Need help to find pressure for Alvelous

[ch5112]

Homework Statement

While a person breaths in, a typical alveolus expands from a radius of 0.13*10^-3m to 0.15*10^-3m. The alveolus behaves as though it has a single surface which is decribed by

P=2$\gamma/r$


1. What is the work done in alveolus, P, while breathing in is 600Pa and while breathing out it is 650Pa
2. Use energy transfer ideas to determine the average tension in the alveolus wall while breathing in. Explain your reasoning steps
3. In fact P increases while breathing in, by a factor of approximately 1.6. What are the implications of the increase for the tension in the alveolous walls? Be quantitive in your answer

Homework Equations

P(alv)=P0+P, where P0=intrapleural space pressure

The Attempt at a Solution

found that,
-the gauge pressure in the alveoli of the lung is denoted by P(alv)
-gauge pressure just outside the lungs, in the intrapleural space is P0, which apporximates the pressure outside the alveoli
-the average ecess pressure in the alveolus, P, while breathing in is 600Pa and while breathing out is 650Pa

 

[member 553083]

1. Work done in alveolus, W=P*V, where V is the volume of the alveolus. V=4/3πr^3When r=0.13*10^-3m, V=4.54*10^-14 m^3When r=0.15*10^-3m, V=7.07*10^-14 m^3When breathing in, W=P*(V2-V1)=600Pa(7.07*10^-14 m^3-4.54*10^-14 m^3)=1.27*10^-14JWhen breathing out, W=P*(V2-V1)=650Pa(7.07*10^-14 m^3-4.54*10^-14 m^3)=1.38*10^-14J2. The average tension in the alveolus wall while breathing in is the sum of the intrapleural space pressure and the average excess pressure in the alveolus. T=(P0+P)/Awhere A is the area of the alveolus wall.A=4πr^2When r=0.13*10^-3m, A=5.12*10^-10 m^2 T=(P0+600Pa)/(5.12*10^-10 m^2)=11,765.1 Pa3. The increase in P by a factor of approxiamtely 1.6 implies that the tension in the alveolus walls increases by a factor of approximately 1.6.T2=T1*1.6=(11,765.1Pa)*1.6=18,816.16Pa