Prac%ce,Week1

Therapeu%csubstanceGases

MetricprefixesPrefix Symbol Factor Factor(sci.nota8on)peta P 1,000,000,000,000,000 ×1015

tera T 1,000,000,000,000 ×1012giga G 1,000,000,000 ×109mega M 1,000,000 ×106kilo k 1,000 ×103

hecto h 100 ×102deca da 10 ×101

(none) (none) 1 ×100deci d 0.1 ×10-1

cen% c 0.01 ×10-2milli m 0.001 ×10-3

micro μ(=u) 0.000001 ×10-6nano n 0.000000001 ×10-9

pico p 0.000000000001 ×10-12

femto f 0.000000000000001 ×10-15

Pharm-relatedquan%%esandunitsQuan8ty SIunit Pharmacology&prac8ce CommentLength m Å,nm,µm,cm 1Å=0.1nmMass kg g,mg,µgTime s s

Temperature(absolute:K,rela%ve:C,F)

K K,°CΔT[°C]=ΔT[K]

T[°C]=T[K]–273.15ConvertFtoC

Amountofsubstance mol mol 6×1023molecules

Area m2 m2Volume m3 1L=1dm3,1mL=1cm3

Density 1kg/m3 1g/cm3=1g/mL DensityofpureH2Ois1g/mL

Concentra%on 1M=1mol/L;nM,µM,… Avoidg/L

Unifiedatomicmass 1Da=1g/mol

Pharm-relatedquan%%esandunitsQuan8ty SIunit Pharmacology&prac8ce CommentSpeed m/s

Accelera%on m/s2Force N=kg×m/s2 Newton,pound,.. Newton

Pressure Pa=N/m21bar=100,000Pa=100kPa1atm=101,325Pa≈1bar1mmHg*=1atm/760

PascalHgismercury

Energy J=N×m

1cal≈4.184J[energy need to heat 1 g H2O by 1°C]

1eV≈1.6×10-19J[Δenergy when 1 electron moves through 1

V potential difference]

Foodcalorie:1Cal=1kcal

Avoidambiguity,use

kcal

*Blood pressure is a gauge pressure, not an absolute pressure –  1 atm = 760 mmHg –  Blood (gauge) pressure of 140 mmHg = absolute P of 900 mmHg ≈ 1.184 atm –  Blood (gauge) pressure of 80 mmHg = absolute P of 840 mmHg ≈ 1.105 atm

Constants•  STP=StandardTemperatureandPressure:

–  T=0°C=273.15K–  P=1bar=100kPa

•  RT=RoomTemperature≈300K–  nottobeconfusedwiththeotherRTbelow

•  AvogadronumberNA=6.022×1023

•  Gasconstant,R:R≈8.314 J /(K⋅mol)R≈5.189×1019 eV/(K⋅mol)R≈0.082 L⋅atm /(K⋅mol)R≈1.9872≈2 cal/(K⋅mol)

•  Boltzmannconstant,kB=R/NA≈1.38×10-23J/K•  “RT”=R×Tisthethermodynamic“currency”

–  atroomtemp.,RT≈0.6kcal/mol≈2.5kJ/mol

•  Airis78%N2,21%O2,0.04%CO2

•  Gravita8onalaccelera8on(g)nearEarth’ssurface,g=9.8m/s2

SIunits

•  Problem:TheSIunitfordensityis§  g/m3

§  0.1Kg/m3§  m/Kg3

§  g/mL§  Kg/m3

•  Answer:Kg/m3

Unitconversion•  Problem:Thedensityofoxygenatroomtemperatureisabout1.3kg/m3.Expressitsdensitying/cm3.§  0.013§  13§  1300§  130§  0.0013

•  Solu3on:§  1kg=103g§  1m3=100×100×100cm3=106cm3

§  1.3kg/m3=1.3×103/106=0.0013g/cm3

Energyunitconversion•  Problem:1Jouleisclosetothefollowingvalue:

§  4.2kcal§  2.092cal§  0.24cal§  8.314cal§  1.035cal§  4.184cal

•  Answer:1Joule=0.24cal;1cal=4.184Joule

KelvinvsCelsius•  Problem:Apa%ent’sbodytemperatureisdeterminedtobe313K.Thepa%entismostlikely§  healthy§  sick§  dead

•  Answer:§  313K≈313-273=40°C§  Thepa%enthasafeverthereforehe/sheissick.

Gasconstant:fromTtoEnergy•  Problem:CalculateRTat0°Cinkcal/mol

•  Solu3on:OfthemanyfacesofthegasconstantR…v R≈8.314 J /(K⋅ mol)v R≈5.189×1019 eV/(K⋅ mol)v R≈0.082 L⋅atm /(K⋅ mol)v R≈1.986 cal/(K⋅ mol)

§  …chooseR≈1.986cal/(K⋅ mol)§  T=0°C≈273K§  ThereforeRT=1.986×273=542.178cal/mol~0.54kcal/mol

•  Answer:0.54kcal/mol§  Atroomtemp,RT~0.6kcal/mol

Sizesoftherapeu%cs(fromatomstoproteinstocells)

•  Problem:Thedistancebetweencentersoftwocovalentlybondedcarbonatomsiscloseto:§  1.5nm§  1.5μm§  1.5ŧ  0.015nm§  150Å

•  Answer:1.5Å,sameas0.15nmor150pm

~1.5A

~1.7A

3 A 4 A

5 A

Non-covalentinterac8ondistanceCovalentbondlengthisshorterthan

Biologicalsizescalecntd.•  Problem:Cellmembraneandthemembranessurroundinginner

cellorganellesarephospholipidbilayersabout____thick

§  5nm§  100pm§  1ŧ  5ŧ  50nm§  5μm

•  Answer:5nm

•  Hint:membraneisabilayersofphospholipids§  Eachlipidhasahead(~5covalentbondstall)andahydrocarbonchain(~

17-20covalentbondslong)§  2×25×1.5A(forthelengthofC-Cbond)=75Å=7.5nm§  Theactualanswerissmaller,becausebondsareconnectedat~120oangles.

Comparative sizes of drugs and targets

Biologicalenergyscale•  Problem:Whatisthebestapproxima%onfortheenergyofa

covalentbond?§  25kcal/molto100kcal/mol§  exactly10.45kcal/mol§  1to2kcal/mol§  about0.1kcal/mol§  1to5kcal/mol

•  Answer:25kcal/molto100kcal/mol

•  Bonusproblem:Whataboutahydrogenbond?

•  Answer:Inbiologicalsewngs,agoodes%mateforafavorablehydrogenbondis~2.5kcal/mol

Concentra%ons,volumes,molaramounts…

•  Concentra%on=molaramount/volume§  measuredinmol/L ≡M,alsomM,μM,nM,pMetc.§  molaramount=volume×concentra%on§  volume=molaramount/concentra%on

•  MW=mass/molaramount§  measureding/mol≡Da§  Mass=MWxmolaramount§  molaramount=mass/MW

Molaramountvsweight•  Problem:Fomepizoleisusedasanan%doteinmethanoland

ethyleneglycolpoisoning.Es%matetheweightofa0.5mmolsampleofFomepizole.§  0.04g§  0.004mg§  0.5μg§  82g§  4.05g

•  Solu3on:§  MW(Fomepizole)is4×12(C)+2×14(N)+6(H)=82§  0.5mmol×82g/mol=0.5×10-3×82g/mol≈0.04g

•  Answer:0.04g

Fomepizoleisacompe%%veinhibitoroftheenzymealcoholdehydrogenase

Molaramountvsweightcntd.•  Problem:Albuministhemostabundantserumproteinandcarrier

forvariousdrugs.Itsconcentra%oninplasmarangesfrom30to50g/L.GiventhatMWforalbuminis67kDa,es%mateitsmolarity.

§  440-740nM§  4-7uM§  440-740uM§  4-7mM§  44-74mM

•  Solu3on:§  33.5g/L÷67,000=5×10-4mol/L=0.5mMor500uM§  Thecorrectrangeincludesthisnumber:440-740uM

Concentra%onvsamount•  Problem:Whatisthetotalmolaramountofacompoundin

0.3mLof1mMsolu%onofthatcompound?§  0.3μmol

§  impossibletotellbecausetheMWofthecompoundisnotgiven

§  3.33mol

§  0.3mg

§  0.3mol

•  Solu3on:§  Molaramount=concentra%on×volume

§  =(1×10-3)×(0.3×10-3)=0.3×10-6=0.3 µmol

Avogadronumber,NA=6×1023

•  Problem:TheapproximatemassofonemoleculeofPencicloviris4.2x10-22g.Calculatethemolecularweightofthedrug.§  252g/mol

§  172g/mol

§  326g/mol

§  472g/mol

§  504g/mol

•  Solu3on:§  MW=4.2×10-22×6×1023~252g/mol

Degrees of freedom (DF) (**advanced) •  DF are the elementary units of a molecular mixture

(“variables”) capable of storing kinetic or potential energy –  1 DF stores energy ½ kT, 1 mole of DF stores energy ½ RT

•  # DF in a molecule depends on the # atoms, physical state… –  # of DF inside a molecule increases with T –  More variables get excited and capable of storing energy

•  DF of a molecule in gas phase: –  # DFtrans = 3 –  # DFrot = 0, 2, or 3 –  # DFvib ≤ 2 × (3 × Nat – 3trans – Nrot):

•  In a fully excited state, a molecule with Nat is described by 3×Nat independent variables ⇒ # of vibrational modes: 3×Nat–3trans–Nrot

•  Each vibrational mode contributes 2 DFs (can store potential and kinetic energy)

–  Some (slow) vibrations ARE excited at 300 K •  E.g. collective motions involving rotatable sp3-sp3 torsional variables

–  Other (fast) vibrations are NOT excited at 300 K •  Bond vibrations in most diatomic gases are not excited •  Bond vibrations in Cl2, Br2, I2 are partially or fully excited

Degrees of freedom in gas phase (**) Overall Movement

and Rotation If ALL vibrations

are excited T = 300K

3

Translational only

3

No vibrations

3

5

# DFtrans=3 # DFrot=2

(3×2–5)×2+5=7

# DFvib=2

5 for most X2 gases

6-7 for Cl2, Br2, I2 (based on Cv,m)

6

# DFtrans=3 # DFrot=3

(3×3–6)×2 + 6 = 12

# DFvib=6

~7 (based on Cv,m)

only 1 of 6 DFvib is excited

6

# DFtrans=3 # DFrot=3

(3×Nat–6)×2+6

Many vibrations!

>> 6

sp3-sp3 torsion vibrations are

excited

Degrees of freedom (**) •  Problem: Determine the number of degrees of freedom of

adenosine triphosphate (ATP) in gas phase at T=300K. §  1 §  2 §  3 §  4 §  5 §  6 §  infinitely many §  the correct answer not given

•  Solution: §  #DFtrans + #DFrot = 3+3 = 6... §  At T=300K, collective motions involving the rotatable

sp3-sp3 bonds are excited… §  Therefore, the total #DF is finite but greater than 6. §  The correct answer is not given.

Degrees of freedom and thermal motions

Athermallyagitatedpep%de,byGregL,CCBY-SA3.0

•  More DF means better ability to store thermal energy –  ⇒ higher heat capacity

•  Monoatomic gas – low heat capacity •  The peptide below – high heat capacity

**(preview)DFandheatcapacity,Cv,m•  #ofDFdeterminesheat

capacityofasubstance:v Eachrota%onal/transla%onalDF

contributesR/2toCv,m

v Each(excited)vibra%onalmodecontributesuptoRtoCv,m

•  Monoatomicgaseshave3DF:Cv,m=3/2R

•  Diatomicgaseshave5DFbelowTvibv  forH2,CO,N2,Tvib>>298K

•  WhenT≥Tvib,vibra%onalDF’sappearv  forCl2,Br2,I2,Tvib≤298K

Gas CV,m, J/(mol·K) CV,m/R # DF He ● 12.5 1.5 3

Ne ● 12.5 1.5 3

Ar ● 12.5 1.5 3

Kr ● 12.5 1.5 3

Xe ● 12.5 1.5 3

H2 ●–● 20.18 2.43 5

CO ●–● 20.2 2.43 5

N2 ●–● 20.8 2.50 5

O2 ●–● 21.03 2.53 5

Cl2 ●–● 24.1 3.06 ~ 6

Br2 (v) ●–● 28.2 3.39 ~ 7

H2O (v)* ●●● 28.49 3.43 ~ 7

CO2 ●-●-● 28.5 3.43 ~ 7

CH4 ●●● ● 27.1 3.26 ~ 6-7

ExperimentallymeasuredCv,mat298K

*ValueforH2Oat373K

Week 1 equations

v = (3RT /m)1/2•  Kine8cenergyvsT

§  1/2mv2=3/2RT§  Rootmeansquarevelocity:

§  Heremisthemassofthesample§  Equipar>>onprinciple:atthermalequilibrium,theenergyisdistributed

equallybetweenallavailabledegreesoffreedom§  Onlytransla%onaldegreesoffreedomma|erforT

•  Graham’slaw:effusionratesvsmolecularmassofthegas§  Rate1/Rate2=(m2/m1)1/2

•  Idealgaslaw§  P×V=n×RT(n=#ofmoles);V=n×RT/P

•  Barometricformula:atmosphericpressureatal%tudeh[m]§  Ph=P0exp(-Mgh/RT)§  HereMismolecularmass[kg/mol]

•  Work=Force×Distance[J]•  Pressure=Force/Area[Pa]

Kine%cenergyvstemperature•  Problem:TheCelsiustemperatureinastorageroomwas

increasedfrom25°Cto50°C.Howmuchdidtheaveragekine%cenergyofmoleculesintheroomchange?Marktheclosestanswer.§  increasedby2½%mes§  decreased§  increasedby8%§  increasedby2%mes§  theincreasecannotbecalculatedwithoutknowingthemolecularmass

•  Solu3on:§  Thekine%cenergyispropor%onaltoKelvintemperature§  KelvinTwas:273+25°C=298K§  KelvinTnow:273+50°C=323K§  Ra%o:323K/298K~1.08§  Thetemperature,aswellastheaveragekine%cenergy,increasedby8%.

•  Note:Rootmeansquarevelocitywillincrease=1.04%mes,i.e.onlyby~4%

Effusionrate•  Problem:Amixtureofoxygen(MW=16)andhelium(MW=4)

escapesthroughaporousmembrane.Whichofthegasesescapesfaster,andhowmuchfaster?§  oxygen,4x§  oxygen,2x§  samerate§  helium,2x§  helium,4x

•  Solu3on:§  UseGraham’slaw:escaperateisinverselypropor%onaltothesquarerootofmass

§  MW(He)is4%meslighter,thereforeitescapes2%mesfaster

Volumeof1moleofgas•  Problem:Es%matethevolumeof1moleofnitrousoxideat0°Candat27°C,assumingtheatmosphericpressure.

•  Solu3on:Idealgaslaw,PV=nRT§  V=RT/P(becausen=1)§  OutofthemanyfacesofR,chooseR=0.082L⋅atm /(K⋅mol)§  ThencanuseP=1atm,andV=RT§  At273K,RT≈22.4L⋅atm /mol;V≈22.4L§  At300K,RT≈24.6L⋅atm /mol;V≈24.6L§  Appliestoanygasthatcanbeapproximatedasideal

Atmosphericpressureateleva%on•  Problem:COPDpa%entsexperienceworseningofsymptoms

(shortnessofbreath)whentheatmosphericpressuredropsbyasli|leas30mmHg(3%).Thisatmosphericpressurecorrespondstowhateleva%onabovethesealevel?Use0.029kg/molformolarmassofair.§  11km§  3km§  1600m(eleva%onofDenver–the“mile-highcity”)§  900m§  <300m

•  Solu3on:§  Ph=P0exp(-Mgh/RT)§  Makesuretousemetricunits(e.g.R=8.314J /(K⋅mol))§  Exp(-Mgh/RT)=0.97;Mgh/RT=-ln0.97;h=-2500/(0.029x9.8xLn0.97)§  h≈268m<300m