Chapter 5 GASES. What we’ve had so far! Different ways of calculating moles of substances Solids: Moles = grams molar mass Liquids:Molarity = moles Liter.

Chapter 5

GASES

What we’ve had so far!

• Different ways of calculating moles of substances

• Solids: Moles = grams molar mass

• Liquids: Molarity = moles Liter

Ideal Gas Law

• In Chapter 5, everything comes down to another way of solving for moles.

PV = nRT

where n = number of moles

Atmospheric Pressure

• Pressure exerted by a gas on its surroundings

Pressure

• Units of Pressure– mm Hg– Atm– Torr (in honor of Evangelista

Torricelli)– Pascal (Pa)

Pressure

• Conversions:

–1 atm = 760 Torr = 760 mm Hg

–1 atm = 101,325 Pa

The Gas Laws

• Boyle’s Law

• PV = k

– the product of pressure (P) and volume (V) of a trapped gas is constant

Pressure vs. Volume

• P is inversely proportional to V

• Since PV = k

• If k = 1,

• then P = 1 or V = 1 V P

Boyle’s Law

• only holds true at very low pressures

• at high pressures, PV is not constant

Application of Boyle’s Law

• commonly used to predict the new volume when pressure is changed

• If: PV = k

• Then: (PV)1 = (PV)2 for same gas at same temperature

Sample Problem on Boyle’s Law

• An aerosol can contains 400 mL of compressed gas at 5.20 atm pressure. When all the gas is sprayed into a large plastic bag, the bag inflates to a volume of 2.14 L. If the T is constant, what is the pressure of gas inside the plastic bag?

Boyle’s Law

• A gas that strictly follows Boyle’s Law is an IDEAL GAS.

Charles’s Law

• filled balloon with H2 and made the first solo balloon flight

• discovered that the volume of a gas at constant P increases linearly with the T of the gas

Charles’s Law

• The V of a gas at constant P is directly proportional to T.

• Charles’s Law Equation:

• V = kT where k is the proportionality constant

Units of Temperature

• Celsius

• Kelvin

• Conversion:

•0 oC = 273 K

Application of Charles’s Law

• To predict new volume of a gas when T is changed

• Thus if : V = kT

• Then: (V1/T1) = (V2/T2)

Sample Problem

• If a gas at 15 oC and 1 atm has a volume of 2.58 L, what volume will this gas occupy at 38 oC and 1 atm? [Note: Convert all T to K]

Avogadro’s Law

• Avogadro’s Postulate: [Chapter 2]

• Equal volumes of gases at the same T and P contain the same number of particles.

Closely obeyed by gases at low P.

Avogadro’s Law

• V = a x n– where n = number of moles– a = proportionality constant

• So for a gas at constant T and P, the volume is directly proportional to the number of moles of gas.

Application of Avogadro’s Law

• To predict changes in V when the number of moles changes.

• V1/n1 = V2/n2

Charles’ Law

• the volume of a gas at constant P increases linearly with the T of the gas

Summary of 3 Laws

• Boyle’s Law: V = K/P[at constant T and n]

• Charles’s Law: V = bT • [at constant P and n]• Avogadro’s Law: V = an

[at constant T and P]

Sample Problem

• A 1-L container contains 2.75 moles of H2 at 400 oC. What would be the pressure in the container at this temperature?

Sample Problem

• 3.5 moles of N2 has a pressure of 3.3 atm at 375 oC. What would be the pressure of the 5.3 moles of gas at 900 oC?

Boyle’s Law

• A gas that strictly follows Boyle’s Law is an IDEAL GAS.

Ideal Gas Law

• expresses behavior that real gases approach at low P’s and high T’s

• An Ideal Gas is a hypothetical substance!

Ideal Gas Law

• Since most gases approach close to ideal behavior anyway, we will assume that the gases we encounter in this course are all ideal gases.

Ideal Gas Law

• PV = nRT

–where R = universal gas constant

– = .08206 L-atm/K

Sample Problem

• A sample of methane gas that has a volume of 3.8 L at 5 oC is heated to 86 oC at constant P. Calculate its new volume.

Another Sample Problem

• A sample of hydrogen gas has a volume of 10.6 L at a T of 0 oC and a P of 2.5 atm. Calculate the moles of hydrogen gas present in this gas sample.

Molar Mass and Density

• Molar Mass = gmRT PV

• Density = Molar Mass x PressureRT

Dalton’s Law of Partial Pressures

• For a mixture of gases in a container, the total P exerted is the sum of the pressures that each gas would exert if it were alone.

• Ptotal = P1 + P2 + P3 …….

Dalton’s Law

• Ptotal = P1 + P2 + P3 …….

• If P1 = n1RT/V P2 = n2RT/V P3= n3RT/V

• Then: Ptotal = (n1RT/V) + (n2RT/V) + (n3RT/V)

Gas Stoichiometry

• PV = nRT

• For 1 mole of an ideal gas at 0 oC, the molar volume at STP is 22.42 L.

• STP = standard T and P • where T = 0 oC and P = 1 atm

Sample Problem

• A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N2 are present?

Kinetic Molecular Theory

• Is a model that attempts to explain the behavior of ideal gases

• Based on speculations about the behavior of the gas particles

Postulates of KMT

• The gas particles are assumed to:

• 1] be so small and have V = 0• 2] be in constant motion.• 3] exert no force on each other• 4] have average KE that is T (in

Kelvin)

KMT on P and V

• Since P 1 (Boyle’s Law) V

• As V increases, P decreases.• As V decreases, P increases.KMT: As V is decreased, P increases

because particles hit walls of container more often.

KMT on P and T

• As T is increased, the speed of the particles increases resulting in more collisions and stronger collisions. Thus P increases.

KMT on V and T

• Charles’s Law: V T

• KMT: As T is increased, P normally increases because particles collide more. To keep P constant, V has to increase to compensate for the increased collisions.

KMT on V and n

• Avogadro’s Law: V n

• KMT: Normally, an increase in n (moles of particles) increases P if V was held constant. To return P back to normal, V has to be increased.