Mass of air breathed during diving

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The discussion focuses on calculating the mass of air a scuba diver breathes during a dive, given the initial and final pressures and temperatures of the air tank. Participants suggest using the ideal gas law, specifically the equations PV=nRT and PV=mRairT, to determine the volume of air before and after the dive. There is confusion regarding the calculations, particularly when the resulting volume exceeds the tank's original capacity. It is emphasized that the mass of air can be derived by calculating the number of moles at the start and end of the dive, then finding the difference. The average molar mass of air is noted as essential for these calculations, with a consensus on using the universal gas constant for clarity.
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Homework Statement



Given: A scuba diver has a tank with an internal volume of 2500 cm^3 air tank which is pressurized to 22 MPa at 27 deg C. After diving for 20 minutes, she notices that the air pressure has reduced to 6 MPa while the temperature of the tank has dropped to 18 deg C.

Find: What is the mass of air that the diver has breathed through her lungs during the dive?


Homework Equations



P2*V2/T2 = P1*V1/T1 or Pv=nRT ?

The Attempt at a Solution



I'm confused which equation we use to find the solution and from there on how to punch in all the numbers to get to the solution.

I know the volume of air is 0.0883 m^3/Kmol, Rair = 0.2870 KJ/Kg*K, and mass = 28.97 kg/kmol
Also R = 8.314 KJ/Kg*K
I'm just really confused on where everything goes into play and why?
 
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You should be able to use the first equation to get the volume of air left after 20 minutes. Once you get that volume then you can use PV=nRT or PV=mRairT
 
rock.freak667 said:
You should be able to use the first equation to get the volume of air left after 20 minutes. Once you get that volume then you can use PV=nRT or PV=mRairT

When I found the volume it came out to be bigger than the original volume which doesn't feel right because I converted the cm^3 to m^3 and then switched to Kelvin for temp and the for the MPa conversion I did 10^3. Doing all this I came out with a volume that was 0.0088 m^3 roughly. Or is the volume supposed to be bigger?
 
The first formula (general transformation of ideal gas) is obtained considering a process involving a fixed amount of gas.
This is not what you have here.
 
nasu said:
The first formula (general transformation of ideal gas) is obtained considering a process involving a fixed amount of gas.
This is not what you have here.
Would I go about it the way rock described but only using the second part PV=mRairT?
 
The volume of air in the tank is unchanged.
Find the mass before and after. And then the difference.
The mass is related to number of moles by n=m/M where M is the molar mass.
 
I would separately calculate n from PV=nRT for the number of moles at the beginning and at the end. That would tell me how many moles of the air mixture were spent. Then multiply by the average molar mass of air (which you can calculate knowing air composition).

My bet is that the average molar mass of air is incorporated into Rair, but I never liked the idea of using separate R values for separate gases. It hides the generality of the ideal gas equation and Avogadro's hypothesis, leaving students confused.
 
The universal R = 8.314 J/K*mol

MW air = 28.97 g/mol

Rair = (8.314 J/K*mol )/(0.02897 kg/mol) = 287 J/K*kg = 0.287 kJ/K*kg
 

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