Scuba Diving and Ideal Gas Law

[Soaring Crane]

Pressure, volume, and temperature of the air in the diver's lungs when the last underwater breath is taken as p1,V1 , and T1, respectively.

Pressure, volume, and temperature of the air in the diver's lungs when the surface is reached to be p2, V2, and , T2 respectively.

Salt water has an average density of around 1.03 g/cm^3 , which translates to an increase in pressure of 1.00 atm for every 10.0 m of depth below the surface. Therefore, at 10.0 m, the pressure on the diver's lungs and body is 2.00 atm.

I solved for other parts:
The average lung capacity for a human is about 6 L. Suppose a scuba diver at a depth of 15 m takes a deep breath and begins to rise to the surface. What is , the volume of the air in his or her lungs when the diver reaches the surface?

Assume that the temperature of the air is a constant 37 (body temperature).

The main idea is that changing the pressure also changes the volume of the gas. Using initial quantities, the pressure at the surface, and the ideal gas law, the final volume of the air can be determined.

Find the equation to calculate final volume. What is the equation for the final volume of the air in the diver's lungs?


Express your answer in terms of p1,V1 , p2 and .
ANSWER: V2 = p_1*V_1/p_2

Express your answer in atmospheres to three significant digits.
ANSWER: p1 = 2.50 atm


Express your answer numerically in liters to three significant digits.
ANSWER: V2 = 15.0 L



Part B.3 Find the pressure at 15 underwater
The initial pressure is related to the depth from which the diver starts the ascent.


The part I need help with is:
If the temperature of air in the diver's lungs is 37 C (98.6 F), how many moles of air n must be released by the time the diver reaches the surface to keep the volume of air in his or her lungs at 6 L ?

I know I must compute the number of moles of air in 6 L at the underwater pressure and again at the surface pressure. The difference is the number of moles of air that must be exhaled. pV = nRT is used (isolate n), but what numbers do I plug in for each scene (diff. p) and the order of the moles for the difference?

Thanks.

 

[Doc Al]

If the diver did not release any air, the volume would have expanded from 6 L to 15 L. So, the diver must release (15-6)/15 of the number of moles of air that she had underwater. So find the number of moles at that point. You have the equation and the data, you just need to convert the units to standard units. Express pressure in Pa, volume in cubic meters, and the temperature in degrees Kelvin.

 

[DaveC426913]

Suppose a scuba diver at a depth of 15 m takes a deep breath and begins to rise to the surface. What is , the volume of the air in his or her lungs when the diver reaches the surface?

Not that this helps you solve your homework, but Jeez, it is unbelievably irresponsible for a high school textbook to propose an experiment which will kill the experimenter sure as a bullet!

Holding your breath and rising to the surface is gauranteed death by embolism.

Has it ever occurred to the writers that students may not make a distinction among which particular things they can learn in school and which they can't? What if a student decided to try this experiment for real?

 

[Soaring Crane]

(15-6)/15 = .6
n = (pV)/(RT), where R = 8.315, T = 37 C = 310 K, p = 2.5 atm *(1.013 x 10 ^5) = 253250 Pa, V = 6 L = 6,000 cm^3 = .006 m^3

n_underwater = [(253250 Pa)(.006 m^3)]/[8.315(310 K)] = 0.58949 mol

0.58949 mol (.6) = 0.3537 mol ?

 

[Doc Al]

Looks OK to me. (Round it off to a reasonable number of significant figures.)

 

[DaveC426913]

Say, Crane, would you mind sending me the publisher's info about that textbook?