The discussion revolves around deriving an expression for the volume of a liquid as a function of pressure, based on the concept of isothermal compressibility. The context includes a homework problem that requires understanding the relationship between volume and pressure changes in liquids.
Participants generally agree on the steps taken to derive the expression for volume as a function of pressure, but there is no explicit consensus on the understanding of differential equations among all participants.
Some participants express uncertainty about their knowledge of differential equations and integration techniques, which may limit their ability to fully engage with the mathematical derivation.
Start out by solving your equation for dV/dP.chickymd said:Homework Statement
When pressure is applied to a liquid, its volume decreases. Assuming that the isothermal compressibility κ=-1/V(δV/δP) is independent of pressure, derive an expression for the volume as a function of pressure.
Homework Equations
The Attempt at a Solution
I don't know where to start, but I know the answer should be V2=V1 exp[-κ(P2-P1)].
Have you learned how to solve a differential equation like this by separation of variables or by integrating factor?chickymd said:δV/δP=-κV
I'm not sure what to do now.
Have you learned how to integrate dV/V?chickymd said:No, I haven't taken differential equations.
Do you know what the derivative of ln(x) with respect to x is equal to?chickymd said:No, I haven't.
OK. This is where we start.chickymd said:It's 1/x
chickymd said:-κV=δV/δP
-κδP=δV/V=δln(V)
I can get this far, but I don't know how to solve it now.
Excellent job! Just get rid of that extra equal sign in the last equation.chickymd said:Do I just integrate both sides from 1 to 2?
∫(from 1 to 2) -κδP=∫(from 1 to 2) δln(V)
-κ(P2-P1)=ln(V2)-ln(V1)=ln(V2/V1)
V2/V1=e^(-κ(P2-P1))
V2=V1=e^(-κ(P2-P1))
Is this correct?