1
$\begingroup$

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 $\mathrm{A = V_1e^{KP_1}}$

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

$A$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.