sure this is what you are asking about? Just in case I corrected the symbol for the entropy to $S.$ Oct 29 '20 at 8:56
• The equality is always true. The equation just describes the relationship between dS, dU and dV that must exist between two closely neighboring (i.e., differentially separated) thermodynamic equilibrium states of a closed system. It has nothing to do with any process, no matter how complicated or tortuous, that took the system from the first state to its neighbor. Oct 29 '20 at 10:24

For a closed system with a single phase of a pure substance, when only expansion work is possible

$$\mathrm dU= -\mathrm p\,\mathrm dV + T\,\mathrm dS, \label{eqn:1}\tag{1}$$

where $$p$$ (the state variable) is the internal pressure and $$T$$ (the state variable) is the internal temperature of the system.

The first law, however, says more generally that

$$\mathrm dU = \mathrm dw + \mathrm dq. \label{eqn:2}\tag{2}$$

Equation \eqref{eqn:2} was derived by applying equation \eqref{eqn:1} to a reversible process.

For the reversible process, thermal and mechanical equilibrium between internal and external pressure means that $$p_\mathrm{ext} = p$$ and $$T_\mathrm{ext} = T,$$ and, combined with the second law, that

$$\mathrm dw = -p_\mathrm{ext}\,\mathrm dV = -p\,\mathrm dV \tag{3}$$

and

$$\mathrm dq = -T_\mathrm{ext}\,\mathrm dS_\mathrm{ext} = T\,\mathrm dS. \tag{4}$$

More generally, the second law says that

$$\mathrm dS_\mathrm{tot} = \mathrm dS_\mathrm{sys} + \mathrm dS_{surr} \ge 0 \tag{5}$$

the equality holding for a reversible process. Dropping the label for the system

$$\mathrm dS \ge -\mathrm dS_\mathrm{surr} \quad\leftrightarrow\quad \mathrm dS \ge \frac{\mathrm dq}{T_\mathrm{surr}}. \tag{6}$$

Therefore for an isothermal process (thermal equilibrium condition applies)

$$\mathrm dS \ge \frac{\mathrm dq}{T}. \tag{7}$$

The inequality applies for instance to an isothermal free expansion.

In a reversible process, the pressure $$p$$ of $$p\,\mathrm dV$$ is the internal pressure of the gas $$p = nRT/V.$$ In an irreversible process, the pressure $$p$$ of $$p\,\mathrm dV$$ is the value of the outer pressure which is usually much bigger than the internal pressure.

The piston is quickly pressed and the gas is quickly compressed. This irreversible compression produces more heat than the reversible compression, where the internal and the external pressure are continuously adjusted to be equal (or nearly equal) to each other.