Gases - PowerPoint PPT Presentation

Gases

The Gas Laws

Labs

#18 Molar Mass of a Volatile Liquid

#19 Calcium Carbonate Analysis: Molar Volume of Carbon Dioxide

• Expansion: Expand indefinitely to fill the space available to them
• Indefinite shape: Fill all parts of container evenly, so have no definite shape of their own
• Compressibility: Most compressible of states of matter
• Mixing: Two or more gases will mix evenly and completely when confined to same container
• Low density: Have much lower densities than liquids and solids (typically about 1/1000 those of liquids/solids)
• Pressure: Exert pressure on their surroundings

• Gas phase at temperature where same substance can also exist in liquid or solid state
• Below critical temperature of substance (vapor can be condensed by increasing pressure without reducing temperature)

• Force generated by collisions of gas particle w/container walls
• Related to velocity of gas particles
• Total force = sum of forces of all collisions each second per unit area
• Force = m x acceleration
• Pressure dependent on
• Gas particle velocity
• Collision frequency

• Collision frequency dependent on
• Gas particle velocity
• Distance to container walls.
• Changing temperature changes collision force, as well as collision frequency
• Collision frequency changed by altering size of container
• Force of collisions is not affected

(force of 1 newton exerted on one square meter of area)

(29.92” Hg)

(psi)

• Used to measure pressure of enclosed gas
• U-tube partially filled with liquid, typically Hg
• One end connected to container of gas being measured
• Other end sealed with vacuum existing above liquid, or open to atmosphere

Closed-end manometer

• Pressure is just difference between two levels (in mm of Hg)-indicates pressure of system attached to apparatus
• Gas connected to one arm
• Space above Hg in other arm is vacuum
• Liquid in tube falls to height (directly proportional to pressure exerted by gas in the tube)
• Since pressure of gas causes liquid levels to be different in height, it is this difference (h) that is measure of gas pressure in container
• Pgas = Ph
• If left to atmosphere, it measures atmospheric pressure-barometer

Open-end manometer

• Used to measure pressure of gas in container
• Difference in Hg levels indicates pressure difference in gas pressure and atmospheric pressure
• Atmospheric pressure pushes mercury in one direction
• Gas in container pushes it in the other direction
• Two ends connected to gases at different pressures
• Closed end to gas in bulb (gas filled)
• Open end to atmosphere
• If pressure of gas higher than atmospheric, Hg level lower in arm connected to gas (Pgas = Pbarometric + h)
• If pressure of gas lower than atmospheric, Hg level higher in arm connected to gas (Pgas = Pbarometric - h)
• If levels are equal, gas at atmospheric pressure

Difference in heights of liquid levels inversely proportional to density of liquid and represents the pressure

Greater density of liquid, smaller difference in height

High density of mercury (13.6 g/mL) allows relatively small monometers to be built

Readings must be corrected for relative densities of fluid used and of mercury (mm Hg = mm fluid)

density fluid

density Hg

A sample of CH4 is confined in a water manometer. The temperature of the system is 30.0 °C and the atmospheric pressure is 98.70 kPa. What is the pressure of the methane gas, if the height of the water in the manometer is 30.0 mm higher on the confined gas side of the manometer than on the open to the atmosphere side. (Density of Hg is 13.534 g/mL).

1) Convert 30.0 mm of H2O to equivalent mm of mercury:

• (30.0 mm) (1.00 g/mL) = (x) (13.534 g/mL) x = 2.21664 mm (I will carry some guard digits.)
• 2) Convert mmHg to kPa:
• 2.21664 mmHg x (101.325 kPa/760.0 mmHg) = 0.29553 kPa
• 3) Determine pressure of enclosed wet CH4:
• At point A in the above graphic, we know this: Patmo. press. = Pwet CH4 + Pthe 30.0 mm water column
• 98.70 kPa = x + 0.29553 kPa
• x = 98.4045 kPa
• 4) Determine pressure of dry CH4:
• From Dalton's Law, we know this: Pwet CH4 = Pdry CH4 + Pwater vapor
• (Water's vapor pressure at 30.0 °C is 31.8 mmHg. Convert it to kPa.)
• 98.4045 kPa = x + 4.23965 kPa
• x = 94.1648 kPa
• Based on provided data, use three significant figures; so 94.2 kPa.

Boyle’s Law

X axis independent variable Y axis dependent variable

Pressure-Volume Relationship

Temperature/# molecules (n) constant

Inversely proportional (one goes up, other goes down)

• V 1/P or
• P = k/T where K is constant
• PV = constant
• P1V1 = P2V2
• Gas that strictly obeysBoyle’s Law is ideal gas (holds precisely for gases at very low temperatures)

• Shows inverse proportion
• If pressure doubled, volume decreased by ½

• Plot of V (P) against 1/P (1/V) gives straight line
• Graph of equation P = k1 x 1/V
• Same relationship holds whether gas is being expanded or contracted

A gas which has a pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant pressure?

• P1 = 1.3 atm
• V1 = 27 L
• P2 = 3.9 atm
• V2 = ?
• P1V1 = P2V2
• 1.3 atm (27 L) = 3.9 atm (X) = 9.0 L

Charles’ Law

Temperature-Volume Relationship

Pressure/n constant

• From extrapolated line, determine T at which ideal gas would have zero volume
• Since ideal gases have infinitely small atoms, only contribution to volume of gas is pressure exerted by moving atoms bumping against walls of container
• If no volume, then no kinetic energy left
• Absolute zero is T at which all KE has been removed
• Does not mean all energy has been removed, merely all KE

http://www.absorblearning.com/media/attachment.action?quick=10o&att=2629

If T↓, V↓ and vs.

• V = kT
• P is constant
• k = proportionality constant
• V/T = constant

A gas at 30oC and 1.00 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60.0oC and 1.00 atm?

• V1 = 0.842 atm
• T1 = 30oC = 303 K
• V2 = ?
• T2 = 60.0oC = 333 K
• V1/T1 = V2/T2
• 0.842/303 K = X/333 K = 0.925 L

Gay Lussac’s Law

Pressure-Temperature Relationship

Volume/n constant

When H burns in O, volume of H consumed is always exactly 2x volume of O

Direct relationship between pressure and temperature (P T)

P/T = constant

Pi/Ti = Pf/Tf

Avogadro’s Law(at low pressures)

Volume-Amount Relationship

Temperature/Pressure constant

Avogadro's law predicts directly proportional relation between # moles of gas and its volume

Helped establish formulas of simple molecules when distinction between atoms and molecules was not clearly understood, particularly existence of diatomic molecules

Once shown that equal volumes of hydrogen and oxygen do not combine in manner depicted in (1), became clear that these elements exist as diatomic molecules and that formula of water must be H2O rather than HO as previously thought

V of gas proportional to # of moles present (V n)

• At constant T/P, V of container must increase as moles of gas increase
• V/n = constant
• Vi/ni = Vf/nf

A 5.20 L sample at 18.0oC and 2.00 atm pressure contains 0.436 moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be?

• V1 = 5.20 L
• n1 = 0.436 mol
• V2 = X
• n2 = 0.436 + 1.27 = 1.70 mol
• V1/n1 = V2/n2
• 5.20 L/0.436 mol = X/1.70 mol = 20.3 L

Generalization applicable to most gases, at pressures up to about 10 atm, and at temperatures above 0°C.

Ideal gas’s behavior agrees with that predicted by ideal gas law.

Ideal Gas Law(Holds closely at P < 1 atm)

• Combined
• If proportionality constant called R
• Rearrange to form ideal gas equation

Pressure (P) in atm

Volume (V) in L

Absolute temperature (T) in K (Charles’ Law)

Amount (number of moles, n)

R (universal ideal gas constant)

0.08206 liter ∙ atm/mole ∙ K or 0.08206 L·atm·K–1·mol–1

8.31 liter ∙ kPa/mole ∙ K

8.31 J/mole ∙ K

8.31 V ∙ C/mole ∙ K

8.31 x 10-7 g ∙ cm2/sec2 ∙ mole ∙ K

6.24 x 104 L ∙ mm Hg/mol ∙ K

1.99 cal/mol ∙ K

Given 3 of 4 variables and calculate 4th

PiVi= niRTi

PfVf nfRTf

Cancel all constants and R, make appropriate substitutions from given data to perform calculation

A sample containing 0.614 moles of a gas at 12.0oC occupies a volume of 12.9 L. What pressure does the gas exert?

• PV = nRT
• P (12.9 L) = (0.614 mol)(0.08206 L atm/K mol)(285 K) = 1.11 atm

Read 5.1-5.3, pp. 189-202

Q pp. 232-234, #29, 31-34, 44

• STP = 0°C (273K) and 1.00 atm pressure (760 mm Hg)
• One mole of gas at STP will occupy 22.42 L
• Allows you to compare gases at STP to each other

What volume will 1.18 mole of O2 occupy at STP?

• PV = nRT
• (1atm)(X) = (1.18 mol)(0.08206 L atm/K mol)(273 K) = 26.4 L
• Alternate way:
• At STP, 1 mole occupies 22.4 L
• Vi/ni = Vf/nf
• 22.4 L/1 mol = X/1.18 mol = 26.4 L

A sample containing 15.0 g of dry ice, CO2(s), is put into a balloon and allowed to sublime according to the following equation: CO2(s)  CO2(g) How big will the balloon be (what is the volume of the balloon) at 22oC and 1.04 atm after all of the dry ice has sublimed?

• 15.0 g CO2 1 mol CO2 = 0.341 mol CO2

44.0 g CO2

• PV = nRT
• (1.04 atm)(X) = (0.341 mol)(0.08206 L atm/K mol)(295K) = 7.94 L

0.500 L of H2(g) are reacted with 0.600 L of O2(g) according to the equation 2H2(g) + O2(g)  2H2O(g). What volume will the H2O occupy at 1.00 atm and 350oC?

• Find limiting reactant:
• 0.500 L H2 1 mol H2 = 0.0223 mol H2/2

22.4 L

• 0.600 L O2 1 mol O2 = 0.0268 mol O2/1

22.4 L

• PV = nRT
• (1atm)(X) = (0.0223 mol)(0.08206 L atm/K mol)(623 K) = 1.14 L

• Gives density which can be determined if P, T and molar mass are known
• Density directly proportional to molar mass
• Density increases as gas pressure increases
• Density decreases as temperature increases

A gas at 34.0oC and 1.75 atm has a density of 3.40 g/L. Calculate the molar mass (M.M.) of the gas.

• M.M. = dRT

P

• M.M. = (3.40 g/L)(0.08206 L atm/K mol)(307K)

(1.75 atm)

• M.M. = 48.9 g/mol

• What are the expected densities of argon, neon and air at STP?
• PV = nRT
• PV = g (RT)

molar mass

• Density = g/V = (Molar mass)P/RT
• density Ar = (39.95 g/mol)(1.00 atm)/(0.0821 L-atm/mol-K)(273 K) = 1.78 g/L
• density Ne = 0.900 g/L
• density air = 1.28 g/L
• molar mass = (0.80)(28 gN2/mol) + (0/20)(32 g O2/mol) = 28.8 g/mol

• V = RT

n P

• (0.0821 L-atm/mol-K)(273K) = 22.4 L/mol

1.00 atm

Read 5.4, pp. 203-206

Q pp. 234-235, #52, 56, 58, 60

Dalton’s Partial Pressure

When two gases are mixed together, gas particles tend to act independently of each other

• Each component gas of mixture of gases uniformly fills containing vessel
• Each component exerts same pressure as it would if it occupied that volume alone
• Total pressure of mixture is sum of individual pressures, called partial pressures, of each component
• Pt = PA + PB + PC + …

Pi/Pt = ni/nt

Pi = ni/nt x Pt

Pi = XiPt

Xi = mole fraction of gas component i

A volume of 2.0 L of He at 46oC and 1.2 atm pressure was added to a vessel that contained 4.5 L of N2 at STP. What is the total pressure and partial pressure of each gas at STP after the He is added?

• Find # moles of He at original conditions-will lead us to finding partial pressure of He at STP.
• PV = nRT
• (1.2 atm)(2.0 L) = n(0.08206 L atm/k mol)(319 K)
• n = 0.0917 mol He
• When gases are combined under STP, partial pressure of He will change while N2 will remain the same since it is already at STP.
• PHeV = nRT
• (X)(4.5 L) = (0.091 mol)(0.08206 L atm/K mol)(273 K)
• P = 0.457 atm
• Total pressure = 1.00 + 0.46 = 1.46 atm

Mole fraction

• Xi = ni = Pi

ntotal Ptotal

• You can use either the number of moles or the pressure of each component of your system to evaluate the mole fraction
• 4.5 L N2 1 mol = 0.201 mol N2

22.4 L

• XN2 = 0.201 mol = 0.69 1.00 atm = 0.69

0.293 mol 1.457 atm

• X He = 0.0917 mol = 0.31 0.457 atm = 0.31

0.293 mol 1.457 atm

Calculate the mole fraction of each gas.

nt = 3.00 mol + 4.00 mol + 1.00 mol = 8.00 mol

Xi = ni/nt

XHe = 3.00 mol/8.00 mol = 0.375

XAr = 4.00 mol/8.00 mol = 0.500

XNe = 1.00 mol/8.00 mol = 0.125

Calculate the partial pressure of each gas.

Pi = XiPi

PHe = 0.375 x 1,200 torr = 450. torr

PAr = 0.500 x 1,200 torr = 600. torr

PNe = 0.125 x 1,200 torr = 150. torr

He-1.2 L, 0.63 atm, 16oC Ne-3.4 L, 2.8 atm, 16oC

Pt = Pgas + PH2O

Consequently, partial pressure of collected gas is Pgas = Pt – PH2O, where partial pressure of water depends on temperature and corresponds to vapor pressure of water

What is the partial pressure of oxygen? Given the vapor pressure of water at 30°C is 31.8 mm Hg.

Pt = PO2 + PH2O

PO2 = Pt – PH2O = 645 mm Hg – 31.8 mm HgPO2 = 613 mm Hg

What are the mole fractions of oxygen and water vapor?

Recall that the partial pressures of O2 and H2O are related to their mole fractions, PO2 = XO2Pt      PH2O = XH2OPt

X O2 = 613/645 = 0.950, and XH2O = 31.8/645 = 0.049

Note also that the sum of the mole fractions is 1.0, within the number of significant figures, given: X O2 + XH2O = 0.950 + 0.049 = 0.999

The vapor pressure of water in air at 28oC is 28.3 torr. Calculate the mole fraction of water in a sample of air at 28oC and 1.03 atm pressure.

• XH2O = PH2O = 28.3 torr = 0.036

Pair 783 torr

The safety air bags in cars are inflated by nitrogen gas generated by the rapid decomposition of sodium azide. If an air bag has a volume of 36 L and is to be filled with nitrogen gas at a pressure of 1.15 atm at a temperature of 26.0oC, how many grams of NaN3 must be decomposed?

2NaN3(s)  2Na(s) + 3N2(g)

• Gas data  mol N2 mol NaN3 g NaN3
• n = PV/RT = (1.15 atm)(36 L) = 1.7 mol N2

(0.0821 L-atm/mol-K)(299K)

• 1.7 mol N2 2 mol NaN3 65.0g NaN3 = 72 g NaN3

3 mol N2 1 mol NaN3

Read 5.5, pp. 206-211

Q pg. 235, #64, 66, 68, 70, 72

Kinetic Molecular Theory of Gases

Describes the properties of gases

• Gases composed of tiny, invisible molecules that are widely separated from one another in empty space
• Gas molecule behave as independent particles
• Volume occupied by molecule considered negligible
• Gas molecules are in constant motion, continuous, random and straight-line motion
• Collisions of particles with walls of container cause pressure exerted by gas
• Molecules collide with one another, but collisions are perfectly elastic (no net loss of energy-exert no force on each other)
• Attractive forces between atoms and/or molecules in gas are negligible
• Average kinetic energy of collection of gas particles assumed to be directly proportional to Kelvin temperature of gas

• Equal #s of molecules of any gas have same average kinetic energy at same temperature
• Greater temperature, greater fraction of molecules moving at higher speeds (higher average kinetic energy of gas molecules)
• Individual molecules move at varying speeds
• Momentum conserved in each collision, but one colliding molecule might be deflected off at high speed while another is nearly stopped
• Result is that molecules at any instant have wide range of speeds

Average KE, e, related to root mean square (rms) speed u

4 molecules in gas sample have speeds of 3.0, 4.5, 5.2, and 8.3 m/s.

Average speed

rms

Because mass of molecules does not increase, rms speed of molecules must increase with increasing temperature

Average kinetic energy of single gas molecule

• KE = ½ mv2
• m = mass of molecule (kg)
• v = speed of molecule (meters/sec)
• KE measured in joules
• Average translational kinetic energy of any kind of molecule in ideal gas is given by:

J = kg· m2/s2

Based on understanding of pressure, molecular meaning of gas laws can be appreciated as follows:

• Kinetic molecular theory agrees
• If V of container decreased, gas particles strike walls of container more frequently, increasing pressure of gas
• If V of container increased, fewer impacts of gas molecules with wall per second, decreasing pressure

• Increasing temperature increases average kinetic energy and speed of gas molecules
• Increases force of each collision w/container wall
• Increases frequency of collisions w/container wall
• Both increase pressure
• If pressure of gas is to remain constant, volume must increase to decrease frequency of collisions with container walls
• If one of the walls is a movable piston, then the volume of the container will expand until balanced by the external force

• Increase in temperature increases kinetic energy of gas particles
• Force of each collision increases
• Frequency of collisions with container walls increases
• When gas is heated in container with fixed volume, gas molecules impact more forcefully with wall, increasing pressure

• As more gas molecules are added to the container, # of impacts per second with wall increases, pressure increases correspondingly
• If V of container is not fixed, then V will increase

Graham’s Law of Effusion

Experiments show that molecules of a gas do not all move at same speed but are distributed over a range.

• Probability distribution that forms basis of kinetic theory of gases
• Explains many fundamental gas properties, including pressure and diffusion

Highest point on curve marks most probable speed-some move with speeds much less than most probable speed, while others move with speeds much greater than this speed

As temp. increases, curve flattens out (more molecules have higher kinetic energies and therefore higher average speeds)

(speed)

• Spread of gas molecules throughout volume
• Much slower than average molecular speed because of collision between molecules

• Mean free path
• Average distance molecule travels between collisions
• Factors affecting it:
• Density (increasing density decreases MFP)
• Radius of molecule (increasing size increases collision frequency, reducing MFP)
• Pressure (increasing pressure increases collision frequency, reducing MFP)
• Temperature (increasing T increases collision frequency, but does not affect MFP)

• Escape of gas through small opening into vacuum
• Rate (in mol/s) proportional to average speed u
• More collisions gas has w/walls of container, higher probability it will hit pinhole and go through it

Gas molecules with higher molar masses will have slower average speeds, while lighter molecules will have higher average speeds

Rates two gases effuse through a pinhole in a container are inversely related to the square roots of the molecular masses of gas particles

Use equation to find relative rates of diffusion of gases, whether evacuated or not)

All gases expand to fill container

Heavier gases diffuse more slowly than lighter ones

KE-average kinetic energy

c = constant that is same for all gases

T = temperature in Kelvin

M = mass of gas

v2 = average of the square of the velocities of the gas molecules

Since cT will be the same for all gases at the same temperature, the average KE of any two gases at the same T will be the same

KE1 = KE2

½ m1V21 = ½ m2v22

root mean square velocity = vrms = √v2

√m1/m2 = vrms2/vrms1

Higher the T, higher vrms

Effusion:

Diffusion:

How many times faster than He would NO2 gas effuse?

• MNO2 = 46.01 g/mol
• MHe = 4.003 g/mol
• √MNO2/MHe= rateHe/RateNO2
• √46.01/4.003 = 3.390
• So He would effuse 3.39 times faster as NO2

• Ideal gas law works very well for most gases, however, it does not work well for gases under high pressures or gases at very low temperature
• Conditions used to condense gases
• Generalized that any gas close to its BP (condensation point) will deviate significantly from ideal gas law
• Kinetic molecular theory makes two assumptions
• Gases have no volume
• They exhibit no attractive or repulsive forces

Cooling gas decreases average KE of its molecules

All gases when cooled enough will condense to liquid

Suggests that intermolecular forces of attraction exist between molecules of real gases

Condensation occurs when average KE is not great enough for molecules to break away from intermolecular attractive forces

When they stick together, there are fewer particles bouncing around and creating P, so real P in nonideal situation will be smaller than P predicted by ideal gas equation

Van der Waals Equation

Developed modification of ideal gas law to deal with nonideal behavior of real gases

Real Gases

Must correct ideal gas behavior when at high pressure(smaller volume) and low temperature(attractive forces become important)

correction for attractive forces

correction for small/finite volume





corrected pressure

corrected volume

Pideal

Videal

• PidealVeffective = nRT
• Van der Waals (real gases)
• P = pressure of gas (atm)
• V = volume of gas (L)
• n = #moles of gas (mol)
• T = absolute temperature (K)
• R = gas constant, 0.0821 L-atm/mol-K
• a = constant, different for each gas, that takes into account the attractive forces between molecules
• b = constant, different for each gas, that takes into account the volume of each molecule

Using the ideal gas law

Using van der Waal’s equation

PV = nRT

(X)(0.2000L) = (0.3000 mol)().08206 L atm/K mol)(248) = 30.55 atm

From table 5.3, a = 0.0341 atm L2/mol2 and b = 0.0237 L mol

(P + 0.0341 x 0.30002/0.20002)(0.2000 – 0.3000 x 0.0237) = 0.3000(0.08206)(248)= 31.60 atm

• A gas (H2, O2) which is not soluble (or only slightly soluble) in water can be collected over water
• Pneumatic trough w/bottle full of water submerged
• Gas from reaction bubbles into bottle, displacing water
• Pressure of gas inside collecting bottle determined to solve ideal gas equation
• When gases are collected over water, there is some water vapor collected
• Pressure of water vapor depends on T only and is obtained from reference table
• Pressure of gas that was generated iscalculated from Dalton’s law of partial pressure Pgas = Patm - Pwater

Read 5.6-5.9, pp. 212-229

Q pp. 235-236, #74, 76, 78, 80, 82, 84, 86, 93

Do one additional exercise and one challenge problem.

Submit quizzes by email to me:

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