# Pressure and Volume

1. Nov 9, 2008

### asleight

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

In the Challenger Deep of the Marianas Trench, the depth of seawater is 10.9 km and the pressure is $$1.16\times10^8$$ Pa (about $$1.15\times10^3$$ atm).

If a cubic meter of water is taken from the surface to this depth, what is the change in its volume? (Normal atmospheric pressure is about $$1.0\times10^5$$ Pa. Assume that for seawater k is $$45.8\times10^{-11}$$/Pa.)

2. Relevant equations

I assumed that $$\rho_0V_0=\rho V$$, but that is not true. So, I don't know what to do.

3. The attempt at a solution

The above was my attempt, resulting in a change in volume of 0.091 m3, which didn't work.

2. Nov 9, 2008

### LowlyPion

They give you the bulk modulus k for the water and the change in pressure and the volume, don't they? So ...

3. Nov 9, 2008

### asleight

$$K=-V\delta\rho/\delta V\rightarrow\Delta V=-V\Delta\rho/k$$?

4. Nov 9, 2008

### LowlyPion

Sorry. I think I misspoke. I think k is the compressibility, which is the reciprocal of the bulk modulus.

5. Nov 9, 2008

### asleight

That sounds better.

6. Nov 9, 2008

### asleight

I got a $$\Delta V > V$$...

7. Nov 9, 2008

### LowlyPion

I don't think so.

$$\Delta v = \Delta p*v*k = (1.15*10^3 Pa)*(1 m^3)*(45.8*10^{-11} /Pa)$$

8. Nov 9, 2008

### asleight

Something's wrong... I've solved and got $$-5.3\times10^{-2}$$. It's completely wrong.

9. Nov 9, 2008

### LowlyPion

That's what I get. And that seems about right at about 5% smaller. (Ignore the typo from the earlier equation, I just wrote in the value I scanned from the problem and switched the atm and Pa values.)

10. Nov 9, 2008

### asleight

$$\Delta V=-k\Delta pV=-\frac{4.58\times10^{-11}}{Pa}\frac{1.16\times10^8 Pa-1.0\times10^5 Pa}{1}\frac{1m^3}{1}=-0.00531m^2$$ right?

11. Nov 9, 2008

### LowlyPion

I think it's 45.8 not 4.58

$$\Delta V=-k*\Delta p*V = -(\frac{45.8\times10^{-11}}{Pa})*(\frac{1.16\times10^8 Pa-1.0\times10^5 Pa}{ })*(\frac{1m^3}{ }) = -0.0531m^3$$