# Isothermal process gives V = A(T)exp(-KP)?

Substance with isothermal compressibility K, where $\mathrm{K = -\frac{1}V(∂V/∂P)_T}$ undergoes constant isothermal, mechanically reversible process from state ($P_1$, $V_1$) to ($P_2$, $V_1$) where V is molar volume.

(a) Starting with the definition of K, show that the path of the process is described by: $\mathrm{V = A(T)e^{-KP}}$

I don't know how to get A in terms of T, the furthest I can get is $\mathrm{A = V_1e^{KP_1}}$

• The key point is "starting with the definition of K"! – user23061 Jul 15 '16 at 19:38
• You have done well so far. Now do the same thing for point 2. Then, eliminate A between the two equations. – Chet Miller Jul 15 '16 at 22:32
• a useful 'trick' is to notice that $\frac{1}V \frac{dV}{dP} = \frac {d ln(V)}{dP}$ – porphyrin Jul 17 '16 at 8:58

$A$ is already an implicit function of $T$, so your answer is essentially correct.
Starting from the definition of $\kappa$ and integrating both sides with respect to $P$, we obtain $$-\int\kappa\,\mathrm{d}P = \int\frac{1}{V}\frac{\partial V}{\partial P}\,\mathrm{d}P,$$ which yields as solution $-\kappa P = \ln V + f(T)$, where $f(T)$ is our "constant" of integration. Rearranging and exponentiating yields $V(P,T) = e^{-f(T)}e^{-\kappa P} = A(T)e^{-\kappa P}$, $A(T) := e^{-f(T)}$, and we are done.