Calculate Volume of Bubble at Surface from Pressure & Temperature: Panama City

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Homework Help Overview

The discussion revolves around calculating the volume of an air bubble released by a scuba diver at a depth of 43.0 m in seawater, considering changes in pressure and temperature as it rises to the surface. The specific gravity of seawater is also mentioned, and the problem involves understanding the relationships between pressure, volume, and temperature in the context of gas laws.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the ideal gas law and the need to correctly calculate pressure at depth using the specific gravity of seawater. There are questions about unit conversions and the correct values to use in calculations.

Discussion Status

Participants are actively engaged in troubleshooting their calculations, with some providing corrections and clarifications regarding the use of units and the formula. There is a recognition of the need to include atmospheric pressure in calculations, and some participants express confusion about the results they are obtaining.

Contextual Notes

There is an emphasis on ensuring that all values are in consistent units, and participants are navigating the complexities of applying the ideal gas law under varying conditions. The original poster mentions using a homework program that is not accepting their answers, indicating potential discrepancies in their calculations.

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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
 
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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!
 
i still get the wrong answer
 
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:
some one please tell me what values i have wrong
 
44176kPa is dead wrong. Try doing g*rho*h again.
 
Last edited:
ok so (1.025x10^3)(9.8)(43)=431935 ...is that right
 
Yes. But remember that it's in Pa not kPa. Convert to kPa before taking the ratio.
 
ok ok let me try
 
  • #10
so now for the entire equation i am getting 7.66 cm^3 but that is wrong
 
  • #11
What is the 'entire equation'? The decrease in pressure and the increase in temperature should both cause it to expand.
 
  • #12
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
Make that 431.935kPa (I TOLD you). And how can v2 be less than 17 if both ratios are larger than 1!
 
  • #14
ok so now that i corrected 431935 to 431.935 i get 76.64 which is not right
 
  • #15
Ok, I'll bite. What is the right answer?
 
  • #16
i don't know
 
  • #17
i am putting this into a homework program and it keeps saying that my answer is not right
 
  • #18
Ah. You didn't add atmospheric pressure to p1 like you did last time. p1=101 kPa+ (roe)gh.
 
  • #19
i am putting the values that i get into a homework program online and it keeps saying that it is wrong...and it won't say what the right answer is
 
  • #20
ok ok let me try
 
  • #21
i get 94.56 cm^3...but i ran out of ties
 
  • #22
Sorry to hear you ran out of time. But the answer is correct.
 
  • #23
thank anyways but now i know how to do it
 

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