Calculate Change in Volume of Seawater at 10.9 km Depth

[asleight]

Homework Statement

In the Challenger Deep of the Marianas Trench, the depth of seawater is 10.9 km and the pressure is [tex]1.16\times10^8[/tex] Pa (about [tex]1.15\times10^3[/tex] 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 [tex]1.0\times10^5[/tex] Pa. Assume that for seawater k is [tex]45.8\times10^{-11}[/tex]/Pa.)

Homework Equations

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

The Attempt at a Solution

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

 

[LowlyPion]

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

 

[asleight]

They give you the bulk modulus k for the water and the change in pressure and the volume, so ...

[tex]K=-V\delta\rho/\delta V\rightarrow\Delta V=-V\Delta\rho/k[/tex]?

 

[LowlyPion]

[tex]K=-V\delta\rho/\delta V\rightarrow\Delta V=-V\Delta\rho/k[/tex]?

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

 

[asleight]

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

That sounds better.

 

[asleight]

I got a [tex]\Delta V > V[/tex]...

 

[LowlyPion]

I got a [tex]\Delta V > V[/tex]...

I don't think so.

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

 

[asleight]

I don't think so.

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

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

 

[LowlyPion]

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

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.)

 

[asleight]

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.)

[tex]\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[/tex] right?

 

[LowlyPion]

[tex]\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[/tex] right?

I think it's 45.8 not 4.58

[tex]\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[/tex]