chapter 5 gases reduced1

Chapter 5- GasesBy

Dr. Mohammad Tariq Saeed

5.1: Existence of gases5.2: Pressure of a gas5.3: The Gas Laws5.4: The Ideal Gas Equation5.5: Gas Stoichiometry5.6: Dalton’s Law of Partial pressure

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5.1 Elements that exist as gases at 250C and 1 atmosphere, Ionic compounds are not, strong attraction

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• Molecular compounds like CO, CO2, HCl, NH3, and CH4 are gases but majority of molecular compounds are liquids or solids at room temp.

• Gases assume the volume and shape of their containers.• Gases are the most compressible state of matter.• Gases will mix evenly and completely when confined to the

same container.• Gases have much lower densities than liquids and solids.

5.2: Physical Characteristics of Gases

NO2 gas

Units of Pressure

Pressure = ForceArea

(force = mass x acceleration)1 pascal (Pa) = 1 N/m2 (SI unit of pressure)

1 atm = 760 mmHg = 760 torr ( Std. Atm pressure

1 atm = 101,325 Pa, = 1.01325 x 102 kPa

1kPa = 1000 Pa

1N = 1kg / m2

Example 5.1The pressure outside a jet plane flying at high altitude falls considerably below standard atmospheric pressure. Therefore, the air inside the cabin must be pressurized to protect the passengers. What is the pressure in atmospheres in the cabin if the barometer reading is 688 mmHg?

Sea level 1 atm

4 miles 0.5 atm

10 miles 0.2 atm

The atmospheric pressure in San Francisco on a certain day was 732 mmHg. What was the pressure in kPa?, 1 atm = 1.0135 x 102 kPa, = 760 mmHg

Example 5.2

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Manometers Used to Measure Gas Pressures other than atmosphereclosed-tube open-tube

Apparatus for Studying the Relationship BetweenPressure and Volume of a Gas

V decreasesAs P (h) increases

Use to measure pressure equal or greater than atmospheric press

Use to measure pressure below atmospheric press

Patm = Ph

Patm = Ph + Patm

Gas Pressure (P gas) less than atmospheric pressure

P gas = P atm - h

Gas Pressure (Pgas) greater than atmospheric pressure

P gas = P atm + h

Gas pressure and Atmospheric pressure

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5.3: Gas Laws:

Pressure and volume are inversely related at constant temperature and moles PV= kAs one goes up, the other goes down

Boyle noted that , if pressure (P) increased at const. temperature, the volume(V) decrease ,

1. Boyle’s Law: Pressure – volume relationship

aP V1

PinitialVinitial=k= PfinalVfinalP = (nRT) x

nRT = constant

P = k x

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Cont’d

Constant temperatureConstant amount of gas

Applications of Boyle’s Law:action of a syringe.action of the diaphragm of our body.deep sea fish die when brought to the surface.

Graphic Representation of Boyle,s law

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Summary of Gas Laws

Boyle’s Law

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As T increases V increases

Variation in Gas Volume with Temperature at Constant Pressure

V T

Charles’ Law: Temperature – Volume relationship

Volume is directly proportional to the temperature,If the volume increase then temperature increase and vice versa.

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Variation of Gas Volume with Temperatureat Constant Pressure, “Charles’s & Gay-Lussac’s Law”

Volume of a gas expand when the gas heated and contract when cooled “V a T”

V = constant x TV1/T1 = V2 /T2

T (K) = t (0C) + 273.15

Temperature must bein Kelvin

Example: Study of temperature-Volume relationship at various pressure. At any given pressure, plot of the volume versus temperature yields a straight line. By extending the line to zero volume, , we find the intercept on the temperature exist to be – 273.15 oC.

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Charles’s Law

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Avogadro’s Law: [Volume – Amount relationship]Avogadro’s complimented the studies of Boyle, Charles, and Gay-Lussac.: He published a hypothesis stating that at the same temperature and pressure , equal volume of all gases contain the same number of molecules (or atoms).

V a number of moles (n)

V = k x n V1 / n1 = V2 / n2

Constant temperature &Constant pressure

n represent the no. of moles and k is the proportionality constant.

3H2 (g) + N2 (g) 2NH3 (g). 3mol 1 mol 2 mol

3H2 (g) + N2 (g) 2NH3 (g). 3 volume 1volume 2 volume

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Avogadro’s Law

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5.4: Ideal Gas Equation

Charles’s law: V a T (at constant n and P)

Avogadro’s law: V a n (at constant P and T)

Boyle’s law: P a (at constant n and T)1V

V a nTP

V = constant x = R nTP

nTP

R is the gas constant

PV = nRTThe molecules of an ideal gas do not attract or repel one another, and their volume is negligible compared with the volume of the container.

Ideal gas equation describe the relationship among the four variables P, V,T, and n. An ideal gas is a hypothetical gas whose pressure–volume –temperature behavior can be completely accounted for by an ideal gas equation.

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The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).

PV = nRT

R = PVnT =

(1 atm)(22.414L)(1 mol)(273.15 K)

R = 0.082057 L • atm / (mol • K)

Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

5.3

Sulfur hexafluoride (SF6) is a colorless and odorless gas. Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5°C.(change to K) R = 0.0821L.atm/K.molP = nRT / V

5.4Calculate the volume (in L) occupied by 7.40 g of NH3 at STP.g mole L Std V= 22.41L

An inflated helium balloon with a volume of 0.55 L at sea level (1.0 atm) is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon? P1V1 = P2 V2

5.5

5.6

Argon is an inert gas used in light bulbs to retard the vaporization of the tungsten filament. A certain light bulb containing argon at 1.20 atm and 18°C is heated to 85°C at constant volume. P1/T1 = P2 / T2 Calculate its final pressure (in atm).

Electric light bulbs are usually filled with argon.

A small bubble rises from the bottom of a lake, where the temperature and pressure are 8°C and 6.4 atm, to the water’s surface, where the temperature is 25°C and the pressure is 1.0 atm. P1V1 / T1 = P2V2 / T2

Calculate the final volume (in mL) of the bubble if its initial volume was 2.1 mL.

5.7

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Density (d) Calculations [n /V = P / RT], n = m/MM

mMVPRT=

m is the mass of the gas in gM is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance

dRTPM = d is the density of the gas in g/L

d = mV = PM

RT

5.8Calculate the density of carbon dioxide (CO2) in grams per liter (g/L) at 0.990 atm and 55°C. d = PM / RT

5.5: Gas Stoichiometry: Use the relationship between amounts( in moles) and masses (in grams) of reactant and products to solve the stoichiometric problems, the reactants and products are gases, then we use amounts (in moles) and volume(V) to solve the such problems.

5.11

Calculate the volume of O2 (in liters) required for the complete combustion of 7.64 L of acetylene (C2H2) measured at the same temperature and pressure.

The reaction of calcium carbide (CaC2) with water produces acetylene (C2H2), a flammable gas.

5.12Sodium azide (NaN3) is used in some automobile air bags. The impact of a collision triggers the decomposition of NaN3 as follows:

The nitrogen gas produced quickly inflates the bag between the driver and the windshield and dashboard. Calculate the volume of N2 generated at 80°C and 823 mmHg by the decomposition of 60.0 g of NaN3., mmHg atm, oC oKg NaN3 mole NaN3 mole N2

An air bag can protect the driver in an automobile collision.

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5.6: Dalton’s Law of Partial Pressures-1801

V and T are constant

P1 Ptotal = P1+ P2

Partial Pressure: that is the pressure of individual gas components in the mixture exerted on the wall of container.

Mole fraction: mole of one gas divided by moles of all gases present in a container

Total gas Pressure: Which states that the total pressure of a mixture of gases is just the sum of the pressure of each gas would exert in a container. .

P1 P2

Ptotal = P1+ P2

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Consider a case in which two gases, A and B, are in a container of volume V.

PA = nARTV

PB = nBRTV

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB

XA = nA

nA + nBXB =

nB

nA + nB

PA = XA PT PB = XB PT

Pi = Xi PT mole fraction (Xi ) = ni

nT

Partial pressure of A = mole fraction of A x Pt

5.14A mixture of gases contains 4.46 moles of neon (Ne), 0.74 mole of argon (Ar), and 2.15 moles of xenon (Xe). Calculate the partial pressures of the gases if the total pressure is 2.00 atm at a certain temperature. PNe/Ar/Xe = XNe/Ar/Xe x PT

Volume of oxygen collected at 24 oC and atm pressure of 762 mmHg is 128 mL. Calculate the mass (in gram) of oxygen gas obtained . The pressure of water vapor at 24 oC is 22.4 mmHg PO2 = PT – PH2O PV = nRT = m/M x RT m = PVM/RT

5.15

• Good Luck• Study from the Book