Question

Typically, when a person coughs, he or she first inhales about 2.30 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.

Answer 1

PV/RT = n

State 1 is just after inhalation and has volume 2.30 L, pressure 1 atm, temp 25 C. State 2 is just before coughing and has volume 1.7 L, temp 37 C.

Fortunately, the number of atoms (ie the number of moles of gas) is the same between state 1 and state 2. Also, the gas constant R is the same between state 1 and sate 2. Therefore we have,

PV/RT (in state 1) = PV/RT (in state 2)

Since the gas constant R is the same in both states we can cancel that out and get,

PV/T (in state 1) = PV/T (in state 2)

the temperature in degrees K

2.3 * 1/(25+273) = 1.7 * P/(37+273)

P = 1.41 atm