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Soaring Crane
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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.
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.
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