# Question c1904

May 16, 2016

The pressure will rise to 1.4 atm, but the lungs will not burst.

#### Explanation:

The average temperature of the human lung is 310 K.

The esophageal temperature of the inhaled air is 301 K.

Since the volume of the lungs is constant, this is an example of a problem involving Gay-Lussac's Law:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {P}_{1} / {T}_{1} = {P}_{2} / {T}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

${P}_{1} = \text{1 atm}$; ${T}_{1} = \text{220 K}$
${P}_{2} = \text{?}$; $\textcolor{w h i t e}{m m l} {T}_{2} = \text{301 K}$

We can rearrange the above formula to get

P_2 = P_1 × T_2/T_1#

${P}_{2} = \text{1 atm" × (301 color(red)(cancel(color(black)("K"))))/(220 color(red)(cancel(color(black)("K")))) = "1.4 atm}$

This is less than the 2 atm burst pressure of the lungs.

The lungs will not burst.