How many particles of air do their lungs contain after inhaling?

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A submarine has run into trouble and is stuck at the bottom of the ocean. Several people are on board and must make their way to the surface without any diving gear. The air pressure aboard the submarine is 3.400 atm. The air temperature inside the submarine is 16.36 °C and you can take body temperature (inside the lungs) to be 37.64 °C.
The first person to leave also takes a breath as deep as possible by exhaling as far as possible (leaving a volume of 1.170 L in their lungs), and then slowly inhaling to increase their lung volume by 4.580 L. His body temperature is also 37.64 °C.

(i)How many particles of air do their lungs contain after inhaling?
 
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You need to demonstrate an attempt at solving this in order to get help here.
 
I used V1/T1=V2/T2
where V1=4.88, T1=309.19k, T2=286.7k and V2 is the unknown.
then I put V2 in the formula PV2=nRT P-3.5atm T-309.19
I got 6.16^e-3 but the answer is 4.92e+23
... far out
 
Why is V2 is unknown? You are given the volume after the exhalation, and you are told by how much it increases after the inhalation.
 
awwwww
I just used PV=nRT
P- atm= 202.658 V=5.13 T=309.19
but its still not right... where went wrong??
 
Why is V = 5.13? And this is 5.13 of what? What units should you be using?
 
the volume of lung? 1.7+4.58m3
then times 6.02*10e23...?
 
Can you image what a cubic meter looks like? Now imagine someone's lungs which are MULTIPLE cubic meters.
 
oh! stupid me!
so that will be 0.00513m3
ummmm in getting 2.44*10^23 ..
 
oh.. that was a typo.. should be 'increase their lung volume by 4.880 L' and so 0.00593 for V
 
no... I am getting 2.81*10^20
... couldn't figure out where has gone wrong...