34 2. OVERALL DESCRIPTION OF THE HUMAN BODY 4. Show that if an average person in

Question

34 2. OVERALL DESCRIPTION OF THE HUMAN BODY 4. Show that if an average person inspires enough air to give 5.7 liters of gas in his lungs, then the apparent density of his body in water wil1 be around 0.985 g/cc. Assume a "true" density (zero gas in lungs) of 1.073 g/cc. 5. If a person is exercising at a rate such that his metabolic rate is six times its basal value, what will his cardiac output be (see Table 2.1)? If his pulse under such conditions is 130 beats per minute, what does this imply about the amount of blood pumped by the heart per beat (i.e., the "stroke volume")?

Explanation / Answer

Question 4 :

First calculate the volume of a body without any air :

Assume total weight of an average person to be 68 kg ( 68000 g ) as given in the table.

Now, density = mass/ volume

--Hence ...volume = mass/ density

Substitute the standard mass, and true density of the body given in the question, to calculate volume of body without air in lungs..

Volume = 68000/1.073 = 63, 373 .7185 ml

---Now in the condition of inhaled air, the total mass of body becomes : total mass + mass of inhaled air.

Hence new mass = 68,000 g + Weight of air ( vol of air * density of air)

= 68000 + ( 5700 ml * 0.001225 g/ml)

= 68000 + 6.9825

= 68006.9825 g

--- Also in case of inhalation , the total volume becomes :

Total volume ( calculated earlier ) + volume of air in lungs ( given in question )

--- Hence total new volume = 63373.7185 + 5700 = 69073.7185 ml

--- Now coming back to old formula

Density : mass/ volume

( Substitute the new mass and new volume )

Density = 68006.9825g / 69073.7185ml

= 0.9845 g/ cc

= 0.985 g / cc round off

  

Question 5 :

As given in the table, the general cardiac output of a person is given by :

Lts of blood = 3.0 + 8 M lts / min

Now, to find the value of variable ( M) by substituting the resting cardiac output in the formula :

5 = 3 .0 + 8 M

Therefore 2 = 8 M

Implies : M = 1/4 or 0.25 , hence at resting M = 0.25

Now (M) being the only variable in the formula , I feel M is the only unit that corresponds to the BMR of the person..

( Sorry to go on assumption here, but I think did not get any other clue in the pages you provided )

---M should be directly proportional to the BMR---

When BMR = basal ....Implies M = 0.25

When BMR = 6 times Basal....Implies M = 6 * 0.25

Hence new M = 1.5

Now come back again to the formula of C.O. and substitute the new (M ) value

C.O = 3.0 + 8 M

= 3.0 + 8 (1.5)

= 3.0+ 12

= 15 Lts

--- Hence the cardiac output when BMR is 6 times Basal level is 15 Litres of blood

- Now coming to Stroke volume

Cardiac Output (CO) = Heart Rate ( HR ) * Stroke Volume ( SV)

HENCE , SV = CO/ HR

At rest SV = 5 000 ml / 65 = 76.92ml/ beat

At 6 times BMR :

SV = 15000 ml ( calculated earlier ) / 130 ( given in Question)

= 115.384ml/ beat

--- Hence in this scenario the stroke volume is increased from 76 at rest to 115 .

This increase can be considered normal acc. to the increase in BMR.

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