Chap 5 Gas Laws and KMT - Palomar College (bar): 1.013 atm Torr (torr) or mm Hg: 1.316 x 10-3 atm...

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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

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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).


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:

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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


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

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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).

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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.



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= +

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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

= =


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

=

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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+ ⎯⎯→


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= =

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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).


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

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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 =


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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.


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

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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


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)

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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.


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


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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∑


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

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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



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?

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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.


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

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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


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

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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


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

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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 )


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

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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


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)

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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


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End Chapter 5 Gas Laws


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Chap 5 Gas Laws and KMT - Palomar College (bar): 1.013 atm Torr (torr) or mm Hg: 1.316 x 10-3 atm...

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