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.
1.53 atm
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)
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).
To find the approximate pressure of the gas inside the lungs just prior to the release, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas (assumed to be constant)
R = ideal gas constant
T = temperature in Kelvin
First, we need to convert the temperatures from degrees Celsius to Kelvin by adding 273.15:
Initial temperature (25 degrees Celsius) = 25 + 273.15 = 298.15 K
Final temperature (37 degrees Celsius) = 37 + 273.15 = 310.15 K
Now, let's calculate the number of moles of gas using the initial conditions:
PV = nRT
(1.00 atm)(2.50 L) = n(0.0821 L·atm/mol·K)(298.15 K)
Solving for n, we get:
n = (1.00 atm)(2.50 L) / (0.0821 L·atm/mol·K)(298.15 K)
n ≈ 0.1016 mol
Next, we can find the pressure of the gas inside the lungs just prior to the release using the final conditions:
PV = nRT
P(1.70 L) = (0.1016 mol)(0.0821 L·atm/mol·K)(310.15 K)
Solving for P, we get:
P = (0.1016 mol)(0.0821 L·atm/mol·K)(310.15 K) / (1.70 L)
P ≈ 3.09 atm
Therefore, the approximate pressure of the gas inside the lungs just prior to the release is approximately 3.09 atmospheres.