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Chap 5 Gas Laws Slide 1
Chapter 5 Gases and Gas Laws
Chem 110
Distinguishing gases from liquids and solids. An Overview of the Physical States of Matter
• Gas volume changes significantly with pressure. – Solid and liquid volumes are not greatly affected by pressure.
• Gas volume changes significantly with temperature. – Gases expand when heated and shrink when cooled. – The volume change is 50 to 100 times greater for gases than for liquids
and solids. • Gases flow very freely. • Gases have relatively low densities. • Gases form a solution in any proportions.
– Gases are freely miscible with each other.
Slide 2 Chap 5 Gas Laws
Chap 5 Gas Laws Slide 3
The Gaseous State Can we discuss the various states of matter from a molecular
standpoint? – Yes! Kinetic Molecular Theory (KMT) is a powerful
concept in discussing just how molecules move, interact, and react!
• Gases are a good place to start: – Least dense state of matter; – Intermolecular forces (IMFs) are minimum
Gases can be very easily described using FOUR variables: Volume (V); Pressure (P); amount (n); and Temperature (T).
Chap 5 Gas Laws Slide 4
Volume No need to describe in depth as concept of volume is straight-
forward. Use liter to quantify (L). A gas will occupy COMPLETELY whatever volume container it is
placed in:
Chap 5 Gas Laws Slide 5
Pressure • More difficult to define and describe. • P = force per unit area gas exerts on the walls of its container;
Measured using a barometer:
Each time wall gets hit the pressure changes.
g g g gF ma m m m hP d hVA A A Vh= = = = = =
d = density; g = 9.807 m/sec2
dHg = 13.59 g/cm3 Chap 5 Gas Laws Slide 6
Pressure units Pressure is measured is a variety of units:
Pascal (Pa): 1 kg m-1sec-2
Atmosphere (atm): 101.325 kPa Bar (bar): 1.013 atm Torr (torr) or mm Hg: 1.316 x 10-3 atm
• Typically, only use pascals if SI units exclusively are needed (commonly needed in energy calculations).
• Atm and torr are most commonly used (Pa is too small) • 1 atm = 760 torr = 760 mm Hg
2
Chap 5 Gas Laws Slide 7 Chap 5 Gas Laws Slide 8
Questions • How does pressure vary with volume? • How does pressure vary with temperature? • How does pressure vary with amount? • How does energy of a gas vary with temperature?
Let’s begin by looking at some of the basic early experiments regarding gases (scientific method).
The Gas Laws
• The gas laws describe the physical behavior of gases in terms of 4 variables: – pressure (P) – temperature (T) – volume (V) – amount (number of moles, n)
• An ideal gas is a gas that exhibits linear relationships among these variables.
• No ideal gas actually exists, but most simple gases behave nearly ideally at ordinary temperatures and pressures.
Slide 9 Chap 5 Gas Laws Chap 5 Gas Laws Slide 10
Boyle’s Law Boyle: studied how P and V were related. • J-tube experiment
(a) When Δh = 0, Pint = Pext = 1 atm (b) Add Hg thru open end, volume of
confined gas decreases; • New pressure goes up as follows:
(mmHg)(atm)
760 mmHg/atmnew initial
hP P= +
Chap 5 Gas Laws Slide 11
Boyles Law: Results • PV = constant at a fixed T • P = c (1/V) • c = 22.414 L-atm (!) • At low P, satisfied exactly; • At 1 atm small corrections
needed; • At high P (> 50 atm)
substantial corrections req’d; xxx
xxxxx
xx
xx
xx
xxx
xx
T1
T2
P
1/V
1 21 1 2 2
2 1
P VPV PVP V
= =
Chap 5 Gas Laws Slide 12
Charles Law • How does T (which is NOT a mechanical property) affect gases? • Charles: at low P all gases expand by the same relative amount
if Ti and Tf are the same for each gas (unlike liquids and solids)
V ∝ t at fixed P and n Extrapolation leads to
absolute zero and Kelvin scale (T);
Charles law is simplified if T (in K) used (true linear relationship)
Since T (in K) = t + 273.15
1 1
2 2T VT V
=
3
Chap 5 Gas Laws Slide 13
Avogadro’s Law Avogadro came up with the idea of ‘combining volumes’. 3 L of hydrogen gas and 1 L of nitrogen gas produces 2 L of ammonia gas
“Equal volumes of different gases at the same temperature and pressure contain equal numbers of particles”.
V ∝ n at fixed P and T V/n = constant
2 2 33 ( ) ( ) 2 ( )H g N g NH g+ ⎯⎯→
Chap 5 Gas Laws Slide 14
Combined Gas Law What we know from the “ABC”s of Gas Laws:
• V ∝ 1/P at fixed n and T (PV = const) • V ∝ T at fixed P and n (V/T = const) • V ∝ n at fixed P and T (V/n = const)
So:
1 1 2 2
1 1 2 2
PV PVn T n T
= 1 1
1 1
constant = RPVn T
=and
–1 –1 –1 –1R 0.082058 L atm mol K 8.3145 J mol K= =
Chap 5 Gas Laws Slide 15
Ideal Gas Law • PV = nRT • R is the universal gas constant • IGL is an equation of state: defining any three states (such as
P, n, and T) allows one to calculate the fourth and final state (here, V);
• Works well as long as one deals with ideal gases, or gases behaving ideally;
• Best ideal gas: monatomic, low pressure, high temperature; most gases are real and deviate from ideality (but not much).
Chap 5 Gas Laws Slide 16
Molar Volume • Avogadro’s hypothesis: equal volumes of gases have equal
numbers of ‘particles’. • Evaluate IGL in terms of volume of a gas and moles:
Note: independent of gas identity! • At SATP (standard ambient temperature and pressure):
Vm = 24.79 L·mol–1 (@ 298.15 K and 1 bar exactly) • At STP (standard temperature and pressure):
Vm = 22.41 L·mol–1 (@ 273.15 K and 1 atm exactly)
Vn= RT
P=Vm Vm is the Molar Volume
Chap 5 Gas Laws Slide 17
Molar Volume Comparisons
At STP (0 °C and 1 atm)
STP vs. SATP (0 °C vs. 25 °C); (1 bar vs. 1 atm)
Molar Mass from the Ideal Gas Law
m
Mn =
PV RT
=
mRT PV M =
Slide 18 Chap 5 Gas Laws
4
Chap 5 Gas Laws Slide 19
Density • Gases are the least dense of all matter; typical to measure and
report in grams per liter of gas; • IGL can be used to derive expression for density and how it
relates to a gas:
MM = massmol
= mn
; so n = mMM
PV = nRT = mMM
RT
P = mV
⎛⎝⎜
⎞⎠⎟
RTMM
= d RTMM
d = P( MM )RT
So, the density of a gas can be used to determine the molar mass and thus the identity of an unknown gas.
Chap 5 Gas Laws Slide 20
Gas Law Problems-1
A sample of SO2 (g) at 80°C is confined to a 3.0 L container and the pressure determined to be 4.5 atm; how many grams of SO2 are in the container?
Since you are being asked about ‘physical’ amounts of the gas, n must be determined first; since PV = nRT then
Since n = m/MM and SO2 has a MM=64 g/mol, n(MM) = m and thus m = 29.8 g of SO2
n = PV
RT= (4.5 atm)(3.0 L)
(0.08206 L ⋅atm ⋅mol–1 ⋅K–1)(353 K)= 0.466 mol
Chap 5 Gas Laws Slide 21
Gas Law Problems-2 An ideal gas at a pressure of 2.50 atm is cooled at constant volume until the pressure drops to 1.00 atm. If the gas is initially at 100. °C, what will the final temperature be?
In this two-state problem, n,V and R are all constant, so P/T = nR/V = constant for both states; therefore
Solving, we get T2 = 149 K (or -124 °C)
P1
T1
=P2
T2
; so T2 =P2T1
P1
= (1.00 atm)(373 K)2.50 atm
Chap 5 Gas Laws Slide 22
Gas Law Problems-3 Into a 500. mL container is placed 1.35 g of an unknown gas and the container is placed into a 25°C constant temperature bath. If the pressure of the gas is determined to be 600. torr, what is the molecular weight of the gas?
Since P(MM) = mRT = (m/V)RT we have all the information we need to solve this problem (remember to convert P to atm and T to kelvins):
Solving, we get MM = 83.7 g/mol
MM = mRT
PV= (1.35 g)(0.08206 L ⋅atm ⋅mol–1 ⋅K–1)(298 K)
(0.789 atm)(0.500 L)
Sample Problem 5.8 Finding the Molar Mass of a Volatile Liquid
PROBLEM: An organic chemist isolates a colorless liquid from a petroleum sample. She places the liquid in a preweighed flask and puts the flask in boiling water, causing the liquid to vaporize and fill the flask with gas. She closes the flask and reweighs it. She obtains the following data:
Calculate the molar mass of the liquid.
Volume (V) of flask = 213 mL T = 100.0°C P = 754 torr mass of flask + gas = 78.416 g mass of flask = 77.834 g
PLAN: The variables V, T and P are given. We find the mass of the gas by subtracting the mass of the flask from the mass of the flask with the gas in it, and use this information to calculate M.
Slide 23 Chap 5 Gas Laws
Sample Problem 5.8
SOLUTION:
m of gas = (78.416 - 77.834) = 0.582 g
M = mRT
PV =
0.582 g x atm·L mol·K
0.0821 x 373 K
0.213 L x 0.992 atm = 84.4 g/mol
1 L
103 mL
V = 213 mL x = 0.213 L T = 100.0°C + 273.15 = 373.2 K
1 atm
760 torr
P = 754 torr x = 0.992 atm
Slide 24 Chap 5 Gas Laws
5
Chap 5 Gas Laws Slide 25
Gas Mixtures • Most gases are found as mixtures; for example, dry air is a
mixture (w/w) of N2 (75.52%), O2 (23.14%), CO2 (0.05%), and Ar (1.29%). Does this affect gas behavior?
• At low pressures (1-2 atm and lower) a mixture of gases that do not react with one another behaves like a single pure gas and can therefore be treated as one in using the IGL.
• Dalton’s Law of Partial Pressures: the total pressure of a mixture of gases is the sum of the partial pressures of its components.
Ptot = PA + PB ++ Pi = Pi∑
Chap 5 Gas Laws Slide 26
Law of Partial Pressures Some uses and modifications:
a) Humid gas calculations
b) Pressure in terms of mole composition:
Ptot = PA + PB ++ Pi =nART
V+
nBRTV
++niRT
V
Ptot = (nA + nB ++ ni )RTV
RTV
=Ptot
(nA + nB ++ ni )=
Ptot
ntot
; so PA =nA
ntot
Ptot
define mole fraction, xi =ni
ntot
then Pi =ni
ntot
Ptot = xi Ptot
Phumid air = Pdry air + Pwater vapor
T(0C) T(0C)
0 5
10 12 14 16 18 20 22 24 26 28 30 35
40 45 50 55 60 65 70 75 80 85 90 95
100
55.3 71.9 92.5
118.0 149.4 187.5 233.7 289.1 355.1 433.6 525.8 633.9 760.0
4.6 6.5 9.2
10.5 12.0 13.6 15.5 17.5 19.8 22.4 25.2 28.3 31.8 42.2
Table 5.2 Vapor Pressure of Water (Pwater) at Different T
P (torr) H2O
P (torr) H2O
Slide 27 Chap 5 Gas Laws Chap 5 Gas Laws Slide 28
Kinetic Molecular Theory The empirical ideal gas law suggests that we should be able to develop a model that explains gas behavior at the molecular level. What exactly is occurring at the microscopic level?
– How do individual particles exert force to give pressure? – Why are gases compressible? And what happens on the
molecular level? – Why does the law of partial pressures work? – What is the temperature effect on molecules? – Why does Avogadro’s law work? Shouldn’t smaller
molecules occupy less space than larger ones? What about pressure considerations?
Chap 5 Gas Laws Slide 29
Molecular Motion in Gases • Gas laws suggest a lot about gases, but not much about their motion
(except that they don’t interact much and their speed seems to increase with temperature). Are all gas processes IGL- derivable?
• Two processes are not: Diffusion: gradual dispersal of one gas thru another (equal pressure
process). Effusion: escape of a gas thru a pinhole (or porous barrier) from a region
of high pressure to one of lower pressure.
Chap 5 Gas Laws Slide 30
Effusion and Size Effusion and diffusion, while different processes, can be explained using similar approaches; effusion is easier to treat quantitatively.
Graham: at constant T, the rate of effusion is inversely proportional to the square root of the molar mass:
This is Grahams Law of Effusion For two gases, we can say the following:
Rate of effusion ∝
1
MM∝ Average speed
effusion RateA
effusion RateB
=N
A
NB
MMB
MMA
=MM
B
MMA
effusion TimeA
effusion TimeB
=MM
A
MMB
6
Chap 5 Gas Laws Slide 31
Effusion and Temperature Effusion experiments at different temperatures reveals that the rate of effusion of a given gas increases as the temperature is raised; specifically:
Therefore:
And so:
Rate of effusion ∝ Average speed ∝ T
effusion Rate T2
effusion Rate T1
=T
2
T1
Average speed ∝
T
MM
Chap 5 Gas Laws Slide 32
Kinetic Molecular Theory: Assumptions
1) A pure gas consists of a large number of identical molecules separated by distances that are great compared to their size (point-charge assumption).
2) Gas molecules are constantly moving in random directions with a distribution of speeds in straight-line trajectories.
3) Gas molecules exert no attractive or repulsive forces on one another between collisions, so between collisions they move in straight lines with constant velocities.
4) Since gas molecules are much, much smaller than the volume they occupy, collisions between two molecules are rare (or, rare compared with collisions with the walls).
5) Collisions of molecules with the walls are elastic: no energy is lost during a collision
Chap 5 Gas Laws Slide 33
Consequences of KMT Out of a rigorous mathematical treatment the following results are obtained: 1. EK = (3/2)RT; thus, the temperature of a gas is a measure of its average
kinetic energy, independent of the identity of the gas; 2. The distribution of molecular speeds can be represented by the Maxwell-
Boltzmann speed distribution:
Once at thermal equilibrium, this distribution will persist indefinitely. 3. Since f(u) is a probability function, three specific speeds can be identified:
f (u) = 4π
m
2πkBT
⎛⎝⎜
⎞⎠⎟
3 2
u2e– ( mu2 / 2 kBT )
u
mp=
2RT
MM; u
ave=
8RT
π ( MM ); and u
rms=
3RT
MM
Chap 5 Gas Laws Slide 34
Maxwell-Boltzmann Distribution
ump = most probable speed; f(u) is at its maximum;
uav = average speed;
urms = root mean square speed; this can be thought of as the ‘typical’ speed used to describe a gas. It is also equal to
2(EK )MM
ump : uav :urms = 1.000 : 1.128: 1.225
Chap 5 Gas Laws Slide 35
Applications of KMT • Zw: the rate of collisions of gas molecules with a section of the wall with area A:
(Graham’s law of effusion explained by this)
• Z1: the frequency of collisions with other molecules; assumes molecules are approx. spherical and sweep out a cylinder in its flight (d = molecular diameter):
• λ: the mean free path; distance traveled between collisions:
Z
w=
1
4
N
V
⎛⎝⎜
⎞⎠⎟
(uave
) A
Z
1= 4
N
V
⎛⎝⎜
⎞⎠⎟
d 2 πRT
MM
λ =
uave
Z1
= 12(π )(d 2 )(N V )
Chap 5 Gas Laws Slide 36
Molecular Questions, KMT Answers
Envoking KMT to answer the questions raised earlier regarding what is occurring at the molecular level:
Pressure: collisions with container cause this; anything that causes them to ⇑ will cause ⇑ P;
Compressibility: ⇓ space and molecules hit walls more often: ⇑ P;
Partial Pressures: Simply an ⇑ in the number of molecules; more molecules to collide with walls (assume NO IMF’s!)
Temp Effect: ⇑ energy and molecules move faster and hit walls more often: ⇑ P;
Avogadro’s: ⇑ molecules, ⇑ collisions; ⇑ V until collisions per surface area gets to where it was before addition
7
Chap 5 Gas Laws Slide 37
Real Gases • The IGL is an equation of state, one that simply relates all states to one
another (P, V, T, and n); how do real gases differ?
• Deviations are revealed in the compressibility factor, z:
z =
PV
RT for 1 mol of a gas
N2 25 C
Chap 5 Gas Laws Slide 38
van der Waals Equation of State
Modifications take into account that forces do exist between molecules and that these are repulsive at short distances and attractive at long distances; that gas molecules do indeed have molecular volumes; and that attractions occur between pairs of molecules.
Effective volume = (V-nb); b is a constant that describes the volume excluded per mole of molecules (units are L per mol); Pressure attractive correction = a(n2/V2); a is a positive constant that depends on the attractive forces of the gases involved.
Now z can be written as follows:
P + a
n2
V 2
⎛⎝⎜
⎞⎠⎟
(V − nb) = nRT or P = nRT
(V − nb)– a n2
V 2
z =
1
1− (nb V )–
an
RTV
If a and b = 0, z = 1 (ideal gas) If a is small and b is appreciable, z > 1 If b is small and a is appreciable, z < 1)
Chap 5 Gas Laws Slide 39
Van der Waals constants a and b
0.034 0.211 1.35 2.32 4.19 0.244 1.39 1.36 6.49 2.25 1.45 3.59 4.17 5.46
He Ne Ar Kr Xe H2 N2 O2 Cl2 CH4 CO CO2 NH3 H2O
0.0237 0.0171 0.0322 0.0398 0.0511 0.0266 0.0391 0.0318 0.0562 0.0428 0.0395 0.0427 0.0371 0.0305
Gas a
atm·L2
mol2 b
L
mol
Table 5.4 Van der Waals Constants for Some Common Gases
Slide 40 Chap 5 Gas Laws
Chap 5 Gas Laws Slide 41
End Chapter 5 Gas Laws
Chap 5 Gas Laws Slide 42