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One liter of alcohol $(1000 \textrm{ cm}^3)$ in a flexible container is carried to the bottom of the sea, where the pressure is $2.6 \times 10^6 \textrm{ N/m}^2$. What will be its volume there?

Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014. Reprinted by permission of Pearson Education Inc., New York.
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Quick Answer: 

$998 \textrm{ cm}^3$

Giancoli 7th Edition, Chapter 9, Problem 44


Chapter 9, Problem 44 is solved.

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Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. The change in volume of this alcohol divided by its original volume is gonna be the negative of 1 over the bulk modulus of alcohol times its change in pressure that it experiences as it goes from what we assume to be originally atmospheric pressure to this pressure under the water. So we can solve for ΔV by multiplying both sides by V naught and we have V naught divided by B times ΔP. So the original volume is 1000 cubic centimeters we divide that by 1.0 times 10 to the 9 newtons per square meter— bulk modulus of alcohol— times the final pressure 2.6 times 10 to the 6 newtons per meter squared at the bottom of the ocean minus the original atmospheric pressure 1.01 times 10 to the 5 newtons per meter squared and that gives negative 2.499 and the units are cubic centimeters since we have cubic centimeters here and this newtons per meter squared cancels with these newtons per meter squared. So the final volume then will be the original plus this change so that's a 1000 minus 2.499 and this is precise to the ones place and so when we do the subtraction, our answer is also precise to the ones place and so we have 998 cubic centimeters.