Consider the following situation.

$1m^3$ of water is in the surface of the ocean. If we get that volume of water, and transport it $11km$ down into the ocean, what is going to be the new density of the water?

After a friend of mine asked this question to me, I considered the compressibility equation, that relates pressure and volume: $$ K = -\frac{\Delta v}{\Delta p \text{ }v} $$ Then I look up for some variables:

The normal atmospheric pressure is: $P_i = 1.0\cdot 10^5 \text{ pa}$

The pressure at $11km$ down into the ocean is close to: $P_f = 1.16 \cdot 10^8 \text{ pa}$

The $K$, compressibility of water, is $K = 45.8 \cdot 10^{-11} \text{ pa}^{-1}$.

Then I applied them into the equation:

$$ -\Delta v = K \cdot v_i \cdot \Delta p\\ -\Delta v = 45.8 \cdot 10^{-11} \cdot (1.16 \cdot 10^8 - 1.0\cdot 10^5)\\ -\Delta v = 7.3 \cdot 10^{-10} v_f = 1 - 7.3 \cdot 10^{-10} $$

After that I look up for the density of the water in normal atmosphere: $d_i = 1.03 \cdot 10^3 \frac{kg}{m^3}$

So for $1m^3$ I'll have $1.03 \cdot 10^3$ kg of water in the surface. Since $d = \frac{m}{v}$, I can get the new density($d_f$):

$$ d_f = \frac{m}{v_f}\\ d_f = \frac{1.03 \cdot 10^3}{1 - 7.3 \cdot 10^{-10}}\\ d_f = \frac{1.03 \cdot 10^3}{9.9\cdot 10^{-1}}\\ d_f = 0.104 \cdot 10^4 = 1.04 \cdot 10^3 $$

So the new density of $1m^3$, $11km$ into the ocean, is going to be $d_f = 1.04 \cdot 10^3$?

Am I correct? Because I think that its compression is too small... I don't know if what I've done is indeed correct. Can someone please correct me or let me know if it's everything correct?

And I know that I've made some approximations ok?


@David Hammen noticed that I've made a mistake at $10^8 - 10^5$. Idk why, I think that's because I was tired, and that's why I've got $1.04\cdot 10^3$ as the new density, when it should be more close to $1.08 \cdot 10^3$.


closed as off-topic by Rob Jeffries, sammy gerbil, ZeroTheHero, Yashas, Jon Custer Mar 27 '17 at 3:50

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  • 1
    $\begingroup$ Hi Bruno, We don't do check my work questions, but I can say, as you know, the Mariana Trench is 11km deep. At the bottom of the trench the water column above exerts a pressure of 1,086 bars (15,750 psi), more than 1,000 times the standard atmospheric pressure at sea level. At this pressure, the density of water is increased by 4.96%, so that 95 litres of water under the pressure of the Challenger Deep would contain the same mass as 100 litres at the surface. en.wikipedia.org/wiki/Mariana_Trench $\endgroup$ – user146020 Mar 26 '17 at 18:07
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    $\begingroup$ I get a density of about $1050$ kg/m$^3$ $\endgroup$ – John Rennie Mar 26 '17 at 18:12
  • $\begingroup$ @Countto10 if there's an increase of $4.96%$ the density on Mariana Trench is $1.08\cdot 10^3$.. Maybe it's because of the approximations that I've done! Anyway, thank you mate! $\endgroup$ – Bruno Reis Mar 26 '17 at 18:33
  • $\begingroup$ @JohnRennie Using the same process that I've used mate? $\endgroup$ – Bruno Reis Mar 26 '17 at 18:34
  • $\begingroup$ @BrunoReis - You should be getting a density of about 1.075 for sea water at the bottom of the Mariana Trench. John Rennie's result is for fresh water. $\endgroup$ – David Hammen Mar 26 '17 at 22:08

Your error is here:

$$\begin{aligned}-\Delta v &= 45.8 \cdot 10^{-11} \cdot (1.16 \cdot 10^8 - 1.0\cdot 10^5)\\ &= 7.3 \cdot 10^{-10}\end{aligned}$$

A quick sanity check says that this should be about $53\cdot 10^{-3}$ rather than $7.3\cdot 10^{-10}$.

  • $\begingroup$ Omg, I don't know why, but I did $10^8 - 10^5 = 10^3$. Maybe it's because I was tired. Thank you mate! $\endgroup$ – Bruno Reis Mar 27 '17 at 4:44

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