I would use PV = nRT and solve for n when inhaling 2.50 L.
Then use PV = nRT again, with n from above, new V (1.70 L) and new T. Solve for new P (in atm).
Typically, when a person coughs, he or she first inhales about 2.50 L of air at 1.00atm and 25 degreesC. The epiglottis and the vocal cords then shut, trapping the air in the lungs, where it is warmed to 37 degrees C and compressed to a volume of about 1.70L 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.
6 answers
1.53 atm
Ashley is wrong
Use this formula to find final pressure before the release:
P1V1/T1 = P2V2/T2
Isolate P2:
P2=P1V1T2/T1V2
= ((1.00atm)(2.50L)(310.15K))/ ((298.15K) (1.70L)
=1.529776.....atm
w/ sig figs = 1.53 atm
(remember to convert *C to kelvin)
P1V1/T1 = P2V2/T2
Isolate P2:
P2=P1V1T2/T1V2
= ((1.00atm)(2.50L)(310.15K))/ ((298.15K) (1.70L)
=1.529776.....atm
w/ sig figs = 1.53 atm
(remember to convert *C to kelvin)
155
1.41 atm. Ur welcome