Unit VI: Gases

Unit VI: GasesUnit VI: Gases

A.A. Characteristics of a Solid, Liquid, and GasCharacteristics of a Solid, Liquid, and Gas

B.B. Real Gases and Ideal GasesReal Gases and Ideal Gases

C.C. Gas LawsGas Laws

D.D. Logical Method Solving of Gas Law ProblemsLogical Method Solving of Gas Law Problems

E.E. Densities and Molar Masses of GasesDensities and Molar Masses of Gases

F.F. The Ideal Gas EquationThe Ideal Gas Equation

G.G. Gas StoichiometryGas Stoichiometry

H.H. Chemical Equations and Gas Law CalculationsChemical Equations and Gas Law Calculations

Characteristics of Solids, Liquids, Characteristics of Solids, Liquids, and Gasesand Gases

The gaseous state is a dispersed state of The gaseous state is a dispersed state of matter, therefore:matter, therefore:

The total volume of a gas is mostly empty spaceThe total volume of a gas is mostly empty space

Gas molecules move with a very high velocity and Gas molecules move with a very high velocity and therefore have a high kinetic energytherefore have a high kinetic energy

Pressure, temperature, and the number of mol of gas Pressure, temperature, and the number of mol of gas determine the volume a sample of gas occupiesdetermine the volume a sample of gas occupies

6-A

Real gases Vs. Ideal gases

At ‘common’ pressures and temperatures gases behave as an ideal gasNo finite sizeNo intermolecular forcesOnly perfectly elastic collisions

The behavior of a gas is dependant on the pressure, temperature, volume, and number of particles (mol) present in a system

Pressure

Pressure is defined as the force per unit area:

P = F / A

In order to measure the pressure of the atmosphere we use a “barometer”

Some common units of pressure

1 atmosphere =14.7 pounds per square inch [psi]760 mm Hg101.325 kiloPascals [kpa]760 torr1.01 bar101,325 Pa [Newton / m2]

Lecture Problem VI-1 (pg 164)

How many atm is a pressure of 345 mm Hg?

345 mm Hg = 0.454 atm1 atm

760 mm Hg

Remember 1 atmosphere is defined as 760 mm Hg

TemperatureThere are several temperature scales:

In the USA we use Fahrenheit and everywhere else they use Celsius. But these are relative scales…..

When dealing with gases we need an absolute scale such as the Rankine or Kelvin scale.

These scales make use of ‘absolute zero’ as a reference.

Just of few of the temperature scales

Temperature Scales

Absolute Zero Water Freezes Water boils -459.67 °F 32 °Farenheit 212 °F

0 °R 491.67 ° Rankine 671.6 °R-273.15 °C 0 °Celsius 100 °C 0 K 273.15 Kelvin 373.15 K

K = °C +273 or °C = K - 273

°C = (°F -32) / 1.8

K = (°F -32) / 1.8 + 273.15

Lecture Problem VI-2 (pg 165)A gas is collected in the laboratory at 34 °C. What

is the temperature on the Kelvin scale?

Room temperature is usually about 72 °F. What is the equivalent temperature in Kelvin?

34°C + 273.15 = 307.15 K

(307 K if we consider sig. figs)

(72°F-32°F) x + 273.15 = 295 K 5 °C

9 °F

VI-C: Gas Laws: Boyle’s Law

Boyle’s law states:For a fixed mass of gas at constant temperature, the volume is inversely proportional to the temperature.

NASA demonstration

V 1/P or V = k/PWhere k is a constant that depends on T and mol of gas present

http://www.grc.nasa.gov/WWW/K-12/airplane/boyle.html

VI-C: Gas Laws: Boyle’s Law

What can we do with Boyle’s Law?

LP VI-3 (pg 168) If we have a gas with an initial volume of 47.5 mL, and a pressure of 911 mm Hg. If the pressure changes to 618 mm Hg what is the new volume of the gas? (assume constant T and mass of gas)

V2 = P1V1

P2

Remember PV= constant

So we can say P1V1 = k = P2V2

V2 = 47.5 mL x = 70.0 mL911 mm Hg

618 mm Hg


If a gas having an initial volume of 47.5 mL undergoes a change in pressure from 911 mm Hg to 618 mm Hg, what is its final volume in mL?

(Assume constant temperature and amount of gas.)

= 70.0 mL

Your turn1. A sample of gas has an initial volume of 58.0 mL and undergoes a

change in pressure from 710 mm Hg to 345 mm Hg. What is the final volume in mL? (119 mL)

2. A fixed mass of gas at 505 torr is compressed to a final pressure of 925 torr and has a final volume of 148 mL. What was the initial volume of the gas in mL? (271 mL)

3. If a sample of gas at 1.25 atm occupies 2.45 L, and is then compressed to a volume of 1.24 L, what is its final pressure in atm? (2.47 atm)

4. 475 mL of gas is compressed to 225 mL and has a final pressure of 375 torr. What was the initial pressure of the gas? (178 torr)

(Assume constant temperature for all problems)

VI-C: Gas Laws: Charles’ Law

Charles’ law states:For a fixed mass of gas at constant pressure, the volume is directly proportional to the temperature.

NASA demonstration

V T or V = kTWhere k is a constant that depends on P and mol of gas present

http://www.grc.nasa.gov/WWW/K-12/airplane/glussac.html

VI-C: Gas Laws: Charles’ Law

What can we do with Charles’ Law?

It is found that a volume of gas changes from 100.0 L to 75.0 L when cooled to a temperature of – 40.0°C. What was the original temperature in °C? (assume constant P and mass of gas)

T1 = T2V1

V2

Remember V/T= constant

So we can say V1/T1 = k = V2/T2

T1 = 233 K x = 311 K 38 °C100.0 L

75.0 L


Nitrogen gas is collected at a temperature of 20.0 °C. Its measured volume is 25.0 L. What volume would it occupy at a temperature of 100.0 °C?

(Assume constant pressure and amount of gas.)

= 31.3 L

Your turn1. A sample of oxygen gas has an initially occupies 58.0 mL at at

temperature of 45 °C. It then is then heated until the volume it occupies is 72.5 mL. What is the final temperature of the oxygen in Celsius? (124 °C)

2. A piston full of gas is heated from 25 °C to 85 °C at constant pressure. If the final volume of the piston was 475 mL, what was the initial volume of the gas in mL? (395 mL)

3. If a 2.25 L sample of gas at 112 °C is cooled at constant pressure until it occupies a volume of 1.45 L, what is the final temperature (in K) of the gas? (248 K)

4. 475 mL of nitrogen gas is heated to 374 K. If the initial temperature was 298 K then what is the final volume of the gas? (596 mL)

(Assume constant pressure for all problems)

VI-C: Gas Laws: Amonton’s Law

The Amonton law states:For a fixed mass of gas at constant volume, the pressure is directly proportional to the temperature.

P T or P = kTWhere k is a constant that depends on V and mol of gas present

VI-C: Gas Laws: Amonton’s Law

What can we do with Amonton’s Law?

If the pressure of a sample of gas at 655 mm Hg and 25.0 ° C is changed to 937 torr by heating at constant volume, what was the final temperature in °C? (assume constant V and mass of gas)

T2 = T1P2

P1

Remember P/T= constant

So we can say P1/T1 = k = P2/T2

T1 = 298.2 K x = 427 K 154 °C937 torr

655 torr


What pressure, in torr, of a fixed volume of gas that is initially at 852 torr and the temperature changes from 95 °C to 15 °C?

(Assume constant volume and amount of gas.)

= 667 torr

Unit VI-C: The Combined Gas Law

In summary we have: Boyle’s Law V 1/P for const. TCharles’ Law V T for const. PAmonton’s Law P T for const. V

But what happens when we have more than two properties of the gas changing?

In other words what happens when the only thing held constant is the amount of gas in our system?

Unit VI-C: The Combined Gas Law

By combining Boyle’s law with either of the other two we can say:

P1V1 P2V2

T1 T2

=

We now have an equation that will let us change all the properties of a system except for the amount of gas present

A very brief introduction to state functions.

For an ideal gas (of fixed mass) its state is completely described by P,V,T.

T

PPa, Va, Ta

Pc, Vc, Tc

Pb, Vb, Tb


A. A 365 mL sample of oxygen gas at 743 mm Hg and 23.0 °C is heated to 68.0 °C and the pressure is increased to 788 mm Hg. What is the final volume, in mL, of the oxygen gas?

= 396 mL

B. A sample of neon gas had a volume of 250.0 mL at 38 °C. At 760.0 torr and 273K it has a volume of 225.0 mL. What was the original pressure of the neon in torr?

= 779 torr

Comparison of gases

How can we compare samples of gas when their properties depend on the conditions at which they measured?

We need a reference point with which to make these comparisons.

We call this reference point:Standard Temperature and Pressure (STP)

STP is defined as:

Exactly 1 atm (760 torr) and 273.15 K (0°C)

Unit VI-D: Logical problem solving

Logical method with Boyle’s Law:

What is the volume, in mL of a 25.0 mL sample of helium at 625 torr if its original pressure was 766 torr?

Conditions Volume Pressure

Initial 25.0 mL 766 torr

Final ? 625 torr

If P then V = 30.6 mL

Unit VI-D: Logical problem solving

Logical method with Charles’ Law:

100.0 mL of oxygen gas is at a temperature of 42°C. What volume, in mL would it occupy at -12°C?

Conditions Volume Temperature

Initial 100.0 mL 42+273 = 315K

Final ? -12 + 273 = 261 K

If T then V = 82.9 mL

Unit VI-D: Logical problem solvingLogical method with the Combined Gas Law:

Consider 44.6 mL of H2 gas at a temperature of 65°C and a pressure of 925 torr. What is the volume in mL at STP ?

Conditions Pressure Volume Temperature

Initial 925 torr 44.6 mL 338 K

Final 760 torr ? 273 K

If P then V

= 43.8 mL

If T then V

One last combined gas law example

Lecture problem VI-7 (pg. 164):At STP the volume of a sample of CO2(g) is 250.0 mL. What volume would it occupy at a temperature of 175.0°C and a pressure of 955 torr?

Conditions Pressure Volume Temperature

Initial 760 torr 250.0 mL 273.15 K

Final 955 torr ? 448.2 K

= 327 mL

Unit VI-E: Density and Molar Mass

As with other phases of matter we can calculate the density of a gas.

We use the units g/L for gases

But the volume of a gas is a function its temperature and pressure! So we must specify the conditions at which we are describing the density

The densities of gases are often given at STP to make comparison easier


The volume that 1 mole of gas occupies is called the “molar volume”

For an Ideal Gas at STP 1 mol of the gas occupies 22.4 L.

Lecture problem VI-8 (pg 166)

Calculate the density, in g/L of chlorine gas at STP.

= 3.17 g/L at STP70.90 g Cl2

1 mol Cl2

1 mol Cl2

22.4 LX



What is the molecular mass of a gas that has a density of 2.99 g/L at STP?

= 67.0 g/mol2.99 g

1 L

22.4 L

1 molX

A common method for finding molecular mass involves vaporizing a known mass of compound and measuring the volume of the resulting gas


But what if we want to know the density of a gas that is NOT at STP?


The molecular mass of a gas at STP is 132.5 g/mol. What is its density at 56 °C and 1250 torr?

= 8.07 g/L

Unit VI-F: The Ideal Gas Equation

We started with the simple gas laws:V = k / P V = k T P = k T

We combined them to form:

P V T

= k or PV=kT


Now we use the relationship between Volume and mols of gas (22.4 L / mol @STP)

PV= kT PV=nRT

Where R is the our new constant

(Called the universal gas constant)


Now we can determine the value of R using the conditions at STP

R= = = 0.0821PV (1 atm) (22.4 L) L atmRT (1 mol) (273 K) mol K

Now if we know any three properties of our system:

(P, V, T, or n) we can find the fourth

Ideal Gas Law practiceLecture Problem VI-14 (page 187)

a. A sample of O2(g) collected at 22°C and 725 mm Hg was found to have a volume of 545 mL. How many grams of oxygen were in the sample?

b. What is the volume of 34.8 g of chlorine gas at 34°C and 735 torr?

0.0216 mol oxygen gas or 0.692 g O2(g)

12.8 L Cl2(g)

Unit VI-G: Gas Stoichiometry

Historically when reactions between gases were considered, volume was the measurement used. Why?

Hint: It has to do with Avogadro’s hypothesis V n

Lecture Problem VI-15 (page 189)Hydrogen combines with nitrogen to produce ammonia, NH3.

How many liters of hydrogen are necessary to react with 4.73 liters of nitrogen if all the gases are at the same temperature and pressure?

14.2 L of hydrogen gas

Unit VI-H: Chemical Equations and Gas Law Calculations!

Lecture Problem VI-16 (page 191-192)a. How many liters of oxygen, measured at STP will result from the decomposition of 4.83g of potassium chlorate into KCl and oxygen gas? (MW of KClO3 =122.55 g/mol)

b. How many mL of hydrogen gas would be formed at 62°C and 544 torr if 0.248 g of Cr reacted with excess HCl(aq) to form aqueous chromium (III) chloride?

1.33 L O2(g)

275 mL H2(g)

Unit VI: Gases

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