Intro Chem Gases Lecture Notes Sp08

8/10/2019 Intro Chem Gases Lecture Notes Sp08

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Introduction to ChemistryChapters 4 and 14

Gas Laws

Some Common Properties of Gases:

The density of gases is much less than that of solids or liquids. Gas molecules must be very far apart compared to liquids and solids.

The relationship among pressure, volume, and temperature of a fixed amount of gas were amongthe earliest quantitative studies in all science.

• Unlike a liquid or solid, a gas always fills its container. As a result, its volume (V) isthe same volume of the container.

• Pressure is a force exerted on a unit area: force

pressurearea

=

1. Gases can be compressed into smaller volumes ; that is,their densities can be increased by applying increased

pressure.

2. Gases exert pressure on their surroundings ; in turn,pressure must be exerted to confine gases.

3. Gases expand without limits , and so gas samples completelyand uniformly occupy the volume of any container.

4. Gases diffuse into one another , and so samples of gas placedin the same container mix completely. Conversely, different gasesin a mixture do not separate on standing.

5. The amounts and properties of gases are described in terms oftemperature, pressure, the volume occupied, and the numberof molecules present . For example, a sample of gas occupies agreater volume when hot than it does when cold at the samepressure, but the number of molecules does not change.

0.005031.591.70CCl4

0.0005880.9980.917H2 O

GasLiquidSolidDensities(g/mL)



2

Barometers are used to measure the pressure the atmospheric pressure (atm).

Atmospheric pressure is measured using a barometer. A barometer can be constructed by filling a long glass tube

(sealed at one end) all the way to the top with Hg. The tube is inverted (without letting air in) into a dish of Hg.

The Hg in the tube falls until the pressure from the mass of theHg in the tube is balanced by the pressure of the atmosphere onthe mercury in the dish.

The pressure of the atmosphere is then expressed by the heightof the Hg in the tube.

Definitions of standard pressure • 76 cm Hg • 760 mm Hg • 760 torr • 1 atmosphere • 101.3 kPa

Because mercury barometers are very common:• gas pressure is often expressed in mmHg• gas pressure is also expressed in torr

1 mmHg = 1 torr

• atmospheric pressure decreases with increasing elevation (less mass of air) At 0 o C, 1.0 atm = 760 mmHg = 760 torr

• The SI unit is the pascal (Pa):1.0 atm = 1.013 x 10 5 Pa

Charles’s Law : The volume of a fixed quantity of gas at constant pressure is directly

proportional to absolute (Kelvin) temperature.

From the height-temperature data:

Height(mmHg) Temperature oC k = V/T

(Kelvin)82 50 0.25375 22 0.25469 0 0.25366 −16 0.256

From this linear relationship:

1 2

1 2

V V k and k T T

= = for a fixed amount

of a gas at constant pressure, the ratio ofthe volume to the absolute temperatureis constant (k).

Therefore: 1 2

1 2

V V T T

=



3

Charles determined that ideal gases (those at normal temperatures and pressures) have thefollowing relationship:

V1 = kT1

V2 = k

T2

From the graph, absolute zero (-273o C) was extrapolated as a result of themeasurements of a gas (at differentpressures) at increasingly lowertemperatures.

Charles’s law states that thevolume of a gas is directlyproportional to the absolutetemperature at constant

pressure.

Gas laws must use the Kelvin scale to be correct.

At Standard temperature O 0 C or 273 K

The relationship between Kelvin and Celsius is:273

C K oT T = +

Example: A sample of hydrogen, H 2 , occupies 1.00 x 10 2 mL at 25.0 oC and 1.00 atm.

What volume would it occupy at 50.0 oC under the same pressure?

T1 = 25 + 273 = 298

T2 = 50 + 273 = 323 1 2 1 22

1 2 1

2

2

V V V T V =

T T T

1.00 10 mL 323 K V =

298 K 108 mL

= ∴

× ×

=

oK = C + 273

therefore, V1 = V 2

T1 T2



4

Boyle’s Law : The volume of a fixed quantity of gas at constant temperature is inverselyproportional to pressure exerted.

Example: A sample of hydrogen, H 2 , occupies 1.00 x 10 2 mL at 25.0 oC and 1.00 atm.

What volume would it occupy at 50.0 oC under the same pressure? Example: At 25 oC a sample of He has a volume of 4.00 x 10 2 mL under a pressure of 7.60

x 10 2 torr. What volume would it occupy under a pressure of 2.00 atm at the same T?

Standard Temperature and Pressure (STP)

Standard P ≡ 1.00000 atm or 101.3 kPa

Standard T ≡ 273.15 K or 0.00 o C

An experiment is conducted inthe apparatus (a) to the left todetermine the relationshipbetween volume and pressure ata fixed amount of gas.

The movable leg of theapparatus is moved either up ordown. The volume of thetrapped gas is determined atdifferent pressures.

The results are recorded in thetable (b).

P1V1 = k and P 2 V2 = k

( )( )

1 1 2 2

1 12

2

2

P V P VP V

VP

760 torr 400 mL

1520 torr 2.00 10 mL

==

=

= ×



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The Combined Gas Law : Combining the laws of Charles and Boyle:

Combined Gas Law Equation

Boyle’s and Charles’ Laws combined into one statement is called the combined gas lawequation.

• Useful when the V, T, and P of a gas are changing.

1 1 2 2

1 2

1 2

1 2

1 2

( ' )

( ' )

PV PV Boyle s Law at constant T

V V Charles Law at constant P

T T

P P at constant volume

T T

=

=

=

Where:

Pressure = atm, mmHg, or torrTemperature = KelvinVolume = liters

Example: A sample of nitrogen gas, N 2 , occupies 7.50 x 10 2 mL at 75.0 0 C under a pressure of

8.10 x 10 2 torr. What volume would it occupy at STP?

Example: A sample of methane, CH 4 , occupies 2.60 x 10 2 mL at 32 oC under a pressure of 0.500atm. At what temperature would it occupy 5.00 x 10 2 mL under a pressure of 1.20 x 10 3 torr?

1 21 1 2 2

1 2

1 1 2 2

1 2

Boyle's Law Charles' Law

V VP V P V

T T

For a given sample of gas: The combined gas law is:

P V P V P Vk

T T T

= =

= =

( )( )( )( )( )

1 2

1 2

1 2

1 1 22

2 1

V =750 mL V =?

T =348 K T =273 K

P =810 torr P =760 torr

P V TSolve for V =

P T810 torr 750 mL 273 K

760 torr 348 K

627 mL

=

=



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Avogadro’s Law: The standard molar volume of gases.

Avogadro theorized that if different gases had the same volume (when the temperature andpressure are the same), then the gases would have the same number of molecules.

Avogadro’s Law:

•

At constant temperature and pressure, the volume of the gas is directly proportional tothe number of moles of the gas.• If the temperature and pressure is the same for equal volumes of gases, then the number

of moles of the gases will be the same.

At constant temperature and pressure:V1 = kn 1

V2 = kn 2

At STP, the standard molar volume of all gases is:

22.4 Lmole

Gas densities: grams gasliter

• The density of a gas is dependent upon the pressure and temperature— mass of gas perunit volume of the gas.

• The number of moles of a gas is not dependent on the pressure and temperature, onlythe volume.

Example: One mole of a gas occupies 36.5 L and its density is 1.36 g/L at a given temperatureand pressure. (a) What is its molar mass? (b) What is its density at STP?

The Ideal Gas Equation : The summation of all the gas laws.

Pv = nRT

R = ideal gas constant

We can determine the value of R from standard conditions. • Recognize that for one mole of a gas at 1.00 atm, and 273 K (STP), the volume is

22.4 L. • Use these values in the ideal gas law.

therefore, V1 = V 2 n 1 n 2

? g 36.5 L 1.36 g( ) 49.6 g/mol

mol mol La = × =

STP

? g 49.6 g 1 mol( ) 2.21 g/L

L mol 22.4 Lb = × =



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0.0821 atm L R mol K = i

i

P = pressure in atmT = temperature in KelvinV = litern = mole

Example: What volume would 50.0 g of ethane, C 2 H6 , occupy at 1.40 x 10 2 oC under a pressure of

1.82 x 10 3 torr? • To use the ideal gas law correctly, it is very important that all of your values be in

the correct units! 1. T = 140 + 273 = 413 K2. P = 1820 torr (1 atm/760 torr) = 2.39 atm3. 50 g (1 mol/30 g) = 1.67 mol

Example: Calculate the number of moles in, and the mass of, an 8.96 L sample of methane, CH 4 ,

measured at standard conditions.

Examples: (We will do these in class)

1) A deep breath of air has a volume of 1.05 L at a pressure of 740. torr and body temperatureof 37.0 oC. Calculate the number of air molecules in the breath.

2). When gases are shipped, many are contained in high-pressure vessels. Suppose oxygen gas isshipped in a 42.0 L steel tank at a pressure of 1.80 x 10 4 kPa and a temperature of 23.0 0C. Whatis the mass of oxygen gas in the steel tank?

3). Calculate the density of sulfur tetrafluoride vapor if the pressure is 560 torr and thetemperature is 100. oC.

( )( )( )( )1.00 atm 22.4 LPV

R =nT 1.00 mol 273 K

L atm0.0821

mol K

=

=

( ) ( )

n R TV =

PL atm

1.67 mol 0.0821 413 K mol K

2.39 atm23.6 L

=

=

( )( )

( )4

4

1.00 atm 8.96 LPVn = 0.400 mol CH

L atmRT 0.0821 273 K mol K

16.0 g? g CH 0.400 mol 6.40 g

mol

= =

= × =



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4.) A unlabeled vessel contains oxygen, or helium, or argon, or nitrogen gas. If a 5.10 g sampleof unknown gas exerts a pressure of 589 torr at 25.5 oC in a 5.75 L container, which of the gasesis in the unlabeled vessel?

Gas Stoichiometry

2 mol KClO 3 yields 2 mol KCl 3 mol O 2

2(122.6g) yields 2(74.6g) 3(32.0g)

Those 3 moles of O 2 can also be thought of as 3(22.4L)or 67.2 L at STP

Example: What volume of oxygen measured at STP, can be produced by the thermaldecomposition of 120.0 g of KClO 3 ?

Ex: An impure sample of potassium nitrate that had a mass of 50.3 g was heated until all of thepotassium nitrate had decomposed? The liberated oxygen occupied 4.22 L at STP. Whatpercentage of the sample was potassium nitrate? Assume that no impurities decomposed toproduce oxygen.

2 KNO 3 (s) → 2 KNO 2 (s) + O 2 (g) 202.2 g 170.22 32.0g

Solution:

The only pure product from the decomposition of potassium nitrate was the oxygen. To determinethe amount of oxygen:

At STP, one mole of oxygen will occupy 22.4 L.

2 3 32

2 3

3

3

2

1.0 2.038.1

75.7%

202.24.22

22.4 1.0 2.0

38.1% 100

50.3

mol O mol KNO g KNO L O

L O mol O mol KNO

g KNO

g

g KNO

x

=

= =

E.g. Heating a 5.913 g sample of ore containing a metal sulfide, in the presence of excessoxygen, produces 1.177L of dry SO 2 , measured at 35.0 oC and 755 torr. Calculate the percentageby mass of the sulfur in the ore.

3 STP 22STP 2 3

3 3 2

STP 2 STP 2

1 mol KClO 22.4 L O3 mol O? L O 120.0 g KClO

122.6 g KClO 2 mol KClO 1 mol O

? L O 32.9 L O

= × × ×

=

2MnO &3 (s) 2 (g)(s)2 KClO 2 KCl + 3 O

∆

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Intro Chem Gases Lecture Notes Sp08

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