Seeing the Unseen• Air is invisible, but we

know it exists:

– Winds blow.

– Flags fly*.

– We breathe.

*Flags can’t fly on the moon because there is no atmosphere – no wind.

Air is a Mixture• Clean air is a

mixture of several pure substances.

• The table shows the composition at sea level for dry air.

• Normal air also contains water vapor - the cause of humidity.

Substance Percentage

Nitrogen (N2) 78.10

Oxygen (O2) 20.90

Argon (Ar) 0.93

Carbon dioxide (CO2)

0.04

Trace gases <0.03

Molecular Interactions in Gases are Negligible

• Gases are mostly empty space: molecules occupy <0.1 % volume.

• 1,000 times less dense than solids and liquids.

• Emptiness allows complete mixing.

• All gases “dissolve” other gases.

Keeping the Genie in the Bottle

• Solids and liquids do not need to be confined in a sealed container.

• Without a sealed container, unconfined gas molecules disperse quickly.

• Gas molecules move quickly – but not all at the same speed.

Going Nowhere Fast

• Molecules move quickly, but take a long time to get anywhere because of collisions with each other.

• Without collisions, a molecule would cross a room in a fraction of a second.

• Collisions can make the journey last several minutes.

KKinetic inetic MMolecular olecular TTheoryheory• Matter is composed of tiny particles (atoms, molecules or ions) with

definite and characteristic sizes that never change.

• The particles are in constant random motion, that is they possess kinetic energy. Ek = 1/2 mv2

• The particles interact with each other through attractive and repulsive forces (electrostatic interactions), that is the possess potential energy. U = mgh

• The velocity of the particles increases as the temperature is increased therefore the average kinetic energy of all the particles in a system depends on the temperature.

• The particles in a system transfer energy from one to another during collisions yet no net energy is lost from the system. The energy of the system is conserved but the energy of the individual particles is continually changing.

Properties of Gases DIFFUSION

Diffusion is the ability of two or more gases to mix spontaneously until a uniform mixture is formed.

Example: A person wearing a lot of perfume walks into an enclosed room, eventually in time, the entire room will smell like the perfume.

EFFUSIONEffusion is the ability of gas particles to pass through a small opening or membrane from a container of higher pressure to a container of lower pressure. The General Rule is: The lighter the gas, the faster it The lighter the gas, the faster it moves.moves.Graham’s Law of Effusion:

Rate of effusion of gas A = √(molar mass B / molar mass A)

Rate of effusion of gas B

The rate of effusion of a gas is inversely proportional to the square root of the molar mass of that gas.

PRESSUREPRESSURE• A physical property of matter that describes

the force particles have on a surface. Pressure is the force per unit area, P = F/A

• Pressure can be measured in:

• atmosphere (atm) atmosphere (atm)

• millimeters of mercury (mmHg)millimeters of mercury (mmHg)

• (torr) after Torricelli, the inventor of the mercury (torr) after Torricelli, the inventor of the mercury barometer (1643) barometer (1643)

• pounds per square inch (psi)pounds per square inch (psi)

1 atm = 760 mmHg = 760 torr = 14.69 psi1 atm = 760 mmHg = 760 torr = 14.69 psi

Got Me Under Pressure

• Gases exert pressure by virtue of molecular collisions with the container surface.

• Gravity makes the air density higher near the earth’s surface.

• Pressure decreases with elevation – air density decreases.

DALTON’S LAW OF PARTIAL DALTON’S LAW OF PARTIAL PRESSURESPRESSURES

If there is more than one gas present in a If there is more than one gas present in a container, each gas contributes to the total container, each gas contributes to the total pressure of the mixture.pressure of the mixture.

PPtotaltotal = P = Pgasgas AA + P + P gasgas BB + P + Pgasgas CC … …If the total pressure of a system was 2.5 atm,

what is the partial pressure of carbon monoxide if the gas mixture also contained 0.4 atm O2 and 1.48 atm of N2?

PPTT - P - PO2O2 - P - PN2N2 = P = PCO CO 2.5 atm - 0.4 atm - 1.48 atm = 2.5 atm - 0.4 atm - 1.48 atm = 0.62 atm0.62 atm

TEMPERATURETEMPERATURE• A physical property of matter that

determines the direction of heat flow.

• Measured on three scales.

• Fahrenheit Fahrenheit ooF Celsius F Celsius ooCC

• Kelvin KKelvin K

• ooF = (1.8 F = (1.8 ooC) + 32 C) + 32 ooC = (C = (ooF - 32)/1.8F - 32)/1.8

• K = K = ooC + 273.15C + 273.15

Standard Temperature and Pressure (STP)

• Standard conditions:– Standard temperature

is 273 K (0ºC).

– Standard pressure is 760 mm Hg.

• At STP, 1 mole of any ideal gas occupies 22.414 L.

Avogadro’s Hypothesis

• Avogadro pictured the moving molecule as occupying a small portion of the larger space apparently occupied by the gas. Thus the “volume” of the gas is related to the spacing between particles and not to the particle size itself.

• Imagine 3 balloons each filled with a different gas (He, Ar, & Xe). These gases are listed in increasing particle size, with Xe being the largest atom. According to Avogadro’s Hypothesis, the balloon filled with one mole of He will occupy that same volume as a balloon filled with one mole of Xe.

• So for a gas, the “volume” and the moles are directly related. V V n n

Practice Problem on Avogadro’s Hypothesis

• A sample of N2 gas at 3.0 atm and 20.0oC is known to occupy a volume of 1.43 L. What volume would a 0.179 mole sample of NH3 gas occupy at the same pressure and temperature?

First calculate the number of moles of nitrogen gas: PV = nRT where P = 3.0 atm, V = 1.43 L, R = 0.082 L-atm/mol-K, and T = 20.0 oC + 273 = 293Kn = PV / RT = (3.0 atm x 1.43L) / (0.082 L-atm/mol-K x 293K) = 0.179 moles of N2

So since the moles of N2 is 0.179 mol and the moles of ammonia is 0.179 mol according to Avogadro’s hypothesis the volume of NH3 at that pressure and that temperature is 1.43 L, the same!!!

EMPIRICAL GAS LAWS

Boyle’s Law P1V1 = P2V2

Charles’ Law V1 / T1 = V2 / T2

Guy-Lussac’s Law P1 / T1 = P2 / T2

Avogadro’s Law V1 / n1 = V2 / n2

Combined Gas Law P1V1 / T1 = P2 V2 / T2

Ideal Gas LawIdeal Gas Law PV = nRT PV = nRTP = pressure (atm) V = volume (L)

n = chemical amount (mol) T = Temperature (K)

R = ideal gas constant = 0.08206 L-atm / mol-K

Lecture PROBLEMSLecture PROBLEMS1. A sample of O2 gas initially at 0oC and 1.0 atm is transferred from a 2-L container to a 1-L container at constant temperature. a) What effect does this change have on the average kinetic energy of the gas molecules? b) What effect does the total number of collisions of O2 molecules with the container walls in a unit time?

2. At constant pressure, a student needed to decrease a volume of 155 mL of Ne gas by 32.0%. To what temperature, (in oC), must the gas be cooled if the initial temperature was 21oC?3. A sample of CO2 gas has a volume of 125.0 L at a pressure of 789 torr and a temperature of 30oC. What will be the temperature if the pressure was increased to 900 torr & the volume decreased to 95.0 L?4. 4. F2 gas, which is dangerously reactive, is shipped in steel containers of 30.0 L capacity, at a pressure of 10.0 atm at 26.0

oC. What should be the volume of the tank if the pressure is increased to 820.0 torr & the temperature is 43.0 oC?

Empirical Gas LawsEmpirical Gas Laws1. At 25oC, a sample of N2 gas under a pressure of 689 mmHg occupies 124

mL in a piston-cylinder arrangement before compression. If the gas is compressed to 75% of its original volume, what must be the new pressure (in atm) at 25oC?

First make a list of the measurements made:

P1=689 mmHg V1 = 124 mL

P2 = ? V2 = 75% V1

From the variables, choose the appropriate equation, in this case Boyle’s Law: P1V1=P2V2

(689 mmHg) (124 mL) = P2 (0.75 x 124 mL)

Solve for P2:

P2 = (689mmHg) (124 mL) / (93 mL) = 919 mmHg

Now convert to atm:

919 mmHg (1 atm / 760 mmHg) = 1.21 atm1.21 atm

Empirical Gas LawsEmpirical Gas Laws2. The gas in a Helium filled ball at 25oC exerts a volume of 4.2 L. If the ball is

placed in a freezer and the volume decreases to 1/8 of its original value, what is the temperature inside the ball?

First make a list of the measurements made:

V1=4.2 atm T1 = 25 oCc + 273.15 = 298.15

V2 = 1/8 P1 T2 = ?

From the variables, choose the appropriate equation, in this case Charles’ Law: V1/T1=V2/T2

(V1) / (298 K) = (1/8 V1) / T2

Solve for T2:

T2 = [(298 K) (1/8 V1)] / (V1) = 298 / 8 = 37.3 K or -235 oC

3. A balloon containing 6.50 grams of NH3 has a volume of 10.30 L at a temperature of 20.0oC and a pressure of 689.2 torr. What would be the pressure of NH3 if the volume decreased to 2.50 L without a change in temperature?

Pressure of NHPressure of NH33 = 2.84 x 10 = 2.84 x 1033 torr. torr.

COMBINED GAS LAW• A gas occupies a volume of 720 mL at 37oC and

640 mmHg pressure. Calculate the volume the gas would occupy at 273 K and 1 atm.

P1V1 / T1 = P2V2 / T2

rearranged to solve for V2 is:

V2 = P1 V1 T2 / P2 T1

V2 = (640 mmHg)(720 mL) (273 K) / (760 mmHg) (310 K)

VV22 = 534 mL = 534 mL

COMBINED GAS LAW• A gas occupies a volume of 720 mL at

37oC and 640 mmHg pressure. – Calculate the pressure if the temperature

is increased to 1000oC & the volume expands to 900 mL.

– Calculate the temperature if the pressure is decreased to 10 torr & the volume is reduced to 500 mL.

PP22 = 2.1 x 10 = 2.1 x 1033 mmHg mmHg

TT22 = 3.4 K or -270 = 3.4 K or -270 ooCC

PRACTICE PROBLEM # 20aPRACTICE PROBLEM # 20a1. You prepared carbon dioxide by adding aqueous HCl to marble chips, calcium carbonate. According to your calculations, you should obtain 79.4 mL of carbon dioxide at 0 oC and 760 mmHg. How many milliliters of gas would you obtain at 27oC at the same pressure?

2. Divers working from a North Sea drilling platform experiences pressures of 50 atm at a depth of 5.0 x 102 m. If a balloon is inflated to a volume of 5.0 L (the volume of a lung) at that depth at a water temperature of 4.0oC, what would the volume of the balloon be on the surface (1.0 atm) at a temperature of 11 oC?

3. What volume would 5.30 L of H2 gas at 0 oC and 760 mmHg occupy if the temperature was increased to 70oF and the pressure to 830 torr?

4. The pressure gauge reads 125 psi on a 0.140-m3 compressed air tank when the gas is at 33.0 oC. To what volume will the contents of the tank expand if they are released to an atmospheric pressure of 751 torr and a temperature of 13oC?

5. A gas has a volume of 397.0 mL at 14.70 atm. What will be its pressure (in torr) if the volume is changed to 4.100 L?

87.3 mL87.3 mL

256 L256 L

5.23 L5.23 L

1.126 m1.126 m33

1082 torr1082 torr

PRACTICE PROBLEM # 20aPRACTICE PROBLEM # 20a6. Which of the following statements is false?a) If the Celsius temperature is doubled, the pressure of a fixed volume of gas would

double.

b) All collisions between gas molecules are perfectly elastic (no energy is lost) according to KMT.

c) The volume of gas is inversely proportional to the temperature of gas present (P constant)

d) Gases are capable of being greatly compressed.

7. Which of the following statements are true?a) In a large container of O2 gas the pressure exerted by the oxygen will be greater at the

bottom of the container.

b) Of the three states of matter, gases are the most compact and the most mobile.

c) The formula of ozone is 3 O2.

d) Molecules of O2 gas and H2 gas at the same temperature will have the same average kinetic energies and the same average velocities.

CC

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