chapter 5 gases. what we’ve had so far! different ways of calculating moles of substances solids:...
TRANSCRIPT
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
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
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?
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
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
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?
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.
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.
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.