Question

The earthworm absorbs oxygen directly through its skin. The worm does have a good circulatory system (with multiple small hearts) that brings the oxygen to all the cells. But the cells are distributed through the worm's volume and the oxygen only gets to come in through the skin-so the surface to volume ratio plays an important role. Let's see how this works. Here are the worm's parameters. A typical specimen of the common earthworm (Lumbricus terrestris) has the following average dimensions: Mass- 3.7 g, Length-12 cm, Width-0.64 cm. The skin of the worm can absorb oxygen at a rate of A = 0.24 μmole (㎛ole = x10-6 moles) per square cm per hour. The body of the worm needs to use approximately B 0.98 pumole (umole x10-6 moles) of oxygen per gram of worm per hour.

question

Answer 1

Surface area of worm is 2 lambda RL = Ara [taking R < < L] Number of moles of oxygen absorbed by worm per hours is N = A times lambda RL here L & R both are in cm. mass of worm is m = P(v compositefunction l) = P lambda R^2 L {Here P is in gm/cm^3} So No. of moles used by worm per hours is N prime = B times m rightwardsdoublearrow N prime = BP lambda R^2 L To survive N prime lessthanorequalto N rightwardsdoublearrow BP lambda R^2 L lessthanorequalto A 2 lambda RL By putting numerical values we can check we her N prime lessthanorequalto N