= Compressibility
{title2=$\kappa_X=-V^{-1}(\partial V/\partial P)_{X,N}$}
{wiki}
Compressibility measures how readily a sample's <volume (thermodynamics)> decreases when its <pressure> increases. For a <simple compressible system> with fixed particle number $N$ and composition, specify the constraint $X$:
$$
\kappa_X=-\frac1V\left(\frac{\partial V}{\partial P}\right)_{X,N},\qquad
\frac{dV}{V}=-\kappa_X\,dP\quad(X,N\text{ fixed}).
$$
Here $P$ denotes <pressure>, also written $p$ elsewhere. The minus sign makes <compressibility> positive on a mechanically stable equilibrium branch. Its units are inverse <pressure>, $\mathrm{Pa^{-1}}$; its reciprocal $K_X=1/\kappa_X$ is the constrained <bulk modulus>, measured in <pascal (unit)> units. Holding <temperature> fixed gives <isothermal compressibility>; holding <entropy> fixed gives <isentropic compressibility> $\kappa_S$ along an <isentropic process>. The latter describes a reversible <adiabatic compression>, since an irreversible <adiabatic process> can increase <entropy>. https://ocw.mit.edu/courses/3-20-materials-at-equilibrium-sma-5111-fall-2003/e402e3f0f555e15ff3af29b5b44b4e57_notes1.pdf[G. Ceder, MIT thermodynamics notes, section 1.14: compressibility]{ref}
For an <ideal gas>, the <ideal gas law> $PV=Nk_BT$ and the <reversible ideal-gas adiabat> give, at the same equilibrium state,
$$
\kappa_T=\frac1P,\qquad
\kappa_S=\frac1{\gamma P},\qquad
\gamma=\frac{C_P}{C_V}>1.
$$
Here $C_P$ and $C_V$ are the whole sample's <heat capacity at constant pressure> and <heat capacity at constant volume>, and $\gamma$ is the <heat capacity ratio>. Constant $\gamma$ is required for the finite-path formula $PV^\gamma=\mathrm{constant}$. Reversible <adiabatic compression> raises the <temperature>, making the <pressure> rise more steeply than in an <isothermal compression>. Thus keeping <entropy> fixed makes the <gas> harder to compress locally. For a classical <monatomic gas> at $100\,\mathrm{kPa}$, $\gamma=5/3$, so $\kappa_T=10^{-5}\,\mathrm{Pa^{-1}}$ and $\kappa_S=6\times10^{-6}\,\mathrm{Pa^{-1}}$; the corresponding values of the <bulk modulus> are $100\,\mathrm{kPa}$ and $166.67\,\mathrm{kPa}$.
For a smooth, stable single <thermodynamic phase> at positive <temperature>, <Maxwell relations> also imply
$$
\frac{\kappa_T}{\kappa_S}=\frac{C_P}{C_V},\qquad
\kappa_T-\kappa_S=\frac{TV\alpha^2}{C_P},
$$
where $\alpha$ is the <coefficient of thermal expansion>. The <ideal gas> result is one example of this broader link between mechanical and thermal responses. The equality $\kappa_T=\kappa_S$ is possible when $\alpha=0$ and the other responses remain finite. These formulas compare equilibrium <partial derivatives>, rather than arbitrary fast or dissipative compression histories. https://ocw.mit.edu/courses/2-43-advanced-thermodynamics-spring-2024/1Q7kBt43LW4oPrh3pxYjs2xxe7HWsC1oR_transcript.pdf[G. P. Beretta, MIT 2.43, Lecture 7: response-function identities]{ref}
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