Chapter 5 review gases
This presentation is the property of its rightful owner.
Sponsored Links
1 / 20

Chapter 5 Review: Gases PowerPoint PPT Presentation

  • Uploaded on
  • Presentation posted in: General

Chapter 5 Review: Gases. Section 1: Pressure. Pressure: the pressing of the particles of a gas against its surroundings Atmospheric pressure is the result of the masses of the gases in Earth’s atmosphere being pulled to Earth due to gravity

Download Presentation

Chapter 5 Review: Gases

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Chapter 5 Review: Gases

Section 1: Pressure

  • Pressure: the pressing of the particles of a gas against its surroundings

  • Atmospheric pressure is the result of the masses of the gases in Earth’s atmosphere being pulled to Earth due to gravity

  • A barometer is a device that measures atmospheric pressure

    • A glass tube filled with mercury is flipped into a container of mercury

    • Atmospheric pressure measured in height of Hg in mm

    • Averages 760 mmHg at sea level

    • Higher in atmosphere = lower height of Hg

  • Since Pressure is the force per unit area:

    • Pressure = force/area

    • Newtons per square meter = pascal (Pa)

Section 1: Pressure

  • Unit Conversions

    • X mm Hg = X torr

    • 760 torr = 1 atm

    • 1 atm = 101,325 Pa

  • The pressure of a gas is measure to be 49 torr. Convert this to both atmospheres and pascals.

Section 2: Gas Laws

  • Boyle’s Law

    • Boyle studied relationship between pressure and volume using a J shaped tube closed at one end.

    • As volume decreases, pressure increases [inverse relationship]

    • PV = k (where k is a constant for a given sample of air at a given temp.)

    • so: P1V1 = P2V2

    • Used for predicting new volume of gas after changing pressure or new pressure of gas after changing volume

    • If the pressure of a 1.53 L sample of gaseous SO2 is changed from 5.6x103 Pa to 1.5x104 Pa, what will be the new volume of the gas?

    • Does the answer make sense?

Section 2: Gas Laws

  • The temperature and the amount of mole of gas are remained constant

  • Gases that strictly obey Boyle’s Law are called ideal gases, but the law is only precise at low pressures. Many deviations occur at higher pressure readings.

  • If you were to plot a P vs. V graph and then a V vs 1/P graph, you would have a hyperbola.

Section 2: Gas Laws

  • Charles’s Law

    • Charles found that volume of gas increases as temperature increases at constant pressure [direct relationship]

    • The volume of all gases at -273oC would theoretically be 0

    • K = oC + 273

    • At 0 Kelvin, the volumes of the gases extrapolate to zero, so anything less than that and the volume would be negative (not possible)

    • 0 K is called absolute zero

    • V = bT where T is in Kelvins and b is a proportionality constant

    • So V/T = b or V1/T1,= V2/T2

    • A sample of a gas at 15oC has a volume of 2.58L. What volume will this gas occupy at 38oC (assuming constant pressure)?

Section 2: Gas Laws

  • Plot V vs. T and you will have a linear line with a positive slope.

Section 2: Gas Laws

  • Avogadro’s Law

    • The same volumes of gas at the same pressure and temperature have the same number of particles.

    • V = an where V is volume of gas, n is the number of moles of gas particles and a is a proportionality constant

    • Number of particles increases, volume increases [directly proportional]

    • V/n = a so V1/n1 = V2/n2

    • An 11.2 L sample of gas contains 0.50 mol N2. At the same temp. and pressure, how many moles of gas would there be in a 20. L sample?

Section 2: Gas Laws

  • Gay Lussac’s Law

  • Gay Lussac studied gases with constant volume and constant amount of moles and found that Pressure is directly related to Kelvin T.

  • P = bT where T is in Kelvins and b is a proportionality constant

  • So P/T = b or P1/T1,= P2/T2

  • Plot P vs T and you will get a linear line with a positive slope

  • note that all of these laws hold true for ideal gases of which there are none so there are only deviations, but we can come close.

  • A 20 L cylinder containing 6 atm of gas at 27 °C. What would the pressure of the gas be if the gas was heated to 77 °C?

Section 3: Ideal Gas Law

  • Boyle’s Law: V = k/P

  • Charles’s Law: V = bT

  • Avogadro’s Law:V = an

  • These relationships can be combined to form the equation:

    • V = R (Tn / P) where R is the combined proportionality constant called the universal gas constant

    • When pressure is in atm and volume is in L, R = 0.08206 L*atm / K*mol

    • More familiar form of ideal gas law: PV = nRT

  • A gas that behaves exactly as the equation is said to behave ideally

  • The ideal gas law expresses the behavior the real gases approach at low temperatures and high pressures

  • So and ideal gas is a hypothetical substance, however most gases behave the equation close enough at pressures below 1 atm that errors are minimal

Section 3: Ideal Gas Law

1. A sample of hydrogen (H2) gas has a volume of 8.56 L at a temperature of 0oC and a pressure of 1.5 atm. Calculate the moles of H2 molecules present in this gas sample.

2. A sample of diborane gas (B2H6) has a pressure of 345 torr at a temperature of -15oC and a volume of 3.48 L. If conditions are changed so that the temperature is 36oC and the pressure is 468 torr, what will be the volume of the sample?

Section 4: Gas Stoichiometry

  • If we have one mole of an ideal gas at STP (0o Celsius and 1 atm) than the volume found using the ideal gas law is

    ((1.000mol) x (.08026 L x atm/K x mol) x (273.2 K)/1.000 atm) = 22.42

    This volume is the molar volume of an ideal gas. Of course, no gas actually has a molar volume of 22.42, but there are close deviations.

  • A sample of methane gas having a volume of 2.80 L at 25o Celsius and 1.65 atm is mixed with a sample of oxygen gas having a volume of 35.0 L 31.0o Celsius and 1.25 atm. The mixture is then combusted. Calculate the volume of carbon dioxide formed at a pressure of 2.50 atm and a temperature of 125o Celsius.

Section 4: Gas Stoichiometry

Molar mass of gas

  • From the measured density, you can find the molar mass of gas using the ideal gas equation.

    n= (grams gas/molar mass) = m/molar mass

    substitute into ideal gas equation-

    (m/molar mass)RT/V = mRT/V(molar mass)

    since mass/volume = density

    P= (dRT)/molar mass

  • The density of a gas was measured at 1.50 stm and 27oC and found to be 1.95 g/L. Calculate the molar mass of the gas.

Section 5: Law of Partial Pressures

  • Dalton’s Law of Partial Pressures, as summarized by dalton:

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

    • PTOTAL = P1 + P2 + P3 + ..... where the subscripts represent individual gases (gas 1, gas 2 and so on)

    • The symbols P1, P2, P3, and so on represent each partial pressure, or the pressure the individual gas would exert on the container if it were alone.

  • Assuming each gas behaves ideally, the partial pressure of each gas can be calculated using the ideal gas law eaquation:

    • P1 = (n1RT) / VP2 = (n2RT) / VP3 = (n3RT) / V…….

  • So:

    • PTOTAL=P1+P2+P3+...= (n1RT)/V + (n2RT)/V + (n3RT)/V + ... = (n1+n2+n3+...)(RT/V) = nTOTAL(RT/V)

Section 5: Law of Partial Pressures

  • Assuming each gas behaves ideally, the partial pressure of each gas can be calculated using the ideal gas law equation:

    • P1 = (n1RT) / VP2 = (n2RT) / VP3 = (n3RT) / V…….

  • So:

    • PTOTAL= P1 +P2 +P3 + … =

    • PTOTAL= (n1RT)/V + (n2RT)/V + (n3RT)/V + ... =

    • PTOTAL= (n1+n2+n3+...)(RT/V) =

    • PTOTAL= nTOTAL(RT/V) → Where nTOTAL is the sum of the numbers of moles of the different gases

  • It’s the number of moles present that’s important, not the volume of the gas particles nor the forces among the particles (for ideal gases).

Section 5: Law of Partial Pressures

Consider the three flasks in the diagram below. Assuming the connecting tubes have negligible volume, what is the partial pressure of each gas and the total pressure when all the stopcocks are opened?

First flask: He --- 200. torr, 1.00L

Second flask: Ne --- 0.400 atm, 1.00L

Third flask: Ar --- 24.0 kPa, 2.00 L

Section 6: Kinetic Molecular Theory of Gases

Ideal gases:

  • Assume that there are no particle interactions- elastic collisions

  • They are infinitely small- point mass. Gases molecules have volume though.

  • Particles are in constant motion. Increased temperature will increase frequency and strength of collision between wall and particles- increased pressure.

  • Average kinetic energy of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.

  • Van der waals equation

Section 7: Effusion and Diffusion

  • Effusion is the term used to describe the rate at which a gas can be transferred through a tiny opening into an evacuated chamber.

    Thomas Graham’s Law

    (ma/mb)1/2= Vb/Va

  • The rate of a gas is inversely proportional to the square root of the mass of its particles.

  • The relative rates of effusion of two gases held at the same pressure and temperature are given by the inverse ratio of the square roots of the masses of the particles.

  • A 3.00 L sample of helium was placed in container fitted with a porous membrane. Half of the helium effused through the membrane in 25 hours. A 3.00 L sample of oxygen was placed in an identical container. How many hours will it take for half of the oxygen to effuse through the membrane?

Section 7: Effusion and Diffusion

  • Diffusion is the rate of the mixing of gases

  • Because so many collisions with oxygen and nitrogen occur as molecules try to diffuse, diffusion is complicated to describe theoretically.

Section 8: Real Gases

No gas follows the ideal gas law exactly, however many gases at low pressure and high temperature come close.

To correct the assumptions made by the Kinetic Molecular Theory, Johannes van der Waals modified the ideal gas law for real gases :

P=((nRT)/(V-nb)) - a(n/V)2


(P + a(n/V)2)x (V-nb) = nRT

This takes into account molecular size(b) and molecular interaction forces(a).

b will be greater for large particles and a is higher for polar molecules.

  • Login