# Compressibility of $1m^3$ of water $11km$ down the ocean! [closed]

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?

UPDATE

@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 CusterMar 27 '17 at 3:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Rob Jeffries, sammy gerbil, ZeroTheHero, Yashas, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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 – user146020 Mar 26 '17 at 18:07
• I get a density of about $1050$ kg/m$^3$ – John Rennie Mar 26 '17 at 18:12
• @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! – Bruno Reis Mar 26 '17 at 18:33
• @JohnRennie Using the same process that I've used mate? – Bruno Reis Mar 26 '17 at 18:34
• @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. – David Hammen Mar 26 '17 at 22:08

\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 $7.3\cdot 10^{-10}$.
• 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! – Bruno Reis Mar 27 '17 at 4:44