# Homework Help: Pressure and scuba

1. Apr 9, 2007

### Rasine

A scuba diver at a depth of 43.0 m below the surface of the sea off the shores of Panama City, where the temperature is 4.0°C, releases an air bubble with volume 17.0 cm3. The bubble rises to the surface where the temperature is 19.0°C. What is the volume of the bubble immediately before it breaks the surface? The specific gravity for seawater is 1.025.

so (p1v1/t1)=(p2v2/t2)

v2=v1(p1/p2)(t2/t1)

p1=101 kPa+ (roe)gh where roe is 1.025 x10^3 Kpa

what is g here?

and i am not sure what units this all should be in

this is what i did....
v2=.17m^3(44176kPa/101kpa)(292.15K/277.15K)

should .17m be in 17 cm^3 or what and what about he kPa can it stay that way

i am confused

2. Apr 9, 2007

### Dick

The units in this case are easy. In v2=v1(p1/p2)(t2/t1), you can see that the units of pressure and temperature simply cancel and the units of v2 are the same as v1. So just put v1=17cm^3. BTW 17cm^3 is NOT the same as .17m^3!!!!!!

3. Apr 9, 2007

### Rasine

i still get the wrong answer

4. Apr 9, 2007

### Dick

And in rho*g*h, rho is not a pressure, it's the density of the water. And g is the acceleration of gravity.

Last edited: Apr 9, 2007
5. Apr 9, 2007

### Rasine

some one please tell me what values i have wrong

6. Apr 9, 2007

### Dick

44176kPa is dead wrong. Try doing g*rho*h again.

Last edited: Apr 9, 2007
7. Apr 9, 2007

### Rasine

ok so (1.025x10^3)(9.8)(43)=431935 ....is that right

8. Apr 9, 2007

### Dick

Yes. But remember that it's in Pa not kPa. Convert to kPa before taking the ratio.

9. Apr 9, 2007

### Rasine

ok ok let me try

10. Apr 9, 2007

### Rasine

so now for the entire equation i am getting 7.66 cm^3 but that is wrong

11. Apr 9, 2007

### Dick

What is the 'entire equation'? The decrease in pressure and the increase in temperature should both cause it to expand.

12. Apr 9, 2007

### Rasine

v2=v1(p1/p2)(t2/t1)

v2=17cm^3(431935kPa/101kpa)(292.15K/277.15K) and for this i am getting 7.66 cm^3

13. Apr 9, 2007

### Dick

Make that 431.935kPa (I TOLD you). And how can v2 be less than 17 if both ratios are larger than 1!

14. Apr 9, 2007

### Rasine

ok so now that i corrected 431935 to 431.935 i get 76.64 which is not right

15. Apr 9, 2007

### Dick

Ok, I'll bite. What is the right answer?

16. Apr 9, 2007

### Rasine

i don't know

17. Apr 9, 2007

### Rasine

i am putting this into a hw program and it keeps saying that my answer is not right

18. Apr 9, 2007

### Dick

Ah. You didn't add atmospheric pressure to p1 like you did last time. p1=101 kPa+ (roe)gh.

19. Apr 9, 2007

### Rasine

i am putting the values that i get into a hw program online and it keeps saying that it is wrong...and it won't say what the right answer is

20. Apr 9, 2007

### Rasine

ok ok let me try

21. Apr 9, 2007

### Rasine

i get 94.56 cm^3.....but i ran out of ties

22. Apr 9, 2007

### Dick

Sorry to hear you ran out of time. But the answer is correct.

23. Apr 9, 2007

### Rasine

thank anyways but now i know how to do it