± Biochemical Applications of the Gas Laws The various gas laws can be used to d

Question

± Biochemical Applications of the Gas Laws

The various gas laws can be used to describe air, which is a mixture of gases. In some cases, these laws have direct application to the air that we breathe.

Part A

How long does it take a person at rest to breathe one mole of air if the person breathes 82.0 mL/s of air that is measured at 25 C and 755 mmHg?

Express your answer numerically in seconds.

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Part B

Typically, when a person coughs, he or she first inhales about 2.00 L of air at 1.00 atm and 25 C. The epiglottis and the vocal cords then shut, trapping the air in the lungs, where it is warmed to 37 C and compressed to a volume of about 1.70 L by the action of the diaphragm and chest muscles. The sudden opening of the epiglottis and vocal cords releases this air explosively. Just prior to this release, what is the approximate pressure of the gas inside the lungs?

Express your answer numerically in atmospheres.

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Part C

Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What percent (by moles) of He is present in a helium-oxygen mixture having a density of 0.518 g/L at 25 C and 721 mmHg?

Express your answer numerically as a percentage.

± Biochemical Applications of the Gas Laws

The various gas laws can be used to describe air, which is a mixture of gases. In some cases, these laws have direct application to the air that we breathe.

Part A

How long does it take a person at rest to breathe one mole of air if the person breathes 82.0 mL/s of air that is measured at 25 C and 755 mmHg?

Express your answer numerically in seconds.

  s  

SubmitHintsMy AnswersGive UpReview Part

Part B

Typically, when a person coughs, he or she first inhales about 2.00 L of air at 1.00 atm and 25 C. The epiglottis and the vocal cords then shut, trapping the air in the lungs, where it is warmed to 37 C and compressed to a volume of about 1.70 L by the action of the diaphragm and chest muscles. The sudden opening of the epiglottis and vocal cords releases this air explosively. Just prior to this release, what is the approximate pressure of the gas inside the lungs?

Express your answer numerically in atmospheres.

  atm  

SubmitHintsMy AnswersGive UpReview Part

Part C

Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What percent (by moles) of He is present in a helium-oxygen mixture having a density of 0.518 g/L at 25 C and 721 mmHg?

Express your answer numerically as a percentage.

  %He  

Explanation / Answer

A)

How long does it take a person at rest to breathe one mole of air if the person breathes 82.0 mL/s of air that is measured at 25 C and 755 mmHg?

PV = nRT

V = nRT/P = (1)(0.082)(25+273)/(755/760) = 24.5978 L

rate is

t = V/v = (24.5978) / (82*10^-3) = 299.973 = 300 seocnds

B

Typically, when a person coughs, he or she first inhales about 2.00 L of air at 1.00 atm and 25 C. The epiglottis and the vocal cords then shut, trapping the air in the lungs, where it is warmed to 37 C and compressed to a volume of about 1.70 L by the action of the diaphragm and chest muscles. The sudden opening of the epiglottis and vocal cords releases this air explosively. Just prior to this release, what is the approximate pressure of the gas inside the lungs?

initially

PV = nRT

1*2 = n*(0.082)(298)

n = 2/(0.082*298) = 0.0818464 mol

then

PV = nRT

P = nRT/V = (0.0818464)(0.082)(37+273)/(1.7) = 1.223 atm

C)

Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What percent (by moles) of He is present in a helium-oxygen mixture having a density of 0.518 g/L at 25 C and 721 mmHg?

PV = nRT

n = PV/(RT) = (721)(1)/(62.3*298) = 0.038835 mol

basis = 1 L

mass = 0.518 g

MW of O2 = 32 g/mol

MW of He = 4 g/mol

Density of O2 --> 0.038835 *32/1 = 1.24272 g/L

Density of He --> 0.038835 *4/1 = 0.15534g/L

the baalnce

x1*D-O2 + (1-x1)D-H2 = 0.518

x1*1.24272 + (1-x1)0.15534 = 0.518

(1.24272 - 0.15534 )x = 0.518 - 0.15534

x = ( 0.518 - 0.15534 ) /((1.24272 - 0.15534 )) = 0.333517

so He = 1-0.333517 = 0.667

66.7%

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