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Teknik kimiaBahasa Inggris : chemical engineering
Ilmu teknik atau rekayasa yang mempelajari pemrosesan bahan mentah menjadi barang jadi
Aplikasi :1. perancangan dan pemeliharaan
proses kimia 2. peralatan proses
Teknik Kimia berhubungan dengan :
a. Produksi bahan kimia baik di skala besar maupun kecil.
b. Membuat produk berskala besar (pabrik), berbeda dengan skala lab.
c. merubah bahan baku menjadi produk bernilai ekonomi lebih tinggi.
d. Sektor : bahan kimia sampai energi, makanan dan minuman, obatobatan.
e. Produk, yang dibuat melalui perubahan kimia (reaksi) atau/dan fisis
(pemisahan).
f. Proses Industri kimia, seperti industri: pupuk, kimia murni, cat, zat
warna, tinta, petrokimia, plastik, resin, sabun, deterjen, parfum,
kosmetik, lemak dan minyak nabati, katalis, gas, minyak bumi,
polimer, kertas, tekstil, makanan dan minuman, bioteknologi, dll.
Understanding how chemical reactors work lies at the heart of almost every chemical processing operation.
Design of the reactor is no routine matter, and many alternatives can be proposed for a process. Reactor design uses information, knowledge and experience from a variety of areas - thermodynamics, chemical kinetics, fluid mechanics, heat and mass transfer, and economics.
CRE is the synthesis of all these factors with the aim of properly designing and understanding the chemical reactor.
What is Chemical Reaction Engineering (CRE) ?
Chemical process
Rawmaterial
SeparationProcess
SeparationProcess
Products
By products
Di dalam proses dan peralatan di industri, rangkaian peralatan menyangkut kedua jenis proses itu, yaitu :
1. unit operation ( satuan operasi) : unit dengan perubahan fisis atau seringkali disebut Operasi Teknik Kimia.
2. unit processes (satuan proses) : unit dengan reaksi kimia.
Unit operation meliputi :
a. transportasi fluida (perpindahan pada proses alir),
b. perpindahan panas dalam alat penukar panas ( heat exchanger),
c. separator,• padat-padat : screening.• Padat-cair : sedimentasi, filtrasi,• Cair – gas : absorpsi, stripper, distilasi,
evaporasi.• Cair-cair : ekstraksi cair-cair, dekantasi, dll.d. pencampuran
Unit Processes meliputi :
a. pembakaran bahan bakar dalam burner, furnace.
b. Reaksi kimia dalam reaktor.
c. Fermentasi.
Ilmu-ilmu yang menjadi dasar dalam
teknik kimia, antara lain adalah:
1. Neraca massa
2. Neraca energi
3. Peristiwa perpindahan massa , energi , momentum
4. Reaksi kimia
5. Termokimia
6. Termodinamika
7. material atau bahan , ekonomi dan manajemen .
Book
• David M Himmelblau “ Prinsip Dasar dan Kalkulasi dalam Teknik Kimia”, jilid 1 dan 2
• Olaf A Hougen dkk. “ Chemical Process Principles”, Part one Material and Energy Balance.
• Richard M Felder dkk. “ Elementary Principles of Chemical Processes “.
• Nicholas P Chopey, “ Handbook of Chemical Engineering Calculations “.
ASAS TEKNIK KIMIA Mol : jumlah tertentu dari molekul, elektron, atau partikel spesifik lainnya
Densitas : Ratio massa persatuan volum
Berat jenis ( spesific gravity )( ref : air )Ratio densitas yang ditanyakan terhadap densitas air.
Volum spesifik :Volum persatuan masssa
Fraksi mol :Mol zat tertentu dibagi dengan jmlah total mol yang ada
Tekanan gauge + tekanan barometrik = tekanan absolut
• Tekanan : gaya persatuan luas
• Tekanan gauge : tekanan yang terbaca pada manometer • Tekanan barometrik : pembacaan tekanan atmosfer
StoikiometriStoikiometri berasal dari kata Yunani,
STOICHEION yang berarti unsur dan METRIA yang berarti ilmu pengukuran . Kemudian difinisi diperluas sebagai berikut :
1. Proses membuat suatu perhitungan yang didasarkan pada rumus-rumus dan persamaan- persamaan berimbang.
2. Ilmu yang mengukur berapa banyak jumlah zat yang dapat dihasilkan dari sejumlah zat tertentu lainnya.
• Reaktan pembatas ( limiting reactan ) reaktan yang ada dalam jumlah stoikiometri
terkecil ( reaktan yang akan hilang pertama kali ).( lebih dari dua reaktan harus menggunakan satu sebagai reaktan pembatas )
Reaktan berlebih ( excess reactan ) reaktan yang terdapat lebih dari pada reaktan
pembatas.( meski jika hanya sebagian dari reaktan pembatas yang sebenarnya bereaksi, kuantitas yang dibutuhkan dan kuantitas yang berlebih didasarkan pada jumlah seluruh reaktan pembatas seolah-olah reaktan tersebut telah bereaksi secara sempurna )
% kelebihan = kelebihan mol : ( mol yang dibutuhkan untuk
bereaksi dengan reaktan pembatas ). 100
Konversi ( conversion ) :
fraksi dari masukkan atau beberapa materi kunci dalam masukkan yang diubah menjadi produk :
Yield ( reaktan tunggal dan produk ) :
Mol ( massa ) produk akhir dibagi dengan mol ( massa ) reaktan awal yang dimasukkan
% konversi A = mol (massa) A yang bereaksi : mol ( massa ) A yang dimasukkan
Daur ulang ( reaksi kimia )
• Konversi fraksi keseluruhan ( overall fraction conversion ) :
• Konversi jalan tunggal ( Single pass or once through conversion ) :
= ( massa ( mol ) reaktan dalam fresh feed – massa ( mol ) reaktan dalam keluaran dari proses keseluruhan ) : massa ( mol ) reaktan dalam fresh feed
= ( massa ( mol ) reaktan yang dimasukkan ke dalam reaktor – massa ( mol ) reaktan yang terdapat dalam reaktor ) : massa ( mol ) reaktan yang dimasukkan ke dalam reaktor
Kesetimbangan materi untuk reaktor
( masukkan melalui batasan sistem ) –
( keluaran melalui batasan sistem ) +
( generasi dalam sistem ) –
( konsumsi dalam sistem ) = akumulasi
Kom-
Ponen
A
B
C
D
Inert
Simbol
A
B
C
D
I
Mula-mula Perubah-an
Zat sisa hasil
reaksi
________ ________
STRATEGI UNTUK MENGANALISIS MASALAH KESETIMBANGAN MATERI ((NERACA MASSA)
( Ann Landers : Masalah yang dikenali adalah masalah yang setengah terpecahkan)
1. Membaca masalah dan menjelaskan apa yang akan dikerjakan 2. Menggambar sketsa proses; mendifinisikan system dengan batas.3. Memberi label dengan symbol aliran dari setiap arus dan komposisi yang
berhubungan dan informasi lain yang tidak diketahui.4. Menaruh semua nilai komposisi dan aliran arus yang diketahui,
menghitung komposisi dan aliran tambahan dari data yang diberikan jika diperlukan.
5. Memilih sebuah basis.6. Membuat sebuah daftar menurut symbol untuk untuk setiap nilai yang
tidak diketahui dari aliran arus dan komposisi.7. Menulis nama-nama dari kumpulan kesetimbangan yang tepat yang akan
dipecahkan; tulis masing-masing kesetimbangan dengan jenis kesetimbangan yang tertulis didekatnya.
8. Menghitung jumlah kesetimbangan independent yang dapat ditulis9. Menyelesaikan persamaan-persamaan tersebut. Setiap perhitungan
harus dibuat pada basi yang konsisten.10. Memeriksa jawaban, ke dalam kesetimbangan materi redundant ( berlebih ), apakah jawaban masuk akal ?.
Material Balance CalculationsThe procedure….– Draw a flowchart
– Choose basis of calculations
– Label unknown stream variables on the flowchart
– Convert known stream volumes or volumetric flow rates to mass or molar basis using densities or gas laws
– Convert all mass and molar unit quantities to one basis
Material Balance CalculationsThe procedure….
– If any given information has not been used in labeling the flowchart, translate it into equations in terms of the unknown variables
– Write material balance equations in an order such that those involve the fewest unknowns are written first
– Solve the equations
Flowcharts• A flowchart is drawn using boxes or other symbols
to represent the process units and lines with arrows to represent inputs and outputs
• It must be fully labeled with values of known and unknown process variables at the locations of the streams
Fresh feed 100 mols C3H8
P1 mols C3H8 P2 mols C3H6 P3 mols H2
product
Q1 mols C3H8 Q2 mols C3H6 Q3 mols H2
Qr1 mols C3H8 Qr2 mols C3H6
100 + Qr1 mols C3H8 Qr2 mols C3H6
Reactor separator
Typical simple flowsheet arrangement
reactorSeparation & purification
Fresh feed(reactants, solvents,reagents, catalysts etc)
product
Recycle of unreacted material
Byproducts/coproductswaste
Recycle and Bypass
• Reasons for Recycle– Recovering and reusing unconsumed reactants– Recovery of catalyst (catalyst : expensive)– Dilution of process stream– Control of process variables– Circulation of working fluid
A + B ABC
A + B C
Recycle unused reactants (A,B)
Recovery of catalyst.
• Many reactors use catalysts to increase the rate of the reaction. Catalysts are usually expensive, and the processes generally include provisions for recovering them from the product stream and recycling them to the reactor. They may be recovered with the unconsumed reactants or recovered separately in special facilities designed for this purpose
Dilution of a process stream.
• Suppose a slurry (a suspension of solids in a liquid) is fed to a filter. If the concentration of solids in the slurry is too high, the slurry is difficult to handle and the filter will not operate properly. Rather than diluting the feed with fresh liquid, a portion of the filtrate can be recycled to dilute the feed to the desired solids concentration.
Control of a process variable.
• Suppose a reaction release an extremely large amount of heat, making the reactor difficult and expensive to control. The rate of heat generation can be reduced by lowering the reactant concentration, which can in turn be accomplished by recycling a portion of the reactor effluent to the inlet. In addition to acting as a diluent for the reactants, the recycled material also serves as a capacitance for the heat released: the greater the mass of the reaction mixture, the lower the temperature to which that mass will be raised by a fixed amount of heat.
Circulation of a working fluid
• The most common example of this application is the refrigeration cycle used in household refrigerators and air conditions. In these devices, a single material is reused indefinitely, with only small makeup quantities being added to the system to replenish working fluid that may be lost through leaks.
particularly useful for reactors, where they allow better control of reactor selectivity when multiple reactions occur.
recycle ratio
• dividing the mass flow of the recycle stream by the mass flow of the "fresh feed" entering the system
• the industrial world, recycle ratios have important consequences for system performance and operating costs.
Bypass
• A fraction of the feed is diverted around the process unit and combined with the output stream.
• Controlling properties and compositions of product stream
Solving Recycle and Bypass Problems
• methods for solving recycle and bypass problems are basically the same
• steady state, there is no buildup or depletion of material within the system or recycle stream of a properly designed and operated process.
• Note the distinction between the fresh feed
to the process and the feed to the reactor, which is the sum of the fresh feed and recycle stream.
When solving, you can write balances (total
material or component) around:
• the entire process structure • the mixing point • the splitter • the processing unit (inside the recycle/bypass)
Only three of these will be independent (the fourth is a linear combination).
Kiat penyelesaian :1. Pilih yang mempunyai data terbanyak ( sistem/sub sistem )
2. Jangan pilih yang meninggalkan variabel pada sub/sistem berikutnya
GAS, UAP, CAIRAN DAN PADATAN
• A gas has no fixed shape or volume, but always spreads out to fill any container.
• There are almost no forces of attraction between the particles so they are completely free of each other.
• The particles are widely spaced and scattered at random throughout the container so there is no order in the system.
• The particles move rapidly in all directions, frequently colliding with each other and the side of the container.
• With increase in temperature, the particles move faster as they gain kinetic energy.
The particle model of a Gas
• The particle model of a Liquid
• A liquid has a fixed volume at a given temperature but its shape is that of the container which holds the liquid.
• There are much greater forces of attraction between the particles in a liquid compared to gases, but not quite as much as in solids.
• Particles quite close together but still arranged at random throughout the container, there is a little close range order as you can get clumps of particles clinging together temporarily.
• Particles moving rapidly in all directions but more frequently collisions with each other than in gases due to shorter distances between particles.
• With increase in temperature, the particles move faster as they gain kinetic energy, so increased collision rates, increased collision energy and increased rate of diffusion.
• The particle model of a Solid
• A solid has a fixed volume and shape at a particular temperature unless physically subjected to some force.
• The greatest forces of attraction are between the particles in a solid and they pack together as tightly as possible in a neat and ordered arrangement.
• The particles are too strongly held together to allow movement from place to place but the particles vibrate about their position in the structure.
• With increase in temperature, the particles vibrate faster and more strongly as they gain kinetic energy.
Changes of State for gas <=> liquid <=> solid • Evaporation and Boiling (liquid to gas)
• On heating particles gain kinetic energy and move faster. • In evaporation* and boiling the highest kinetic energy molecules can ‘escape’
from the attractive forces of the other liquid particles. • The particles lose any order and become completely free to form a gas or vapour. • Energy is needed to overcome the attractive forces in the liquid and is taken in
from the surroundings. • This means heat is taken in, so evaporation and boiling are endothermic
processes (ΔH +ve). • If the temperature is high enough boiling takes place. • Boiling is rapid evaporation anywhere in the bulk liquid and at a fixed
temperature called the boiling point and requires continuous addition of heat. • * Evaporation takes place more slowly than boiling at any temperature between
the melting point and boiling point, and only from the surface, and results in the liquid becoming cooler due to loss of higher kinetic energy particles.
Condensing (gas to liquid)• On cooling, gas particles lose kinetic energy
and eventually become attracted together to form a liquid.
• There is an increase in order as the particles are much closer together and can form clumps of molecules.
• The process requires heat to be lost to the surroundings i.e. heat given out, so condensation is exothermic (ΔH -ve). – This is why steam has such a scalding effect, its not
just hot, but you get extra heat transfer to your skin due to the exothermic condensation on your surface!
Distillation
• Simple and fractional distillation involve the processes of boiling and condensation and are described on the Elements, Compounds and Mixtures
Melting (solid to liquid)
• When a solid is heated the particles vibrate more strongly as they gain kinetic energy and the particle attractive forces are weakened.
• Eventually, at the melting point, the attractive forces are too weak to hold the particles in the structure together in an ordered way and so the solid melts.
• The particles become free to move around and lose their ordered arrangement.
• Energy is needed to overcome the attractive forces and give the particles increased kinetic energy of vibration.
• So heat is taken in from the surroundings and melting is an endothermic process (ΔH +ve).
Freezing (liquid to solid)
• On cooling, liquid particles lose kinetic energy and so can become more strongly attracted to each other.
• Eventually at the freezing point the forces of attraction are sufficient to remove any remaining freedom and the particles come together to form the ordered solid arrangement.
• Since heat must be removed to the surroundings, so strange as it may seem, freezing is an exothermic process (ΔH -ve).
Cooling and Heating Curvesand the comparative energy changes of state changes
gas <=> liquid <=> solid Cooling curve • A cooling curve summarises the changes:
gas ==> liquid ==> solid
Heating curve:
• A heating curve summarises the changes:
solid ==> liquid ==> gas
Sublimation
• This is when a solid, on heating, directly changes into a gas without melting, AND the gas on cooling re-forms a solid directly without condensing to a liquid. They usually involve just a physical change BUT its not always that simple!
Dissolving solids, solutions and
miscible/immiscible liquids
• In general: solute + solvent ==> solution • The solid loses all its regular structure and the individual solid particles (molecules or
ions) are now completely free from each other and randomly mix with the original liquid particles, and all particles can move around at random.
• This describes salt dissolving in water, sugar dissolving in tea or wax dissolving in a hydrocarbon solvent like white spirit.
• It does not usually involve a chemical reaction, so it is generally an example of a physical change.
• Whatever the changes in volume of the solid + liquid, compared to the final solution, the Law of Conservation of Mass applies.
• You cannot make mass or lose mass, but just change it into another form.
+
If two liquids completely mix
• called miscible liquids because they fully dissolve in each other
+
the two liquids do NOT mix
• form two separate layers and are known as immiscible liquids
Gas Laws• A large number of experiments have determined that 4
variables are sufficient to define the physical condition (or state) of a gas: the gas laws.
Boyle’s Law, Charles’ Law, Avogadro’s hypothesis
Robert Boyle: (1627-1691) the first modern chemist, known as the father of chemistry.
His 1661 book The Sceptical Chymist marks the introduction of the scientific method, a definition of elements and compounds and a refutation of alchemy and magic potions.
Boyle biography
Boyle’s Law
• Boyle investigated the variation of the volume occupied by a gas as the pressure exerted upon it was altered and noted that the volume of a fixed quantity of gas, at constant temperature is inversely proportional to the pressure
constantor 1
constant PVp
V
Boyle’s Law(valid at low pressures)
The pressure-volume dependence of a fixed amount of perfect gas at different temperatures. Each curve is a hyperbola (pV = constant) and is called an isotherm.
Charles’ Law
• A century later, a French scientist, Jacques Charles discovered that the volume of a fixed amount of gas, as constant pressure, is proportional to the absolute temperature. Cool a balloon, or a sealed plastic bottle, to verify this!
constantor constant T
VTV
It was recognised (by William Thomson, Lord Kelvin, a Belfast born physicist) that if the graph was extrapolated to zero volume, an absolute zero of -273.15 oC is obtained.
Charles Law
The variation of the volume of a fixed amount of gas with the temperature constant. Note that in each case they extrapolate to zero volume at -273.15 C. The lines are isotherms.
Avogadro’s Law• Relationship between quantity of gas and volume established by
Gay-Lussac (balloon science!) and Avogadro in the 19th Century.Result was Avogadro’s hypothesis: equal volumes of gases at
the same temperature and pressure contain equal numbers of molecules
Experiments show that 22.4L of gas at 0oC and 1atm (STP), or 24.8L of gas at 298.15 K and 1 bar (SATP), contains 6.022 x 1023 molecules (Avogadro’s number, NA)Avogadro’s law: volume of gas at constant temperature and pressure is proportional to the number of moles of gas (n)
constant nV Remember:
1 mole = Avogadro’s number of objects
Putting it all together
nRTPV
P
nTRV
P
nTV
nVTVP
V
, ,1
Boyle, Charles, Avogadro
Combine
Call proportionality constant R
(gas constant)
Ideal Gas Equation
An “ideal” gas exhibits certain theoretical properties. Specifically, an ideal gas …
• Obeys all of the gas laws under all conditions.• Does not condense into a liquid when cooled.• Shows perfectly straight lines when its V and T &
P and T relationships are plotted on a graph.In reality, there are no gases that fit this definition
perfectly. We assume that gases are ideal to simplify our calculations.
A note on units and dimensional analysisSI unit for R is J/mol.K or m3.Pa/mol.K (R=8.315 of these units)Need to use the units of Pa for pressure and m3(=1000L) for volume in
any calculation.
Alternatively you can use units of kPa and L.
If you wish to use atm and L (as in USA and Textbook) R=0.0826 L.atm/mol/K.
Always use absolute temperature scale (K)
Solution
What is the value of R when the STP value for P is 760 mmHg?
R = PV = (760 mm Hg) (22.4 L)
nT (1mol) (273K)
= 62.4 L-mm Hg mol-K
Learning Check
Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mmHg) in the tank in the dentist office?
Solution
Set up data for 3 of the 4 gas variables
Adjust to match the units of R
V = 20.0 L - 20.0 L
T = 23°C + 273 - 296 K
n = 2.86 mol- 2.86 mol
P = ? ?
Rearrange ideal gas law for unknown P
P = nRT
V
Substitute values of n, R, T and V and solve for P
P = (2.86 mol)(62.4L-mmHg)(296 K)
(20.0 L) (K-mol)
= 2.64 x 103 mm Hg
Density of a Gas
Calculate the density in g/L of O2 gas at STP. From STP, we know the P and T.
P = 1.00 atm T = 273 K
Rearrange the ideal gas equation for moles/L
PV = nRT PV = nRT P = n
RTV RTV RT V
Substitute
(1.00 atm ) mol-K = 0.0446 mol O2/L
(0.0821 L-atm) (273 K)
Change moles/L to g/L
0.0446 mol O2 x 32.0 g O2 = 1.43 g/L
1 L 1 mol O2
Therefore the density of O2 gas at STP is
1.43 grams per liter
Sample problemsSample problemsHow many moles of H2 is in a 3.1 L sample of
H2 measured at 300 kPa and 20°C?PV = nRT
(300 kPa)(3.1 L) = n (8.31 kPa•L/K•mol)(293 K)
(8.31 kPa•L/K•mol)(293 K)(300 kPa)(3.1 L)
= n = 0.38 mol
How many grams of O2 are in a 315 mL container that has a pressure of 12 atm at 25°C?
P = 300 kPa, V = 3.1 L, T = 293 K
PV = nRT
(8.31 kPa•L/K•mol)(298 K)(1215.9 kPa)(0.315 L) = n = 0.1547 mol
P= 1215.9 kPa, V= 0.315 L, T= 298 K
0.1547 mol x 32 g/mol = 4.95 g
Exercises
1. What is the volume of 1 mole of an ideal gas
under standard temperature and pressure (STP)?
2. How many moles (g) of CO2 is liberated into a 250mL flask when a pressure of 1.3atm is found upon heating calcium carbonate to 31oC?
3. If a metal cylinder holds 50L of oxygen at 18.5atm and 21oC, what volume will the gas occupy at 1atm and same T?
4. a) Calculate the volume of 4.50 mol of SO2(g) measured at STP.
b) What volume would this occupy at 25°C and 150 kPa? (solve this 2 ways)
5. How many moles of CO2(g) is in a 5.6 L sample of CO2 measured at STP?
6. How many grams of Cl2(g) can be stored in a 10.0 L container at 1000 kPa and 30°C?
7. At 150°C and 100 kPa, 1.00 L of a compound has a mass of 2.506 g. Calculate its molar mass.
8. 98 mL of an unknown gas weighs 0.087 g at SATP. Calculate the molar mass of the gas. Can you determine the identity of this unknown gas?
Gas Mixtures--Partial Pressure
• So far: pure gases
• Many gases are actually mixtures of two or more gases:– air: O2, N2 , H2O, etc
• How do mixtures of gases behave?
Gas Mixtures--Partial Pressure
P= 8 psi
N2 (g)
P= 6 psi
O2 (g)
P= 9 psi
CO2 (g)
Gas Mixtures--Partial PressureWhat happens when you put all three samples of gas together into one container (the same size container as each was in alone)?
•The gases form a homogeneous mixture.
•The pressure in the container increases, V and T stay the same
– How do you know what the new pressure will be?
P
Dalton’s Law of Partial Pressure
Pressure exerted by mixed gas is sum of the partial pressures
p = pA+pB+ …
Recall mole fraction
xJ=nJ/n n = nA+nB+
Partial pressure
pJ= xJp
Gas Mixtures--Partial Pressure
• Ptotal = PO + PN + PCO
• So for this example:
Ptotal = 6 psi + 8 psi + 9 psi
= 23 psi
2 2 2
Partial Pressure
• In other words, at constant T and V,
– Ptotal depends only on the total number of moles of gas present
– Ptotal is independent of the type (or types) of gases present.
Partial Pressure
• The partial pressure of a gas in a mixture can be found:
PA = XA Ptotal
where PA = partial pressure of gas A
XA = mole fraction of gas A
Ptotal = total pressure of mixture
Partial Pressure Calculation
A mixture of gases contains 0.51 mol N2, 0.28 mol H2, and 0.52 mol NH3. If the total pressure of the mixture is 2.35 atm, what is the partial pressure of H2?
PH2 = XH2
Ptotal
XH2=
0.28 mol0.28 mol + 0.51 mol + 0.52 mol
= 0.21
PH2 = 0.21 x 2.35 = 0.50 atm
Partial Pressure-Mole Fraction
• When describing a mixture of gases, it is useful to know the relative amount of each type of gas.
• Mole fraction (X): a dimensionless number that expresses the ratio of the number of moles of one component compared to the total number of moles in a mixture.
Daltons’ Law of Partial Pressures
The % of gases in air Partial pressure (STP)
78.08% N2 593.4 mmHg
20.95% O2 159.2 mmHg
0.94% Ar 7.1 mmHg
0.03% CO2 0.2 mmHg
PAIR = PN + PO + PAr + PCO = 760 mmHg 2 2 2
Total Pressure 760 mm Hg
Learning Check
A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air?
1) 35.6 2) 156 3) 760
B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?
1) 557 2) 9.14 3) 0.109
Solution
A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air?
2) 156
B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?
1) 557
Gas Mixtures
Composition of a Gas Mixture: Mass and Mole Fractions
• Mass Fraction (mf): The ratio of mass of a component to the mass of the mixture is called the mass fraction.
• Mole Fraction (y): The ratio of the mole number of a component to the mole number of the mixture is called the mole fraction.
m
ii m
mmf
m
i
N
Ny
k
iim mm
1
k
iim NN
1
• The mass of a substance can be expressed in terms of the mole number N and molar mass M of the substance as
• The molar mass and the Gas constant of a mixture can be expressed as
NMm
k
iii
m
mm My
N
mM
1 m
um M
RR
Dalton’s Law of additive Pressures
• The pressure of a gas mixture is equal to the sum of the pressures each gas would exert if it existed alone at the mixture temperature and volume
k
immim VTPP
1
,
Amagat’s Law of additive Volume
• The volume of a gas mixture is equal to the sum of the volume each gas would occupy if it existed alone at the mixture temperature and pressure
k
immim PTVV
1
,
P-v-T Behavior of Ideal Gas Mixtures
• For ideal gases ,the above two laws are identical and give identical results.
• Pi is called the component pressure and Vi the component volume.
• The ratio Pi/Pm is called the pressure fraction and the ratio Vi/Vm is called the volume fraction of component i.
Ideal Gas Mixtures
• Partial Pressure: The quantity yiPm is called the partial pressure
• Partial Volume: The quantity yiVm is called the partial volume
• For ideal gas mixture, the mole fraction, the pressure fraction and the volume fraction of a component are identical
mii PyP
mii VyV
mixture of n gases,
• partial pressure of a gas in a mixture is the pressure it would exert if alone in the container.
Exercise:
A gaseous mixture is made from 6.00g oxygen and 9.00g methane placed in a 15L vessel at 0oC. What is the partial pressure of each gas and the total pressure in the vessel?
Contoh 1.
• Dua puluh kaki kubik ( ft3 ) nitrogen pada 300 psig dan 100oF dan 30 ft3 oksigen pada 200 psig dan 340oF dimasukkan ke dalam tangki bervolum 15 ft3. Tangki selanjutnya didinginkan menjadi 70oF. Hitung tekanan parsial masing-masing komponen dalam tangki tersebut. Asumsi gas adalah ideal.
• Jawab.
Persoalan dapat diselesaikan dengan menentukan jumlah mol oksigen dan nitrogen dan penyelesaian untuk tekanan total pada kondisi akhir. Selanjutnya tekanan parsial dapat dihitung dengan menggunakan mol fraksi.
Real Gases
• Due to intermolecular forces, the Ideal gas law breaks down
• Understanding deviations from ideal gas behavior teach us about molecular interactions
• Van der Waals Eqn and Virial Expansion– Coefficients relate to energy of interaction
between molecules
Real Gases
• Real gases do not completely follow the ideal gas law.
• In kinetic molecular theory, the following assumptions are made:– gas molecules occupy no space– gas molecules have no attraction for each
other
Real Gases
• Neither assumption is correct.
– Real gas molecules have a finite volume.
– Real gas molecules do attract each other.
Real Gases
• The greatest deviation from ideal gas behavior occurs at:high pressure
higher density of gas molecules– Molecules are closer together so:
» finite volume of gas molecules more important» attraction between molecules more important
Real Gases
In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.
Curves for 1 mol of gas
Real Gases
– Low temperatureAttractive forces between molecules becomes more
important.
– Average kinetic energy decreases.
– Gas molecules have less energy to overcome attractive forces.
Real Gases
Even the same gas (e.g. nitrogen) will show wildly different behavior under high pressure at different temperatures.
Compression Factor
RT
VpZ
gasperfectvolumemolar
volumemolarZ
Variation of Z with Pressure
At 0 C
Low pressures, Z=1, all gases ideal
At high pressures, Vreal >Videal ; Z>1 repulsive forces dominate
Low pressure, Vreal <Videal, Z< 1
attractive forces dominate
Low temperatures -> molecules moving less rapidly more influenced by attractive forces
REAL GASES
Deviations from ideal gas law
WHY?1. Molecules have volume
2. Molecules have attractive forces (intermolecular)
1. V-nb
2. -a(n/V)2
Van der Waals Equation of State2
V
na
nbV
nRTP
Van der Waals Equation (1873)2
V
na
nbV
nRTp2
mm V
a
bV
RTp
a reflects how strongly molecules attract each other
b corrects for the molecule’s size
Van der Waal’s constants
van der Waals Coefficients
Gas a (Pa m3) b(m3/mol)
Helium 3.46 x 10-3 23.71 x 10-6
Neon 2.12 x 10-2 17.10 x 10-6
Hydrogen 2.45 x 10-2 26.61 x 10-6
Carbon dioxide 3.96 x 10-1 42.69 x 10-6
Water vapor 5.47 x 10-1 30.52 x 10-6
a correlates with boiling point
b can be used to estimate molecular radii
“Derivation” of vdw Eq. State
Repulsive interactions cause molecules to behave as impenetrable spheres
Molecules restricted to smaller volume V-nb, where nb is volume molecules take up
Pressure depends on frequency of collisions with walls and force of each collision – both reduced by attractive forces proportional to molar concentration (n/V)
Pressure is then reduced according to a(n/V)2
Critical ConstantsThe `critical isotherm', the isotherm at the critical temperature Tc, is at 31.04 C. The critical point is marked with a star.
Critical pressure Pc
Critical molar volume Vc
Tc, Pc,Vc critical constants
If you compress along Tc, liquid does not appear – these highly dense materials are called “supercritical fluids” and their properties are a subject of intense current research
Other Equations of State
• Redlich-Kwong Peng-Robinson
Both are quantitative in region where gas liquefies
• Berthelot,Dieterici and others with more than ten parameters can give good fits !!!
• with seven free parameters, you can describe an elephant …
)(2/1 BVVT
A
BV
RTp
mmm
)()(
mmmm VVVV
RTp
Principle of Corresponding States
All gases have the same properties if they are compared at corresponding conditions
Define reduced variables
For homework you will write the vdw eqn in terms of the reduced variables
cRcRcR VVVTTTPPP /;/,/
Compression factor plotted using reduced variables. Different curves are different TR
Virial Equation of State
...)()(
12
32 V
TB
V
TB
RT
VPZ VV
...)()(1 232 PPBPPB
RT
VPZ PP
)(2 TB V = 0 at Boyle temperature
Most fundamental and theoretically sound
Polynomial expansion – Viral Expansion
Used to summarize P, V, T data
Also allow derivation of exact correspondence between virial coefficients and intermolecular interactions
P=101.325 kPa, V=5.6 L, T=273 K PV = nRT
(101.3 kPa)(5.6 L) = n (8.31 kPa•L/K•mol)(273 K)
5. Moles of CO2 is in a 5.6 L at STP?
(8.31 kPa•L/K•mol)(273 K)(101.325 kPa)(5.6 L)
= n = 0.25 mol
4. a) Volume of 4.50 mol of SO2 at STP.
P= 101.3 kPa, n= 4.50 mol, T= 273 K PV=nRT
(101.3 kPa)(V)=(4.5 mol)(8.31 kPa•L/K•mol)(273 K)
(101.3 kPa)
(4.50 mol)(8.31 kPa•L/K•mol)(273 K)V = = 100.8 L
4. b) Volume at 25°C and 150 kPa (two ways)?Given: P = 150 kPa, n = 4.50 mol, T = 298 K
(150 kPa)
(4.50 mol)(8.31 kPa•L/K•mol)(298 K)V = = 74.3 L
From a): P = 101.3 kPa, V = 100.8 L, T = 273 KNow P = 150 kPa, V = ?, T = 298 K
P1V1
T1
=P2V2
T2
(101.3 kPa)(100 L)(273 K)
=(150 kPa)(V2)
(298 K)(101.3 kPa)(100.8 L)(298 K)
(273 K)(150 kPa)=(V2) = 74.3 L
6. How many grams of Cl2(g) can be stored in a 10.0 L container at 1000 kPa and 30°C?
PV = nRT
(8.31 kPa•L/K•mol)(303 K)(1000 kPa)(10.0 L) = n = 3.97 mol
P= 1000 kPa, V= 10.0 L, T= 303 K
3.97 mol x 70.9 g/mol = 282 g
7. At 150°C and 100 kPa, 1.00 L of a compound has a mass of 2.506 g. Calculate molar mass.
PV = nRT
(8.31 kPa•L/K•mol)(423 K)(100 kPa)(1.00 L) = n = 0.02845 mol
P= 100 kPa, V= 1.00 L, T= 423 K
g/mol = 2.506 g / 0.02845 mol = 88.1 g/mol
8. 98 mL of an unknown gas weighs 0.081 g at SATP. Calculate the molar mass.
PV = nRT
(8.31 kPa•L/K•mol)(298 K)(100 kPa)(0.098 L) = n = 0.00396 mol
P= 100 kPa, V= 0.098 L, T= 298 K
g/mol = 0.081 g / 0.00396 mol = 20.47 g/mol
It’s probably neon (neon has a molar mass of 20.18 g/mol)
• ( P + a / Vm2 )( Vm - b ) = R T
• P = tekanan
• Vm =volum molar
• R = konstanta gas ideal
• T = suhu
