Unit 2: Gases Mr. Anthony Gates. Quick Reminders Gas: Uniformly fills any container. Mixes completely with any other gas Exerts pressure on its surroundings.

Unit 2: GasesMr. Anthony Gates

Quick Reminders

• Gas:

• Uniformly fills any container.

• Mixes completely with any other gas

• Exerts pressure on its surroundings.• Pressure:

• is equal to force/unit area• 1 standard atmosphere = 1 atm • 1 atm = 760 mm Hg• 1 atm = 760 torr• 1 atm = 101.325 kPa

Barometer and Manometer

• Barometer measures pressure as the external pressure applies force to the mercury in the dish causing the mercury in the tube to rise.

• Manometers measure the pressure through the height changes the mercury undergoes.

Practice

a) 750 mmHg

b) 750torr

c) 0.985 atm

d) 14.5 psi

Boyle’s Law

• If temperature is constant, pressure and volume are inversely related. The product of pressure and volume (under constant temperature) is equal to some constant k.

• V=k/P

• P1V1=P2V2

• Holds precisely at very low temperatures.

Ideal Gases

• A gas that strictly obeys Boyle’s Law is called an Ideal gas.

• Figure 5.5: Plotting Boyle's data from Table 5.1. (a) A plot of P versus V shows that the volume doubles as the pressure is halved. (b) A plot of V versus 1/P gives a straight line. The slope of this line equals the value of the constant k.

Charles’ Law

• When pressure is constant, the temperature and volume are directly proportional.

• V=b•T (“b” is some proportionality constant)

• (when pressure is constant)

Graphing Charles’ Law

Avogadro’s Law

• When temperature and pressure are constant, volume is directly proportional to the amount of moles of gas.

• V=a•n (“a” is a proportionality constant)

• Example: balloons

Ideal Gas Law

• PV=nRT• R = proportionality constant = 0.08206 L atm mol

• P = pressure in atm• V = volume in liters• n = moles• T = temperature in Kelvins• Holds closely at P < 1 atm

AP Understandings

• Ideal Gases exhibit specific mathematical relationships among the number of particles present, the temperature, the pressure and the volume.

• Graphical representations of the relationships between P, V, and T are useful to describe behavior.

Practice

• A sample Hydrogen gas (H2) has a volume of 8.56L at a temperature of 0˚C and a pressure of 1.5atm. Calculate the moles of H2 molecules present in this gas sample.

• 0.57 mol

Standard Temperature and Pressure

“STP”P = 1 atmosphere

T = CThe molar volume of an ideal gas is 22.42 liters at STP

Molar Volume

1mol (gas) = 22.4L under STP

The use of stoichiometry with gases also has the potential for laboratory experimentation, particularly with respect to the experimental determination of molar mass of a gas.

Dalton’s Law of Partial Pressures

• For mixtures of gases in a mixture,

• Ptotal= P1+P2+P3+…

• The partial pressure of a gas is dependent on the number of moles of particles.

• In a mixture of ideal gases, the pressure exerted by each component (the partial pressure) is independent of the other components. Therefore, the total pressure is the sum of the partial pressures.

Mole Fractions

• Mole Fraction• The ratio of the number of moles of a given

component in a mixture to the total number of moles in the mixture

• For an ideal gas, the mole fraction (x):

x1 = n1 = P1

ntotal Ptotal

Dalton’s Law of Partial Pressures

• Example: Oxygen gas is collected over water at 28°C. The total pressure of the sample is 5.5 atm. At 28 °C, the vapor pressure of water is 1.2 atm. What pressure is the oxygen gas exerting?

• Ptotal = PO2 + PH2O

• 5.5 = x + 1.2• X = 4.3 atm

AP Practice

• A sample of dolomitic limestone containing only CaCO3 and MgCO3 was analyzed.

• (a) When a 0.2800 gram sample of this limestone was decomposed by heating, 75.0 milliliters of CO2 at 750 mm Hg and 20°C were evolved. How many grams of CO2 were produced? Justify your answer.

• (b) Write equations for the decomposition of both carbonates described above.

• (c) It was also determined that the initial sample contained 0.0448 gram of calcium. What percent of the limestone by mass was CaCO3?

• (d) How many grams of the magnesium-containing product were present in the sample in (a) after it had been heated, assuming the reaction went to completion? Justify your answer.

Kinetic Molecular Theory

1. Particles are so small compared to the distance between them, that the volume of the individual particles can be assumed to be zero

2. The particles are in constant motion. The collisions between the particles and the walls of the container are the cause of the pressure exerted by the gas.

3. The particles are assumed to exert no forces on each other; they are assumed neither to attract or repel each other

4. The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.

KMT explains it all…

•As volume decreases, pressure will increase because the gas particles will hit the wall more often.•Hence… V=1/P

continued…

•As the temperature increases, the kinetic energy increases and so the particles will move faster and hit the wall more often. Since the volume is constant, the pressure of the gas must increase. •Hence… P=T

Even more…

• If pressure is kept constant, an increase in temperature will cause an increase in kinetic energy. This causes the particles to speed up and hit the walls of the container more often and thus cause the volume to increase. •Hence… V=T

Still more…

• If temperature and pressure are kept constant, an increase in the number of particles would cause particles to collide more often and thus more collisions with the wall of the container. As a result the volume of the container would expand.•Hence… V=n

And finally…

•KMT assumes that all particles are independent of each other and that they have no volume, thus the combined pressures of each would equal the total pressure of the system. •Hence… PT=P1+P2+P3+…

KMT and the meaning of temperature (see appendix 2 for clarification)

• Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.)

(K E )3

2av g RT

Continued…

Average kinetic energy equation is derived from the KE equation of a single particle.

KE=1/2mv2

Root Mean Square Velocity

• The symbol u2 means the average of the squares of the particle velocities.

• The square root of u2 is called the root mean square velocity and is symbolized by u rms.

2 3

8.3145

in Kelvin

= mass of one mole of gas particles in kg

rms

RTu u

MR J mol K

T

M

RMS velocity cont.

• Root Mean Square Velocity • Equals the average of the squares of the velocities of the

particles (urms )• urms = 3RT

M

• Implies that velocity of a gas is dependent on mass and temperature

• Velocity of gases is determined as an average• M = mass of one mole of gas particles in kg!• R = 8.3145 J/K•mol!

• joule = kg•m2/s2

Practice Makes Perfect

• Find the Root Mean Square Value.

KEave= (3/2)(8.3145J/molK)(315K)=3928.6J/mol

urms==298.8m/s

More practice

• urms= = 1064.9m/s

Maxwell-Boltzmann Distribution

• Temperature is a measure of the average kinetic energies of the particles.

• At any given temperature, there is a wide distribution of kinetic energies.

Maxwell-Boltzmann continued…

• Figure 5.21: A plot of the relative number of N2 molecules that have a given velocity at three temperatures.

• Along with varying kinetic energies, at any given temperature, there is also a wide distribution of velocities.

Maxwell-Boltzmann cont…

• As the temperature rises, the curve flattens and thus a larger distribution of particles has a high kinetic energy.

Effusion vs. Diffusion

• Effusion: the rate at which a gas escapes through a tiny orifice and into an evacuated chamber.

• Diffusion: the rate at which gases mix together.

Effusion vs. Diffusion

Practice

• • The molar mass is 158.98g/mol

Real Gases… because the world isn’t perfect

• van der Waals created a correction factor to account for the fact that real gases do take up space.

• Ideal: P = nRT/V

• Real: P = nRT/(V-nb)• n is the number of moles of a gas• b is a correction factor determined by experimental results

Reality… uugh

• To account for the attractive forces of the particles, another correction factor was created:

• Preal=Pideal-a(n/V)2

• a is a correction factor derived from experimental observations.

Real Gas Equation

Corrected Pressure Corrected Volume

Correction factors