Question

On a chilly 12∘C day, you quickly take a deep breath--all your lungs can hold, 4.0 L. The air warms to your body temperature of 37∘C.

If the air starts at a pressure of 1.0 atm, and you hold the volume of your lungs constant (a good approximation) and the number of molecules in your lungs stays constant as well (also a good approximation), what is the increase in pressure inside your lungs?

Answer 1

\(1 \mathrm{~atm}=101325 \mathrm{~Pa}\)

Volume of air your lungs can hold \(=\mathrm{V}=4 \mathrm{~L}\)

Initial pressure of the air \(=\mathrm{P}_{1}=1 \mathrm{~atm}=1 \times(101325) \mathrm{Pa}=101325 \mathrm{~Pa}\)

Initial temperature of the air \(=T_{1}=12^{\circ} \mathrm{C}=12+273 \mathrm{~K}=285 \mathrm{~K}\)

Pressure of the air in your lungs \(=\mathrm{P}_{2}\)

Temperature of the air in your lungs \(=T_{2}=37^{\circ} \mathrm{C}=37+273 \mathrm{~K}=310 \mathrm{~K}\)

The volume of your lungs remains constant therefore the volume of the air remains constant.

\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

\(\frac{101325}{285}=\frac{P_{2}}{310}\)

\(P_{2}=110213 \mathrm{~Pa}\)

Increase in pressure in your lungs \(=\Delta \mathrm{P}\)

\(\Delta P=P_{2}-P_{1}\)

\(\Delta P=110213-101325\)

\(\Delta \mathrm{P}=8888 \mathrm{~Pa}\)

Converting from Pascals to atm,

\(\Delta P=\frac{8888}{101325}\)

\(\Delta \mathrm{P}=8.77 \times 10^{-2} \mathrm{~atm}\)

Increase in pressure in your lungs \(=8.77 \times 10^{-2} \mathrm{~atm}\)