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Chapter 13

GasesChapter 13

Section 13.1: Pressure

Pressure refers to an amount of force applied over a given area.

Area

Force Pressure

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Pressure is directly proportional to force and inversely proportional to area.

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Barometer – Device that measures atmospheric pressure.

o A column of mercury, 760 mm high (~30 in.), supported by air pressure.

PressureThe SI derived unit for force is the Newton (N).

The SI derived unit for area is the square meter (m2).

The SI derived unit for pressure is the Pascal (Pa).

1 Pa = 1 N/1 m2.

1.000 atm = 101,325 Pa = 101.3 kPa = 760.0 mm Hg = 760.0 torr = 14.69 psi (Write these conversion factors on your PT)

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Section 13.2: Pressure and Volume – Boyle’s Law

An Irish scientist named Robert Boyle (1627-1691) conducted a series of experiments on the P-V relationship.

Simply stated, Boyle’s Law says that theinitial pressure times the initial volume is equal to the product of the final pressure and final volume.

P1V1 = P2V2

1 = initial values and 2 = final values

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Figure 13.5: A plot of P versus V from Boyle’s data.

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Figure 13.6: Illustration of Boyle’s law.

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Boyle’s Law: ExplanationGas pressure is a result of the particles colliding with the walls of their container.

As the “container” becomes smaller, the collisions over any given area will increase.

The increase in the number of collisions translates to an increase in pressure.

If the size of the container is cut in half, the number of collisions (and the pressure) doubles.

The number of particles (as well as temperature) MUST be held constant.

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Boyle’s Law: Practice1) A sample of helium gas has a pressure of

3.54 atm in a container with a volume of 23.1 L. This sample is transferred to a new container and the pressure is measured to be 1.87 atm. What is the volume of the new container?

2) A steel tank of argon has a pressure of 34.6 atm. If all of the argon is transferred to a new tank with a volume of 456 L, the pressure is measured to be 2.94 atm. What was the volume of the original tank?

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Section 13.3: Volume & Temperature – Charles’ Law

In the 1780’s a French physicist named Jacques Charles (1746-1823) noticed that as the temperature of a gas increases, so does its volume (at constant pressure).

V/T = k (where k is a constant)

If we relate the initial temperature and volume (T1 & V1) to the final (T2 & V2), we get:

V1/T1 = V2/T2

This is a direct relationship.13-10

Figure 13.7: Plots of V (L) vs. T (°C) for several gases.

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Kelvin Temperature Scale

Lord Kelvin found the volume always decreased to zero at -273ºC.

He devised a new “absolute” scale for temperature where there were no negative numbers.

He called “0 Kelvin” (not degrees Kelvin) “absolute zero.”

From now on when dealing with gases we will ONLY use Kelvin temperatures.

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Figure 13.8: Plots of V vs. T using the Kelvin scale.

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Charles’ Law: Practice1) A 2.45 L sample of nitrogen gas is

collected at 273 K and heated to 325 K. Calculate the volume of the nitrogen gas at 325 K (assume constant pressure).

2) A sample of oxygen gas has a volume of 4.55 L at 25 C. Calculate the volume of the oxygen gas when the temperature is raised to 45 C (assume constant pressure).

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Temperature-Pressure RelationshipsBoyle’s Law said P & V are inversely related:P1V1 = P2V2

Charles’ Law said V & T are directly related:V1/ T1 = V2/ T2

How do you think P & T are related?What happens to the gas inside an aerosol can when you throw it into a fire?That’s right . . . As the temperature increases, the pressure increases.

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This is analogous to Charles’ Law.

P/T = k (where k is a constant)

If we relate the initial pressure and temperature (P1 & T1) to the final (P2 & T2), we get:

P1/T1 = P2/T2

This is a direct relationship.

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Temperature-Pressure Relationships

The Combined Gas Law

If we take all three of these different relationships and put them all together, we get the “Combined Gas Law:”

(P1V1) = (P2V2)

T1 T2

Remember: 1’s are always initial values and 2’s are always final.

Use only Kelvin for temperature.

Once you know this, you can use it on any problem.

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Practice Problems

1. What pressure (in atm) is required to compress 1.00 L of gas at 760 mm Hg pressure to a volume of 50.0mL?

2. A sample of oxygen gas has a volume of 4.55 L at 25 C. Calculate the volume of the oxygen gas when the temperature is raised to 45 C (assume constant pressure).

3. A helium balloon has a volume of 2.30L at 23.5 C and a pressure of 1.00 atm at sea level. The balloon is released and floats upward. At a certain height the atomosperic pressure is 0.810 atm and the temp. is 12C. Calculate the volume of the balloon.

Section 13.4: Volume & Moles – Avogadro’s Law

What is the relationship between the volume of gas and the number of molecules present in the gas sample?

In 1811 Italian scientist Amadeo Avogadro postulated that, at constant temperature and pressure, the volume is directly proportional to the number of moles of gas.

Avogadro’s Law states:

V1/n1 = V2/n2 (n=# of moles)13-19

Figure 13.9: The relationship between volume V and number of moles

n.

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Avogadro’s Law: Practice1) If 2.55 mol of helium gas occupies a

volume of 59.5 L at a particular temperature and pressure, what volume does 7.83 mol of helium occupy under the same conditions?

2) If 4.35 g of neon gas occupies a volume of 15.0 L at a particular temperature and pressure, what volume does 2.00 g of neon gas occupy under the same conditions?

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Section 13.5: The Ideal Gas Law

When all four important characteristics of a gas (P, V, T, & n) are combined into one equation, we get what is known as the “Ideal Gas Law:”

PV=nRT

R is the universal gas constant

R = 0.08206 L atm/K mol

Since R is always the same, we need to know three of the other four variables to solve any given ideal gas problem.Note the units for R; all units of the variables must match (T in K, V in L, etc.)

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Ideal Gas Law Practice Problems1) A sample of neon gas has a volume of

3.45 L at 25 C and a pressure of 565 torr. Calculate the number of moles of neon present in this gas sample.

2) A 0.250 mol sample of argon gas has a volume of 9.00 L at a pressure of 875 mm Hg. What is the temperature (in C) of the gas?

3) What volume does 4.24g of nitrogen gas occupy at 58.2 C and 2.04atm?

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The Ideal Gas Law

As the name states, this equation illustrates the behavior of gases that behave ideally.

Not all gases behave ideally and therefore their behavior cannot be accurately predicted by this equation.

Most gases behave ideally when P is 1 atm or lower and T is 273 K or higher.

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Section 13.6: Dalton’s Law of Partial Pressures

Many gases, air for example, are a mixture of components.

Studies of these mixtures have shown that each component behaves independently of the others.

In 1803, scientist John Dalton noted: “For a mixture of gases in a container, the total pressure exerted is the sum of the partial pressures of the gases present.”

This is Dalton’s Law of Partial Pressures.13-25

Dalton’s Law of Partial Pressures

Partial Pressure – The pressure that the gas would exert if it were alone in a container.

Ptotal = P1 + P2 + P3 . . .

P1, P2, & P3 represent the partial pressures.

Assuming each gas behaves ideally, we can calculate the partial pressure of each component using the ideal gas law:

P1=(n1RT)/V, P2=(n2RT)/V, P3=(n3RT)/V13-26

Dalton’s Law of Partial PressuresPtotal=P1+P2+P3=(n1RT)/V+(n2RT)/V+(n3RT)/V

=(n1+n2+n3)(RT/V)

=ntotal(RT/V)

ntotal is the sum of the number of moles of the gases in the mixture.

For ideal gases that means ntotal is the total number of particles in the mixture and that their identity does not matter.

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Figure 13.10: When two gases are present, the total pressure is the sum of the

partial pressures of the gases.

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Partial Pressure Practice Problems1) A 5.00 g sample of helium gas is added

to a 5.00 g sample of neon in a 2.50 L container at 27 C. Calculate the partial pressure of each gas and the total pressure.

2) Equal masses of oxygen and nitrogen gas are present in a container. Which gas exerts the larger partial pressure? By what factor?

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Section 13.7: Laws & Models–A Review

Let’s distinguish between a law & a model:

Laws allow us to predict behavior.

Models explain why things behave the way they do (laws do not).

Consider gravity – we can predict something will fall (law), but not explain why gravity works the way it does (we’d need a model for that).

We have an “Ideal Gas Law,” but there’s no such thing as an ideal gas.

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Section 13.8: The Kinetic Molecular Theory (KMT) of Gases

The Kinetic Molecular Theory (KMT) is a simple model that attempts to explain the behavior of an ideal gas.

The next slide shows the assumptions (postulates) that are made by this theory.

Remember that KE = ½mv2.

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Postulates of KMT

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Section 13.9: Implications of the KMT

Temperature:

As Temperature (T) increases Kinetic Energy (KE) increases.

As KE increases particles collide with the container walls more often and with greater force.

T (in Kelvins) is directly proportional to KE.

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Implications of the KMT

Pressure & Temperature Relationship:

Increased number and force of collisions translates to increased Pressure (P) at constant V.

So, As T increases, P increases (a direct relationship).

Volume & Temperature Relationship:

If P is constant and T increases, V must increase.

Again, a direct relationship (Charles’ Law). 13-34

Section 13.11: Gas Stoichiometry

Using PV=nRT we can define the volume occupied by one mole of ANY gas at STP:

V=(nRT)/P

V=(1 mol)(0.08206L atm/K mol)(273K)/1 atm

V=22.4 L

The value of 22.4 L/mol (@STP) is called the molar volume.

It is true for ANY ideal gas (or real gas behaving ideally as they do under STP conditions).

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Gas Stoichiometry Practice Problems

1) A sample of argon gas has a volume of 3.45 L at STP. What is the mass of the argon?

2) A sample of hydrogen gas occupies a volume of 15.0 L at STP. What volume will this sample occupy at 22°C and 2.50 atm?

3) When subjected to an electric current, water decomposes to hydrogen and oxygen gas: 2H2O(l) 2H2(g) +O2(g). If 25.0 g of water is decomposed, what volume of oxygen gas is produced at STP?

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