Expression for volume as a function of pressure

[chickymd]

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)].

 

[Chestermiller]

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)].

Start out by solving your equation for dV/dP.

 

[chickymd]

δV/δP=-κV

I'm not sure what to do now.

 

[Chestermiller]

δV/δP=-κV

I'm not sure what to do now.

Have you learned how to solve a differential equation like this by separation of variables or by integrating factor?

 

[chickymd]

No, I haven't taken differential equations.

 

[Chestermiller]

No, I haven't taken differential equations.

Have you learned how to integrate dV/V?

 

[chickymd]

No, I haven't.

 

[Chestermiller]

No, I haven't.

Do you know what the derivative of ln(x) with respect to x is equal to?

 

[chickymd]

It's 1/x

 

[Chestermiller]

It's 1/x

OK. This is where we start.
If \frac{dln(x)}{dx}=\frac{1}{x}
then
dln(x)=\frac{dx}{x}
Is this OK with you so far?

 

[chickymd]

Yes, I'm following.
Does that mean dV/V=dln(V)?

 

[chickymd]

-κV=δV/δP
-κδP=δV/V=δln(V)

I can get this far, but I don't know how to solve it now.

 

[chickymd]

-κV=δV/δP
-κδP=δV/V=δln(V)

I can get this far, but I don't know how to solve it now.

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?

 

[Chestermiller]

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?

Excellent job! Just get rid of that extra equal sign in the last equation.

 

[chickymd]

Thanks for the help (and for catching my typo).