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 form 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.

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)

1 atm = 760 mmHg = 760 torr1 atm = 760 mmHg = 760 torr

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

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 = nRTPV = nRTP = pressure (atm) V = volume (L)

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

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

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 gases in a rigid Helium filled ball at 25oC exerts a pressure of 4.2 atm. If the

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

First make a list of the measurements made:

P1=4.2 atm T1 = 25 oCc + 273.15 = 298.15

P2 = 1/8 P1 T2 = ?

From the variables, choose the appropriate equation, in this case Guy Lussac’s Law: P1/T1=P2/T2

(P1) / (298 K) = (1/8 P1) / T2

Solve for T2:

T2 = [(298 K) (1/8 P1)] / (P1) = 298 / 8 = 37.3 K or -235 oC

3. A balloon containing 6.50 moles of NH3 has a volume of 550L at a certain temperature and pressure. How many grams of NH3 would have to be removed from the balloon in order for the volume to decrease to 250 L under the same conditions?

Mass of NHMass of NH33 removed = 60.3 g of NH removed = 60.3 g of NH33..

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

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!!!

At STP, gas molecules are so far apart that for 1 mole of gas, the overall volume does not change.

STP : P = 1 atm & T = 273 K

GAS WEIGHT MOLARVOLUME

H2 2.0 g/mol 22.4 L/mol

N2 28.0 g/mol 22.4 L/mol

Xe 131.3 g/mol 22.4 L/mol

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 STP.

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

IDEAL GAS LAWQ. What is the pressure inside a gas balloon if it

filled with 852 g of Xe gas at 25.0oC and occupies a volume of 7.00 L?

P = ? 852 g Xe ( 1 mol / 131 g) = 6.50 mol

V = 7.00 L T = 25oC + 273 = 298 K

P = nRT V

P = (6.50 mol) (0.082 L-atm / mol-K) (298 K) 7.00 L

P = 22.7 atm

IDEAL GAS LAWWhat would be the temperature of 100 g

of Ar gas contained in a 500 L sealed container at 0.8976 atm.

How much would a balloon weigh if it contained 40.0 L of O2 gas at 987 mmHg and 45.3 oC?

TT22 = 1914 = 1914 ooCC

mass Omass O22 = 63.7 g = 63.7 g

DENSITY OF A GASThe density of a gas at STP can be

calculated by dSTP = molar mass/molar volume

Calculate the density of hydrogen sulfite gas at STP.

Identify an unknown homonuclear diatomic gas that was found to have a density of 3.165 g/L at STP.

d d (STP)(STP) = (82 g/mol) / 22.4 L/mol) = 3.66 g/L = (82 g/mol) / 22.4 L/mol) = 3.66 g/L

ClCl22

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.

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

STOICHIOMETRY & THE GAS STOICHIOMETRY & THE GAS LAWSLAWS

1.. Write a balanced chemical equation Write a balanced chemical equation

2. Convert to moles (if gas, use PV=nRT or Molar Volume)2. Convert to moles (if gas, use PV=nRT or Molar Volume)

3. Use the mole ratio to convert from moles of “A” to moles of “B”.3. Use the mole ratio to convert from moles of “A” to moles of “B”.

4. Convert moles of “B” to desired measurement, if a gas use PV=nRT4. Convert moles of “B” to desired measurement, if a gas use PV=nRT ..

1. What volume of O2 is needed to combust 348.0 L of C3H8?

CC33HH88 + 5 O + 5 O22 3 CO 3 CO22 + 4 H + 4 H22OO

Due to Avogadro’s Hypothesis, the moles of a gas are directly Due to Avogadro’s Hypothesis, the moles of a gas are directly related to the volume of a gas therefore it is possible to use the related to the volume of a gas therefore it is possible to use the mole ratio on volumes of gas.mole ratio on volumes of gas.

348.0 L C348.0 L C33HH88 (5 mol O (5 mol O2 2 / 1 mol C/ 1 mol C33HH88) = ) = 1740 L O1740 L O22

STOICHIOMETRY & THE GAS STOICHIOMETRY & THE GAS LAWSLAWS

2. How many grams of CO2 is produced from 348.0 L of C3H8 if the temperature is 40.0oC and the pressure is 654 torr?

CC33HH88 + 5 O + 5 O22 3 CO 3 CO22 + 4 H + 4 H22OO

P = 654 torr (1 atm / 760 torr) = 0.861 atm

T = 40oC + 273 = 313 K

PV / RT = n

= (0.861 atm) (348.0 L) /(0.082 L-atm/mol-K) (313 K)

= 11.67 mol of C3H8

11.67 mol C3H8 (3 mol CO2 / 1 mol C3H8) = 35.02 mol CO2

35.02 mol CO2 (44 g / 1 mol) = 1541 g of CO1541 g of CO22

STOICHIOMETRY & THE GAS STOICHIOMETRY & THE GAS LAWSLAWS

3. In lab, you decomposed potassium chlorate into oxygen and potassium chloride. What volume of O2 at STP can be formed from 3.65 g of potassium chlorate?

2 KClO2 KClO33 3 O 3 O22 + 2 KCl + 2 KCl

3.65 g (1 mol / 122.6g) = 0.02977 mol KClO3.65 g (1 mol / 122.6g) = 0.02977 mol KClO33

0.02977 mol KClO0.02977 mol KClO3 3 (3 mol O(3 mol O22 / 2 mol KClO / 2 mol KClO33) = 0.04466 mol O) = 0.04466 mol O22

0.04466 mol O0.04466 mol O22 ( 22.4 L / 1 mol) = ( 22.4 L / 1 mol) = 1.00 L1.00 L

PRACTICE PROBLEMS # 20PRACTICE PROBLEMS # 201. Both hydrogen and helium have been used as buoyant gas in blimps. If a small leak were to occur, which gas would effuse more rapidly and by what factor?

2. At STP, 560 mL of a gas has a mass of 1.08 g. What is the molecular weight of the gas?

3. 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?

4. A 50.0 L cylinder of nitrogen has a pressure of 17.1 atm at 23oC. What is the mass of nitrogen in the cylinder?

5. When a 2.0 L bottle of concentrated HCl was spilled, 3.0 kg of CaCO3 was required to neutralize the spill.

CaCO3 (s) + 2HCl (aq) CaCl2 (aq) + H2O (l) + CO2 (g)What volume of CO2 gas was released by the neutralization at 735 mmHg and

20 oC?

Hydrogen effuses first by a factor of 1.41

43.2 g/mol43.2 g/mol

87.3 mL87.3 mL

986 g986 g

745 L745 L

Group Study Problem # 20Group Study Problem # 201. Measured at 65 oC and 500.0 torr, the mass of 3.21 L of a gas is 3.5 g. What is the molar mass of this gas

2. A 100.0 mL sample of air is analyzed and found to contain 0.835 g N2, 0.0640 g CO2 and 0.197 g O2 at 35 oC. What is the total pressure of the sample and the partial pressure of each component?

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

4. 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?

5. Hydrogen gas is produced by the complete reaction of 8.34 g of aluminum metal with an excess of gaseous hydrogen sulfate. How many liters of hydrogen will be produced if the temperature is 50.0 oC and the pressure is 0.950 atm?

Group Study ProblemGroup Study ProblemAnswers:Answers:1. 45.9 g/mol1. 45.9 g/mol

2. 2. PPNN22=7.53 atm, P=7.53 atm, POO22=1.55 atm, =1.55 atm, PPCOCO22=0.380 atm; =0.380 atm; PPTT = 9.46 atm = 9.46 atm

3. 5.23 L3. 5.23 L

4. 256 L4. 256 L

5. 12.9 L5. 12.9 L