1 GASES Paul Gilletti, Ph.D. Mesa Community College.

1

GASES

Paul Gilletti, Ph.D.

Mesa Community College

2

Gases (Vapors)

Gases expand to fill any container.

Therefore, gases are highlycompressible.

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Kinetic Molecular Theory (of an Ideal Gas):

1. Gases are composed of molecules or atoms whose size is negligible compared to the average distance between them. (Most of the space in the gas container is empty.)2. Gas molecules move randomly in straight lines in all directions at various speeds.

3. The forces of attraction or repulsion between gas molecules are very weak or negligible (except during collisions)

4. Collisions between gas molecules are considered elastic.

5. The average kinetic energy of a molecule is proportional to the absolute temperature.

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Pressure and Volume: Boyle’s Law

How is the pressure applied to a gas related to its volume?

Piston

Gas molecules

Let’s apply pressure

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Pressure and Volume: Boyle’s Law

How is the pressure applied to a gas related to its volume?

Piston

Gas molecules

Piston

Gas molecules

Boyle’s Law: P1V1 = P2V2

Volume is inversely proportional to applied pressure.

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The Harder we Pushthe smaller the gas

volume gets!

Boyle’s Law: P1V1 = P2V2

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molecules of air

1

2

3

Where is the pressure the greatest?

We live in “sea of air”

Why does a diver get the bends?

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Pressure: force per unit area of surface

Units

lbs per in2 (psi)

mm of Hg (torr)

atmospheres (atm)

1 atm = 760 mm of Hg =760 torr = 14.70 psi = 101.325 kPa

Pascal (Pa)

Pairs of these can be used as conversion factors.

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Temperature and Volume: Charles’s Law

How is the volume of a gas related to its temperature?

gas molecules

moveable mass(constant pressure)

What happens if heat is applied to the gas?

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gas molecules


Why did the volume change?

What happens to the average speed of the gas molecules?

.

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gas molecules


The volume of a gas is directly proportional to its Temperature (temperature must be in Kelvin)

Charles’s Law: V1/T1 = V2/T2

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Combined Gas Law (Boyle and Charles):

T

VP

T

VP

2

22

1

11 T must be in Kelvin

Can be rearranged to:

P1V1T2 = P2V2T1

A combined gas law problem can be recognized by having two sets of conditions.

Note: if one set of parameters is unchanged that termwill cancel on each side.

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A balloon contains helium gas with a volume of 2.60 L at 25 oC and 768 mmHg. If the balloon ascends to an altitude where the helium pressure is 590 mmHg and the temperature is 15 oC, what is the volume of the balloon?

What type of problemis this?

There are 2 sets of conditions.

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A balloon contains helium gas with a volume of 2.60 L at 25 oC and 768 mmHg. If the balloon ascends to an altitude where the helium pressure is 590 mmHg and the temperature is 15 oC, what is the volume of the balloon?

P1V1T2 = P2V2T1

P1=V1=T1=P2=V2=T2=

768 torr2.60 L

25 + 273 = 298 K

590 torr

15 + 273 = 288 K?

12

2112 TP

TVP V

= (768 torr)(2.60 L)(288 K) (590 torr)(298 K)

= 3.27 L

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Ideal Gases and the Ideal Gas Law:

PV = nRT

Temperature in K*gas constant 0.0821 L•atm = 62.37 L•torr mol•K mol•Kmoles of gasvolume in Lpressure in units to match *R units

Note: there is only one set of conditions.

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Avogadro’s Law:

Equal volumes of any two gases (ideal) at the same temperature and pressure contain the same number of molecules (they also occupy equal volumes).

STPPressure 1 atm (760 mm Hg)

Temperature 0oC (273 K)

Standard

At STP one mole of ideal gas occupies 22.4 L

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A 12.25 L cylinder contains 75.5 g of neon at 24.5 oC. Determine the pressure in the cylinder.

What type of problemis this?

Only one set of conditions

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A 12.25 L cylinder contains 75.5 g of neon at 24.5 oC. Determine the pressure in the cylinder.

PV = nRT

P =V =

n =

R =

T =

?

12.25 L

75.5 g = mol20.18 g

mol 3.74

62.4 L•torr mol•K24.5 + 273 = 297.5 K

P = nRT V

= (3.74 mol)(62.4L•torr)(297.5K) (12.25 L) mol•K

= 5667.7 torr

= 5670 torr

How many atmospheres is this?

atm 7.46 torr 760

atm 1torr 5667.7

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What is the density of carbon dioxide gas at 25 oC and 725 mmHg pressure?

Density = g/L = g L

so if we can find g and L, division will work!

P =V =

n =

R =

T =

725mmHg

62.4 L• torr mol•K

25 + 273 = 298 K

What do wedo now?

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Density = g/L = g L

so if we can find g and L division will work!

P =V =

n =

R =

T =

725mmHg

62.4 L•torr mol•K

25 + 273 = 298 K

Two variables! Let’s pick an amount for one and calculate the other!

Let’s choose 1 mol of CO2 and find the number of Liters.

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Density = g/L = g L

so if we can find g and L division will work!

P =V =

n =

R =

T =

725mmHg


25 + 273 = 298 K

1.0 mol (44.0 g)

V = nRT P

= (1 mol) (62.4 L•torr) (298 K) ( mol•K ) (725 torr)

= 25.6 L

NOW: 1.72___________ = g L

44.0 g

25.6 L

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A 2.50 gram sample of a solid was vaporized in a 505 mL vessel. If the vapor pressure of the solid was 755 mmHg at 155 oC, what is the molecular weight of the solid?

molecular weight ~ molar mass = g/mol = g mol

..so if we can find grams and moles and divide....

P =V =

n =

R =

T =

755 torr0.505 L

...we already have grams!! We’re halfway there!


155 + 273 = 428 K

n = PV RT

= 755 torr | 0.505 L | mol•K_____|______ | 62.4 L•torr | 428 K

= 0.01428 mol

NOW: 2.50 g = g 0.01428 mol mol

175.1

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So Density is g/L (g ÷ L) and

molar mass is g/mol (g ÷ mol).

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Dalton’s Law of Partial Pressures:

Ptotal = P1 + P2 + P3 +...

He H2 N2

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Dalton’s Law of Partial Pressures:

Ptotal = P1 + P2 + P3 +...

Since they are considered to be ideal gases, the attractions and repulsions between molecules are ignored. ... and...

PV=nRT

so: PV = (n1 + n2 + n3)RT

or:

V

RTn

V

RTn

V

RTnP 321total

We also refer to mole fractions: /VRTn

/VRTn

P

P

n

n

total

1

total

1

total

1

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To find the gas pressure, the pressure of the watervapor must be subtracted from the total pressure.

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A 250.0 mL flask contains 1.00 mg of He and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres.The total pressure is due to the partial pressures of each of these gases.

so:

V

RT)nn(PPP HHeHHetotal 22

For He:

_____________________ = mol He1.00 x 10-3 g He4.00 g

mol 2.50 x 10-4

For H2:

______________________ = mol H22.00 x 10-3 g H2

2.016 gmol 9.92 x 10-4

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A 250.0 mL flask contains 1.00 mg of He and and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres.

so:

V


For He: _____________________ = mol He1.00 x 10-3 g He4.00 g

mol 2.50 x 10-4

For H2: ______________________ = mol H22.00 x 10-3 g H2

2.016 gmol 9.92 x 10-4

And: Ptotal = (2.50 x 10-4 + 9.92 x 10-4)(RT/V)

= (0.001242 mol)(0.0821 L•atm)(25 + 273)K mol•K (0.2500 L)

Ptotal= 0.1216 atm

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A 250.0 mL flask contains 1.00 mg of He and and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres.

so:

V


For He: _____________________ = mol He1.00 x 10-3 g He4.00 g

mol 2.50 x 10-4

For H2: ______________________ = mol H22.00 x 10-3 g H2

2.016 gmol 9.92 x 10-4

Calculate the pressure due just to He (you have 37 seconds):

V

RTn P He

He = 0.0245 atm

and Phydrogen= ? 0.1216 - 0.0245 = 0.0971 atm

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Magnesium is an active metal that replaces hydrogen from an acid by the following reaction:

Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

How many g of Mg are needed to produce 5.0 L of H2 at a temperature of 25 oC and a pressure of 745 mmHg?

Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) 5.0 L? g

Hint: find moles of H2 using PV = nRT then work as a stoichiometry problem.

n = PV RT

n = 0.20 mol

=____________________________________745 mmHg 5.0 L62.4 L•mmHg

mol•K298 K

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Magnesium is an active metal that replaces hydrogen from an acid by the following reaction:

Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

How many g of Mg are needed to produce 5.0 L of H2 at a temperature of 25 oC and a pressure of 745 mmHg?

Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) 5.0 L? g

0.20 mol

____________________________________ = g Mg0.20 mol H2

1 mol H2

1 mol Mgmol Mg

24.3 g Mg 4.87

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Molecular Speeds:

K.E. = ½ mv2

Average kinetic energy of a gas molecule: = ½ m2

Where = the rms (root-mean-square) speed of the molecules at each temperature.

From kinetic-molecular theory: At any given temperature the molecules of all gases have the same average kinetic energy.

Which molecules travelfaster, big or little?

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At room temperature, the average speed of an N2 molecule is ........ 1150 mi/hr

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Molecular diffusion and effusion:

Diffusion: “gas molecules spreading out to fill a room are diffusing.”

Its not easy since an average gasmolecule at room temperature and

pressure will experience about10 billion collisions per second!

It only travels about 60 nm between collisions!

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Effusion: “A Helium filled balloon loses He by effusion.”

Small hole or poreescaping molecule

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Which molecules will effuse faster from this semiporouscontainer?

Graham’s Law of effusion: effusion rate is inversely proportional to the square root of its molar mass.

For 2 gases:

1

2

2

1

2

1

r

r

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1

2

2

1

2

1

r

r

r = rate of effusion

u = root mean speed (~average speed) of molecules

M = molar mass

Compare the rates of effusion of He and N2.

652004

028

2

..

.

r

r

N

He He effuses 2.65 times as fastas N2.

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N2

He

Which balloon will lose pressure sooner?

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N2 He

Which balloon will lose pressure sooner?

Big molecules Little molecules(escape more easily)

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Real Gases: When do gases become non-ideal?

Temperature: low

Pressure: high

As they approach theliquid state, attractions

between moleculesincrease and theybecome less ideal.

van der Waal’s equation is one equation used to treatnon-ideal gases.

nRTnb -VV

n P

2

2

a

a and b are constants found in tables for each gas.

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CO boiling Pt. 81K

SiH4 boiling Pt. 161K

PH3 boiling Pt. 188K

C3H8 boiling Pt. 231K

Which gas would deviate the most from the ideal gas law at room temperature (25oC)?

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100K

200K

300K

CO(l) boiling Pt. 81K

SiH4(l) boiling Pt. 161K

PH3(l) boiling Pt. 188K

C3H8(l) boiling Pt. 231

298K

Which gas would deviate the most from the ideal gas law at room temperature (25oC)?

1 GASES Paul Gilletti, Ph.D. Mesa Community College.

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