A scuba tank has a volume of 10 L and is filled with air to a pressure of 250 at

Question

A scuba tank has a volume of 10 L and is filled with air to a pressure of 250 atm at a temperature of 25 degrees C. We will approximate the behavior of air as that of nitrogen gas (N2). For nitrogen, Pc = 33.5 atm and Tc = 126 K. The tank dispenses air to the diver through a regulator, delivering 0.5 L of air per breath, at a pressure of 2 atm and temperature of 25 degrees C. The diver takes 20 breaths per minute. How long can the diver stay under water?

How do you solve this problem is the answer is around 120 minutes?

Explanation / Answer

Moles of air required for 0.5L of air per breath at 2atm and 25 deg.c (25+273.15=298.15)

n= 2*0.5/(0.08206*298.15)= 0.040873 mol/per breath

number of breaths the diver takes 20 breaths/minutes

moles of air required.per minuted = 0.040873*20 mol/minute=0.817455 moles/minute

From

Ideal gas law can't be used at a pressure of 250atm hence based on crticia properites

compressibilty factor will have to be determined

Tr= T/Tc= (25+273.15)/ 126= 2.36627

Pr= P/Pc= 250/33.5 =7.462687

From thermodynamic tables, Zo and Z1 need to be calculated

Zo =1.1 and Z1= 0.32

for Nitrogen the accentirc factor =0.040

Z=Z0+acentric factor*Z1

Z= 1.1+acentric factor*0.040 =1.1+0.32*0.040 =1.1128

From PV= nZRT

n= PV/ZRT= 250*10/(1.1128*0.08206*298.15)= 91.82 moles

moles consumed per minute =0.817455

the diver can stay for 91.82 moles/ 0.817456 moles.minute =112minutes

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