# Expression for volume as a function of pressure

1. Jan 21, 2014

### chickymd

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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)].

2. Jan 21, 2014

### Staff: Mentor

Start out by solving your equation for dV/dP.

3. Jan 22, 2014

### chickymd

δV/δP=-κV

I'm not sure what to do now.

4. Jan 22, 2014

### Staff: Mentor

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

5. Jan 22, 2014

### chickymd

No, I haven't taken differential equations.

6. Jan 22, 2014

### Staff: Mentor

Have you learned how to integrate dV/V?

7. Jan 22, 2014

### chickymd

No, I haven't.

8. Jan 22, 2014

### Staff: Mentor

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

9. Jan 22, 2014

### chickymd

It's 1/x

10. Jan 22, 2014

### Staff: Mentor

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?

11. Jan 22, 2014

### chickymd

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

12. Jan 22, 2014

### chickymd

-κV=δV/δP
-κδP=δV/V=δln(V)

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

13. Jan 22, 2014

### chickymd

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?

14. Jan 22, 2014

### Staff: Mentor

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

15. Jan 22, 2014

### chickymd

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