Calculate Change in Volume of Seawater at 10.9 km Depth

In summary, the depth of seawater in the Challenger Deep of the Marianas Trench is 10.9 km with a pressure of 1.16\times10^8 Pa. The change in volume of a cubic meter of water taken from the surface to this depth is -0.00531 m^3, calculated using the bulk modulus and compressibility of seawater. There was some initial confusion about the values and equations, but the correct solution was determined to be -0.0531 m^3.
  • #1
asleight
152
0

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.
 
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  • #2
They give you the bulk modulus k for the water and the change in pressure and the volume, don't they? So ...
 
  • #3
LowlyPion said:
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]?
 
  • #4
asleight said:
[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.
 
  • #5
LowlyPion said:
Sorry. I think I misspoke. I think k is the compressibility, which is the reciprocal of the bulk modulus.

That sounds better.
 
  • #6
I got a [tex]\Delta V > V[/tex]...
 
  • #7
asleight said:
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]
 
  • #8
LowlyPion said:
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.
 
  • #9
asleight said:
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.)
 
  • #10
LowlyPion said:
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?
 
  • #11
asleight said:
[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]
 

FAQ: Calculate Change in Volume of Seawater at 10.9 km Depth

1. How do you calculate the change in volume of seawater at 10.9 km depth?

To calculate the change in volume of seawater at 10.9 km depth, you can use the formula: ΔV = Vf - Vi, where ΔV is the change in volume, Vf is the final volume, and Vi is the initial volume. This formula takes into account the change in pressure at 10.9 km depth and can be calculated using the ideal gas law.

2. What is the ideal gas law?

The ideal gas law is a fundamental equation in thermodynamics that relates the pressure, volume, temperature, and number of moles of an ideal gas. It can be expressed as PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

3. How do you determine the pressure at 10.9 km depth in seawater?

The pressure at 10.9 km depth in seawater can be determined using the equation P = ρgh, where P is the pressure, ρ is the density of seawater (which can be found in a table), g is the acceleration due to gravity, and h is the depth. By plugging in the values for seawater at 10.9 km depth, you can calculate the pressure.

4. What factors can affect the change in volume of seawater at 10.9 km depth?

Several factors can affect the change in volume of seawater at 10.9 km depth, including temperature, pressure, and salinity. These factors can all impact the density of seawater, which in turn affects the volume. Additionally, changes in the composition of seawater, such as the presence of dissolved gases, can also affect the change in volume.

5. Why is it important to calculate the change in volume of seawater at 10.9 km depth?

Calculating the change in volume of seawater at 10.9 km depth is important for several reasons. Firstly, it can help us understand the dynamics of the ocean and how different factors can affect its properties. It is also important for industries such as deep-sea exploration and oil drilling, where accurate calculations of pressure and volume are necessary for safety and efficiency. Understanding the change in volume of seawater at different depths can also provide valuable information for climate studies and predicting the effects of global warming on the ocean.

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