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gases chemistry i chapter 14 chemistry i honors chapter 13
GASESChemistry I – Chapter 14Chemistry I Honors – Chapter 13

SAVE PAPER AND INK!!! When you print out the notes on PowerPoint, print "Handouts" instead of "Slides" in the print setup. Also, turn off the backgrounds (Tools>Options>Print>UNcheck "Background Printing")!

general properties of gases
General Properties of Gases
  • There is a lot of “free” space in a gas.
  • Gases can be expanded infinitely.
  • Gases fill containers uniformly and completely.
  • Gases diffuse and mix rapidly.
importance of gases
Importance of Gases
  • Airbags fill with N2 gas in an accident.
  • Gas is generated by the decomposition of sodium azide, NaN3.
  • 2 NaN3 ---> 2 Na + 3 N2
slide5
Our Atmosphere: Gravity holds the atmosphere close to the Earth. It is divided into various layers according to differences in temperature:
  • The Troposphere:
    • We live in the Troposphere
    • It contains more than 75% of the atmosphere’s mass
  • The Stratosphere:
    • Most of the clouds and rain are in this layer
    • It extends up to 48km above the earth!
  • The ozone layer is between the Stratosphere and Mesosphere
  • The Mesosphere:
    • This layer extends to 80km above the Earth!
  • The Thermosphere
    • Here the air is very thin
    • Over 99.99% of the atmosphere lies below this layer
    • It is comprised of:
slide8
The Ionosphere:Reflects radiowaves back to Earth so signals can be sent around the world
  • The Exosphere:This layer begins at ~480km above the Earth and fades away into space
slide9
Did you know...
  • About 99% of the atmosphere is made up of oxygen (21%) and nitrogen (78%).
  • The remainder is made up of Ar (argon), CO2and very small amounts of H2, NH3, O3(ozone), CH4, CO, He, Ne, Kr and Xe. also gases such as SO2, NO2 are put into the air from vehicles & industry
  • About 20 % of the Earth's population breathes severely contaminated air, largely CO2 and SO2 resulting from industrial processes. This increases the number of respiratory conditions, especially amongst children and elders. 13 % of the British children experience asthma caused by air contamination
properties of gases
Properties of Gases

Gas properties can be modeled using math. Model depends on—

  • V = volume of the gas (L)
  • T = temperature (K)
    • ALL temperatures in the entire chapter MUST be in Kelvin!!! No Exceptions!
  • n = amount (moles)
  • P = pressure (atmospheres)
temperature
Temperature
  • Average kinetic energy of the particles of a substance
  • Without particles there is no temperature
  • No temperature in a vacuum
  • Must be measured in Kelvin to avoid negative values while doing calculations C° + 273 = K
pressure
PRESSURE:

A FORCE per unit area

  • Measured in pascals (Pa) = 1 N/m2

= 100 g / m2

  • A normal day has a pressure of 100 kPa

= 100 x 1000 Pa x 100 g / m2

1 kPa

= 10 000 kg / m2

Atmospheric pressure varies with altitude

    • the lower the altitude, the longer and heavier is the column of air above an area of the earth.
pressure1
Pressure

Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643)

Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). (Just like a straw in a soft drink)

P of Hg pushing down related to

  • Hg density
  • column height
pressure2
Pressure

Column height measures Pressure of atmosphere

  • * 1 standard atmosphere (atm)

= 760 mm Hg (or torr)

= 29.92 inches Hg

= 14.7 pounds/in2 (psi)

= * 101.3 kPa (SI unit is PASCAL)

= about 34 feet of water!

* Memorize these!

pressure conversions
Pressure Conversions
  • The air pressure in a cave underwater is 2.30 atm, what is the pressure in kPa?

101.3 kPa

B. What is 475 mm Hg expressed in kPa?

101.3 kPa

760 mm Hg

2.30 atm x

= 233 kPa

1 atm

475 mm Hg x

= 63.31 kPa

pressure conversions1
Pressure Conversions
  • The air pressure in a cave underwater is 2.30 atm, what is the pressure in kPa?

101.3 kPa

B. What is 475 mm Hg expressed in kPa?

101.3 kPa

760 mm Hg

2.30 atm x

= 233 kPa

1 atm

475 mm Hg x

= 63.31 kPa

pressure conversions2
Pressure Conversions

A. What is 3.71 atm expressed in kPa?

B. The pressure of a tire is measured as 830 torr.

What is this pressure in atm?

slide19
Crushed Can Demo:

Why did the can implode? What happened to the pressure inside once the temperature decreased?

Lower temperature = slower moving particles

= less collisions with the wall

T P

Did the # of particles/moles change?

Did the air pressure change(# particles colliding)?

Did the volume change?

Only because the pressure outside is much greater then inside – the ouside pressure crushed the can.

http://phet.colorado.edu/en/simulation/gas-properties

T α P

gay lussac s law
Gay-Lussac’s Law

If n and V are constant, then P α T

P and T are directly proportional.

P1 P2

=

T1 T2

  • If temperature goes up, the pressure goes up!

P1=T1

P2 T2

Joseph Louis Gay-Lussac (1778-1850)

Charles’ Law http://www.youtube.com/watch?v=tPH57yp0x1U

gas pressure temperature and kinetic molecular theory
Gas Pressure, Temperature, and Kinetic Molecular Theory

As the temperature increases the particles move faster  increasing # of collisions = ↑ pressure

P proportional to T

a gas has a pressure of 3 0 atm at 127 c what is its pressure at 227 c
A gas has a pressure of 3.0 atm at 127º C. What is its pressure at 227º C?

T1 = 127°C + 273 = 400K

P1 = 3.0 atm

T2 = 327°C + 273 = 600K

P2 = ?

What law applies to this situation?

T and P – Gay-Lussac’s Law:

3.0 atm=P2P2= 3.0 atm x 600 K P2 = 4.50 atm

400K 600K 400 K

P1=P2

T1 T2

charles s law
Charles’s Law

If n and P are constant, then V α T

V and T are directly proportional.

V1 V2

=

T1 T2

  • If temperature goes up, the volume goes up!

V1=T1

V2 T2

Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.

slide24

Charles’s original balloon

Modern long-distance balloon

charles s law1
Charles’s Law

0 K-273 °C

Boyle’s Law – Vacuum Demo - http://www.youtube.com/watch?v=N5xft2fIqQU

a gas has a volume of 3 0 l at 127 c what is its volume at 227 c
A gas has a volume of 3.0 L at 127°C. What is its volume at 227 °C?

T1 = 127°C + 273 = 400K

V1 = 3.0 L

T2 = 227°C + 273 = 500K

V2 = ?

Which law applies to this situation?

T and V - Charles’s Law:

3.0 L = V2V2 = 3.0 L x 500 K V2 = 3.75 L

400 K 500 K 400 K

V1=V2

T1 T2

boyle s law
Boyle’s Law

P α 1/V

This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down.

P1V1 = P2 V2

Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.

slide28

As Pressure Increases

Volume

decreases

boyle s law and kinetic molecular theory
Boyle’s Law and Kinetic Molecular Theory

P proportional to 1/V

Smaller volume means the particles are forced closer together – increasing collisions with each other and with the walls of the container.

boyle s law1
Boyle’s Law

A bicycle pump is a good example of Boyle’s law.

As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.

boyle s law2

Volume

Pressure

Boyle’s Law:

-

PV = k

Temperature is constant

a gas has a volume of 3 0 l at 2 atm what is its volume at 4 atm
A gas has a volume of 3.0 L at 2 atm. What is its volume at 4 atm?

P1 = 2 atm

V1 = 3.0 L

P2 = 4 atm

V2 = ?

Which law applies to this situation?

P and V = Boyle’s Law

2 atm x 3.0 L = 4 atm x V2

V2 = 2 atm x 3.0 L

4 atm

V2 =1.5 L

P1 V1 = P2 V2

combined gas law
Combined Gas Law
  • The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION!

V1 P1 V2 P2

=

T1 T2

combined gas law1
Combined Gas Law

If you only need one of the gas laws, you can cover up the item that is constant and you will get that gas law!

=

P1

V1

P2

Boyle’s Law

Charles’ Law

Gay-Lussac’s Law

V2

T1

T2

combined gas law problem
Combined Gas Law Problem

A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

Given:

P1 = 0.800 atm V1 = 180 mL T1 = 29°C + 273

= 302 K

P2 = 3.20 atm V2= 90 mL T2 = ??

calculation
Calculation

P1 = 0.800 atm V1 = 180 mL T1 = 302 K

P2 = 3.20 atm V2= 90 mL T2 = ??

P1 V1 P2 V2

= P1 V1T2 = P2 V2 T1

T1T2

T2 = P2 V2 T1

P1 V1

T2 = 3.20 atm x 90.0 mL x 302 K 0.800 atm x 180.0 mL

T2 = 604 K - 273 = 331 °C

= 604 K

learning check
Learning Check

A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

one more practice problem
One More Practice Problem

A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

and now we pause for this commercial message from stp
And now, we pause for this commercial message from STP

STP in chemistry stands for Standard Temperature and Pressure

STP allows us to compare amounts of gases between different pressures and temperatures

Standard Pressure = 1 atm (or an equivalent)

Standard Temperature = 0 deg C (273 K)

try this one
Try This One

A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?

ideal gases avogadro s theory and the ideal gas law
Ideal Gases, Avogadro’s Theory and the Ideal Gas Law

What are three things that make the behaviour of a gas ideal?

  • How do they collide?

2) Does their size matter?

3) How do they move?

slide44

What are three things that make the behaviour of a gas ideal?

  • How do they collide?

ELASTICALLY  no attraction,

no energy loss, perfect bounces

2) Does their size matter? NO

too small, spaces between too big

3) How do they move?

straight lines and randomly

avogadro s theory

twice as many molecules

Avogadro’s Theory

Equal volumes of gases at the same

T and P have the same number ofmolecules.

  • Volume is proportional to moles
  • V and n are directly related
  • V α n

twice the volume

ideal gas law
Ideal Gas Law:

Generally speaking, an ideal gas obeys all gas laws perfectly under all conditions:

1) Charles’s Law V α T (V / T = k )

2) Gay-Lussac’s Law P α T (P / T = k )

3) Boyle’s Law V α 1 / p ( V x P = k )

4) Avogadro’s Theory V α n

In other words, V α n α T α 1/p

All these variables can be combined to include an equation with moles, n using a constant… V = constant x n x T x 1/p

The constant is the letter R

ideal gas law1
Ideal Gas Law:

The Constant R:

  • Is called the Ideal GAS CONSTANT
  • Relates P, T, V, and moles
  • Is found by substituting in known values for P, T, V and n (moles) Ex. You could sub in values at STP: T = 273 K, P = 1 atm, 1 molegas= 22.4 L

R with different units : These #’s are always provided .

  • 8.314 J / K • mol or 8.314 J K-1 mol-1
  • 0.0821 L•atm/K•mol mainly used
  • 8.31451 Pa m3 K-1 mol-1
  • 62.364 L Torr K-1 mol-1
ideal gas law2
IDEAL GAS LAW

V= R x n x T x 1/p

Multiply both sides by p, rearrange n…

Brings together gas properties.

Can be derived from experiment and theory.

BE SURE YOU KNOW THIS EQUATION!

P V = n R T

http://jersey.uoregon.edu/vlab/Piston/index.html

using pv nrt
Using PV = nRT

P = Pressure

V = Volume

T = Temperature

N = number of moles

R is the constant called the Ideal Gas Constant

R = 0.0821

L • atm

Mol • K

slide51

REMINDER:

1 standard atmosphere (atm)

1 atm = 760 mm Hg (or torr)

= 101.3 kPa (SI unit is PASCAL)

Ex. 202.6 kPa = 202.6 kPa

101.3 kPa / atm

= 2.000 atm

Ex. 1040 torr = 1040 torr 760 mm Hg / atm

= 1.50 atm

using pv nrt1
Using PV = nRT

How much N2 is required to fill a small room with a volume of 27,000 L to 745 mm Hg at 25 oC?

Solution

1. Given: convert to proper units

V = 27 000 L

T = 25 oC + 273 = 298 K

P = 745 mm Hg (760 mm Hg/atm) = 0.98 atm

And we always know R = 0.0821 L•atm / mol•K

using pv nrt2
Using PV = nRT

How much N2 is required to fill a small room with a volume of 27 000 L to P = 745 mm Hg at 25 oC?

Solution

1. Divide both sides by RT to solve for n. 2. Now plug in those values and solve for the unknown.PV = nRT

n = PV / RT

RT RT

n = 1.1 x 103 mol (or about 30 kg of gas)

learning check1
Learning Check

EX. 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 in atm in the tank in the dentist office?

Given: n=2.86 mol, V=20.0L T=23°C + 273 K P=?PV = nRT P = nRT / V P = 2.86 mol (0.0821 L atm/mol K)(296 K) 20.0L

P = 3.48 atm

using pv nrt when given mass
Using PV = nRT When Given Mass

0.78 grams of hydrogen gas is produced at 22°C and 125 kPa. What volume of hydrogen is expected?

Given: T = 22°C + 273 = 295 K,

P = 125 kPa = 125 kPa/101.3 = 1.23 atm n = find n using molar mass first

n = massPV = nRT

Mm V = nRT/P

n = 0.78g = 0.0386mol(0.0821 L atm/mol K)295K 2.02g/mol 1.23 atm

= 0.0386 mol = 0.76 L

slide56
A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

Step 1) find moles using PV = nRT

n = 0.201 mol

Step 2) find mass using m = n x Mm

m = 40.3 g

deviations from ideal gas law
Deviations from Ideal Gas Law
  • Real molecules havevolume.

The ideal gas consumes the entire amount of available volume. It does not account for the volume of the molecules themselves.

  • There areintermolecular forces.

An ideal gas assumes there are no attractions between molecules. Attractions slow down the molecules and reduce the amount of collisions.

    • Otherwise a gas could not condense to become a liquid.
gases in the air
Gases in the Air

The % of gases in air Partial pressure (STP)

78.08% N2 593.4 mm Hg

20.95% O2 159.2 mm Hg

0.94% Ar 7.1 mm Hg

0.03% CO2 0.2 mm Hg

PAIR = PN + PO + PAr + PCO = 760 mm Hg

2 2 2

Total Pressure 760 mm Hg

dalton s law of partial pressures
Dalton’s Law of Partial Pressures
  • 2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
          • 0.32 atm 0.16 atm

What is the total pressure in the flask?

Ptotal in gas mixture = PA + PB + ...

Therefore,

Ptotal = PH2O + PO2 = 0.48 atm

Dalton’s Law: total P is sum ofPARTIALpressures.

dalton s law
Dalton’s Law

John Dalton

1766-1844

health note
Health Note

When a scuba diver is several hundred feet under water, the high pressures cause N2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O2 in scuba tanks used for deep descents.

collecting a gas over water
Collecting a gas “over water”
  • Gases, since they mix with other gases readily, must be collected in an environment where mixing can not occur. The easiest way to do this is under water because water displaces the air. So when a gas is collected “over water”, that means the container is filled with water and the gas is bubbled through the water into the container. Thus, the pressure inside the container is from the gas AND the water vapor. This is where Dalton’s Law of Partial Pressures becomes useful.
solve this
Solve This!

A student collects some hydrogen gas over water at 20 degrees C and 768 torr. What is the pressure of the H2 gas?

768 torr – 17.5 torr = 750.5 torr

gas density

Low

density

High

density

GAS DENSITY

22.4 L of ANY gas AT STP = 1 mole

gases and stoichiometry
Gases and Stoichiometry

2 H2O2 (l) ---> 2 H2O (g) + O2 (g)

Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP?

Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

gases and stoichiometry1
Gases and Stoichiometry

2 H2O2 (l) ---> 2 H2O (g) + O2 (g)

Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP?

Solution

1.1 gH2O2 1 mol H2O2 1 mol O2 22.4 L O2

34 g H2O2 2 mol H2O2 1 mol O2

= 0.36 L O2 at STP

gas stoichiometry practice
Gas Stoichiometry: Practice!

A. What is the volume at STP of 4.00 g of CH4?

B. How many grams of He are present in 8.0 L of gas at STP?

what if it s not at stp
What if it’s NOT at STP?
  • 1. Do the problem like it was at STP. (V1)
  • 2. Convert from STP (V1, P1, T1) to the stated conditions (P2, T2)
try this one1
Try this one!

How many L of O2 are needed to react 28.0 g NH3 at24°C and 0.950 atm?

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

gas diffusion and effusion
diffusionis the gradual mixing of molecules of different gases.

effusionis the movement of molecules through a small hole into an empty container.

HONORS only

GAS DIFFUSION AND EFFUSION
gas diffusion and effusion1

HONORS only

GAS DIFFUSION AND EFFUSION

Graham’s law governs effusion and diffusion of gas molecules.

Rate of effusion is inversely proportional to its molar mass.

Thomas Graham, 1805-1869. Professor in Glasgow and London.

gas diffusion and effusion2

HONORS only

GAS DIFFUSION AND EFFUSION

Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is

  • proportional to T
  • inversely proportional to M.

Therefore, He effuses more rapidly than O2 at same T.

He

gas diffusion relation of mass to rate of diffusion

HONORS only

Gas Diffusionrelation of mass to rate of diffusion
  • HCl and NH3 diffuse from opposite ends of tube.
  • Gases meet to form NH4Cl
  • HCl heavier than NH3
  • Therefore, NH4Cl forms closer to HCl end of tube.