# Air sacs fish problem

1. Nov 2, 2007

I don't know why I can't set up this proportion correctly.

A fish maintains its depth in fresh water by adjusting air contents in air sacs in its body. With its air sacs fully collapsed the fish has a density of 1.08 g/cm^3.

To what fraction of its expanded body volume does it need to inflate its air sacs to reduce its density to that of the water.

I can't seem to set this up. It is asking for $$\frac{V_{empty}}{V_{max}}$$ I know that $$\rho =m/V$$ $$\rho_w=\frac{1g}{cm^3}$$ and $$\rho_{fish}=1.08\frac{g}{cm^3}$$

I am not sure what my problem is right now. Anyine have a hint?

Casey

Am I an idiot for not getting this?

Last edited: Nov 2, 2007
2. Nov 2, 2007

### saket

Work with two variables. Yo should get it.

3. Nov 2, 2007

I'm sorry? I don't follow.

Casey

4. Nov 2, 2007

### saket

Okay one way could be:
Assume final volume (after expansion) be y, and the volume of air inhaled be x. You need to find x/y.
Set up relations from given data.

Last edited: Nov 2, 2007
5. Nov 2, 2007

The volume of air inhaled IS the volume of the sacs after expansion isn't it?

Casey

6. Nov 2, 2007

### saket

Assuming density of air to be negligible, mbefore expansion = mafter expansion.
Relate x and y using densities and above assumption.

7. Nov 2, 2007

### saket

yes!

8. Nov 2, 2007

Okay I know that the answer is .074 from the back of the book.

I used $$\frac{m/V}{m/V'}=\frac{1}{1.08}$$ which gets me$$\frac{V'}{V}= .9259$$ but that is not quite it..I have to subtract that number from 1 to get .074. So I am clearly misunderstanding the question.

Why do I need to do 1-V'/V ? Is it because V'/V is the part of the sacs IN USE and the entire air sac is 1?

Casey

Last edited: Nov 2, 2007
9. Nov 2, 2007

### saket

Think and realize, what is your V and V' in the formula?
It is mis-typed I guess.. not 11.08.. it is 1.08

Last edited: Nov 2, 2007
10. Nov 2, 2007

V' is the volume after inhalation and V is volume before inhalation.

oops I forgot \frac. I edited it. I used 1/1.08 since density of h20= 1g/cm^3

11. Nov 2, 2007

### saket

Really?? If V' is total volume after inhalation, and V before it.. how come V'/V is less than 1?? After inhalation, volume of the fish should increase isnt it??

12. Nov 2, 2007

Pkay...let me start over.
$$\rho$$ before inhale=1.08 $$\rho'$$ after inhale =1
$$\Rightarrow \frac{m/V}{m/V'}=\frac{1.08}{1}$$
$$\Rightarrow \frac{V}{V'}=\frac{1}{1.08}=.9259$$

This is the fraction of V'/V

13. Nov 2, 2007

### saket

14. Nov 2, 2007

I am asked "To what fraction of its expanded body volume does it need to inflate its air sacs to reduce its density to that of the water.

I am still trying to make the conection between V'/V-->1-(V'/V)

Can you put 1-(V'/V) into words.....I would like to understand this better for future problems.

Thanks for your help thus far!

Casey

15. Nov 2, 2007

### saket

It is not 1 - (V'/V) .. rather it is 1 - (V/V'). (Note that V/V' = 0.9259, see your second last post.)
Now 1 - (V/V') = (V' - V)/V'.
V' = Final Volume of the fish after inhaling (i.e., expanded body volume)
V' - V = Final Volume of the fish (after inhaling) - Initial Volume of the fish (before inhaling) = Volume of air inhaled.

So, it is 1 - (V/V'), that you were asked!

16. Nov 2, 2007