J. A. CAMPBELL H a n s y Mudd College Claremont, Cdifornio 91711
Questions 438. Rleasurcmcnts show that the gas density is 1.0 X 10%olecules in the exosphere (out,cr spacc above -700 Iim, T E 1200"Ii). The average mean free path exceeds 13,000 Iim, thc radius of thc carth. Calculate P in atm. 439. Raising rats in an atmosphere whcrc He has been subst,ituted for N, leads to an incrcase in food intalcc and oxygm 'onsumption from 23 to 28°C but, almost no effect at 33% Similar cxpcrimcnts \\-it,h man show much smallcr changes. Suggest a possible interpretation in terms of the molecular propcrtics of Hc compared to RTI for t.he cffccts in rats and their lcsscr effect in man. 440. "Caisson discasc" is due to gas bubbles in tbc bloodstream formed vhcn a man worliing at high pressure movcs too rapidly to a. l o m r pressurc for thc dissolved gases (principally N2) to cscapc through the lungs. Calculate the volume of N2 which might bc released in a decp sea diver working at 250 meters if he suddenly rose to the surface. His blood fluid volumc is about 3.5 1, and the solubility of Npin human blood is about 12 cm3/l atm. 441. Azotobacter chrooeoceum bacteria grow in air on a nitrogen-free medium obtaining all their nitrogen by "fixing" the air. Calculate the volume of air rcquircd to provide nitrogen for 11 of aqueous culturc which will finally contain bacteria at a concentration of 0.M mg dry weight per cm3, having a nitrogen content of 7% dry weight.
442. Resting muscle consumes O2 at about 2 ml/ (kg. min), but. a maximally worlting musclc may use 100 times this and rcquirc a 50-fold increase in blood circulation. Maximal O2 intake in a fit subject is 4-5 l/min. What do thcse figurcs suggcst must occur in a runner operating at, maximal activity in a 440-yd foot race? 443. Puricwitsch in 1900 mcasured thc respiratory quotient (C02 produced/Ot consumed) for Aspergillus organisms groxing on sucrose or dextrose and on tartaric acid. He obtained values of 1 and 1.6, rcspcctivdy. Intcrprct thcse values. 444. An acre of corn, in a thrcc-month growing scason, releases about 350 tons of moisture into the atmosphere, and forms about 200 tons of csrbonaccous material, about 40% carbon by weight,. What volume of air must be "processed" per day to acquirc thc necessary carbon from the atmospheric Cop? How docs this compare to thc volume of gaseous water produced? 445. About 30% of thc oxygcn in thc lungs is cxtractcd per inhalation, and about 0.21 1 of oxygcn are required to gencratc 1 6cal of body cncrgy in humans. How much air is inhaled per breath on average? Questions (plus possible, but certainly not uniquely sstisfactory, answers) requiring no more than a concurrent first-year, college level course, a data handbook, and s willingness to apply fundamental chemical ideas to the systems which surround us (or even are inside us) constitute this column. Contributions fos oassibla inclusion are solicited. Initiated in the Januarv. 1072 issue of THE JOURNAL.
Answers A38. P = nBT/V = 1.0 X 108 (molecules/om3) X 82 (cm3 X atm/mole K ) X 1200 Kl6.0 X 1OaY(molecnles/n~ole) = 1.6 X 10FLSatm. Even the best vacuums on earth are seldom less than l U P L atm, thus one must anticipate the possibility of problems in space due t o the hirhvacuum. For example, adsorbed gas films or ordinary lubricants may no longer be effective between sliding parts in cameras and other machines. A39. We presume that substituting He for Nz is principally a change from molecules of wt 28 to those of wt 4 with a r e s u l t ing rise in average molecular speed, since the translational energy (muz/2) 1,einainsconstant. T h ~ the s atoms will collide more frequently with the a n i d (v'\/Z8J4 = 2.6 tim- as often). They will carry away less energy per collision (6/7 as much according t o their relative gaseous heat capacities). Tho net = 1.9, or twice as high eficct is 2.6 of heat of loss. This will be most serious when the ;tl.ea. to mass is biggest (larger in a rat than in man) and when t,he dinerenee in 1' between animal and surroundings is large.
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The effect disappears as tho two temperatures hecomc equal. Thus the r a t must take in more food and Oz to generate t,he additional energy to onset the increased rate of heat loss, hut there iq less eflect in man. Of course, man could readily adjusl the amount of clothing. A40. 3.5 1 X 12 (cmvl atm) X I250 rn H20/1:3.T,(mH?O/m Hg)1/0.76 m Hg/atm = 1.0 X 10'hcma. This is considerable volume compared t o t.he 3500 cmJ of blood, and the hnhbles form in the veins, arteries, and capillaries causing "vapor lock," i.e., the gas does not move as readily as the liquid blood. intense ,,%in ~h~ deerease in blood flow T~leusual cure (effect,iveif applied very soon after ,.allid decompression) is to repressurize t,he patient to tho N., then to decompress very slowly. The attacks can be minimized by substit,uting He (much less soluble in blood) for Ns, but this leads t o problcn~swith heat loss and to vocal communication since a larynx full of He emits a high pitched voice of a stmnge quality.
A41. 0.84(mg/om" X 0.07 (g N/g dry bacteria) X lOOO(cm3/l) = 60(mg Nsll) required for bacteria. At room conditions (to 1 sig. fig.) molar volume = 20(l/mole), volume and wt %of N will hot,h he 80, and f wt, of air is 30(g/mole), so volume of air required is
V
0.06(g N2/1 culture) X 20(l/mole air)/[30(g air/mole air) X 0.8(g Nx/g air)] = 0.05(1 air11 culture), or 50 cm3
=
This assumes fixation of 100% of the N2 which may be impos sible due to the equilibria and/or rates involved. A42. Note 1 sig. fig. Assume a 70 kg runnor (154 lb). Oxygen consumption will he: Oz consumption = 2(ml/kg min) X 70 kg X 100 (factor for maximal work) = 1.4 X 10' (ml/min) = 14 llmin. But m a i m a l intake is 5 l/min. The 440-yd race requires less than one minute, so the runner will acquire sn oxygen deficit of about 7-8 1 and will have to breathe heavily for 2 or 3 min after the race to rebuild his oxygen balance. A43. Sucrose and dextrose are carbohydrates, empirical Their oxidation is described by the equaformula C,(H1O),. nOx = nCOl mHnO. One mole of COXis tion C.(HnO), produced per mole of OXconsumed and the respiratory ratio is 1.0. Tartaric acid must (assrming only C, H, 0 are present) Its oxidation to COXwould he have a formula C,(H20),H..
+
+
Clsn(H,O), H,
+ no2
=
1.6n C01
+ mHgO + (p12)HgO
Applying mass balances For C: no information, 1.6 n =1.6n. For H : 2 m For 0 : m
+p
+ 2n
+ p, so no information again = 3.2 n + m + (p/2), or p -2.4 n =
2m
=
that is, there are leqs than enough H's in tartaric acid to oonvert its original 0 into H&. Let us try to calculate an empirical formula for tartaric acid. C,H,O,
+ (z/1.6)02 = zCOz + (y12)HaO
Now 2,y, and z must be integers and ( ~ / 1 . 6 )most be integral or an integer plus 0.5. Any multiple of 4 satisfies this condition for z. There are no other possible values. Using an 0 balance L (2z/1.6) = % ( ~ 1 2 ) . Assume z = 4; z
+
+
+
+
z/1.6 = 8 (y/2), r - y/2 = 3. There are many solutione here hut y must he at least 2 and always an even number, and 8 any integer larger than 3. Possible empirical formulas, based on C,, are: C4H20,, C4H,0s CIH80a,etc. [The actual formula is C,OsHa giving the equat,ion: CIOIH~ 2.5 0% = 4CO. 3He0.1 A&.' C& ebnstitutes 0.03% by volume of the atmosphers and O.0.5% by weight.
+
+
Amt of C needed = 200 (tons cornlacre yr) X OAO(C/corn) X 454Wlb) X 2000 (Ib/ton)/l2(s/ . .. -. mole) = 6 x 10' (moles carhon/acre vr) . . in 3 months or 90 days = 7 X 10Vmoles carbon/acre day) Molar volume a t ambient conditions . ~ 3 1. 0
V of air needed
= =
[30(1 air/mole air>/0.0003(C/air)] X 7 X lo3 (moles Clacre day) 7 X 10B(1 airlacre day)
1 ft3 =30 1, SO V = 7 X 108 (I/&cred&y)/30(l/ft8)= 2 x 10' (fta/acreday) of air must he processed assuming (incorrectly) that 100% of the COZis extracted.
V of H~0r.j
350 (tonslacre X yr) X 2000(lh/ton) X 454 (gllh) X 30 (limole) X (760/30)/90(days/yr) X 18 (g/mole) = 1.5 x 108 (1 of gaseous waterlacre day) =
This is somewhat less than the air processed which means the processed air can easily sweep sway the waler produced without getting saturated. The saturated vapor pressure far H.0 is about 30 mm or (301160) X 100 = 4% by volume. A45. Experiment on yourself to determine breathing rate. I t will be about 20 hreathslmin. Air is 20Y0 by volume O1. Daily requirement = 3 X lo3 kcal. 3 X 10S(kcal/dav) . . . . X OX1 Ol/kcal) X 5(1 a&/1 OX) V(l/hreath) = 20(hreaths/min) X GO(min, hr) X 24(hr/ davi X 0.3 retained
Volume 49, Number 9, September 1972
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