1. In the figure below, a solution containing lung extract
exhibits hysteresis in is surface tension...
1. In the figure below, a solution containing lung extract exhibits hysteresis in is surface tension versus area relationship. This means the surface tension is higher during inflation of the lung than deflation By recalling that mechanical work can be expressed as f pdV, show that the work required to inflate all the alveoli in the lung against the effects of surface tension can be written as a. work-I ơdA where σ is the surface tension coefficient, A is the aggregate surface area of all alveoli in the lung, and the integral is carried out from minimum surface area (start of inspiration) to maximum surface area (end of inspiration). To show the result, you may assume that the pressure outside the alveoli is constant and equal to 0 (gauge). You will find it convenient to recall that the volume and surface area of a sphere are V- R and s -4R2. This mplies that dv-4rR2dR and ds - 8aRdR Assume that the surface tension versus area curve for the entire lung over one breathing cycle can be approximated by the shape shown the second figure below Using this information, determine how much energy is dissipated in surface tension hysteresis during one breathing cycle b. Lung Detergent act Plasa Water 100 82 m2. 80m 75 m2 - 10 20 30 450 60 70 Surfaco tension (dynesicm) σ (dynes/cm) 50