gases chang chapter 5. chapter 5 outline gas characteristics pressure the gas laws density and molar...

Gases

Chang Chapter 5

Chapter 5 Outline

• Gas Characteristics• Pressure• The Gas Laws• Density and Molar Mass of a Gas• Dalton’s Law of Partial Pressure• Kinetic Molecular Theory of Gases• Molecular Effusion and Diffusion• Deviations from Ideal Gas Behavior

Gases Characteristics

• Which substances are most likely to be gases at 25oC and 1 atm?–Not ionic compounds–Most likely molecular compounds

• Properties of Gases–Assume shape and volume of container–Easy to compress–Mix with other gases completely–Have low densities

Pressure

• Pressure = Force = m ∙ a

Area area» Units = Pascals (Pa) = Newtons

m2

–Atmospheric pressure varies with altitude

–Pressure is measured using a barometer

–Units of Pressure• 1 atm = 760 torr = 101,325 Pa

= 760 mmHg (UNIT FACTORS)

The Gas Laws

• Boyle’s Law (1661)–Relates Pressure and Volume of Gases–P a 1/V or PV = k1 or P = k1/V

– The pressure of a fixed amount of gas at constant temperature is inversely proportional to the volume of the gas.

The Gas Laws

• Charles’ and Gay-Lussac’s Law–Relates Temperature and Volume of Gases–V a T or V = k2∙T (T must be in K)

– The volume of a fixed amount of gas at constant pressure is directly proportional to the absolute temperature of the gas

The Gas Laws

• Avogadro’s Law–Relates the Volume of a Gas with the

Amount of Gas–V a n or V=k3∙n

– At constant pressure and temperature the volume of gas is directly proportional to the number of moles of gas present

The Gas Laws

• Combing all 3 gas laws yields the IDEAL GAS EQUATION–PV = k1 and V = k2∙T and V = k3∙n

• PV = k1k2k3nT

• k1k2k3 = constant = R = 0.8206 L∙atm/mol∙K

• PV = nRT

The Gas Laws

• Example: What volume of gas would be occupied by 1.00 mol N2 at STP?

• STP = standard temperature (0oC) and pressure (1 atm)

• UNIT FACTOR!!!!

The Gas Laws

• Example: A sample of CH4 in a 2.00 L vial at 25oC and 0.750 atm is compressed and heated until T=75oC and the new volume is 100L. What is the new pressure?

Density and Molar Mass of Gas

• PV = mRT therefore n/V = P/RT

• n= m/M (substitute)

• m/MV = P/RT

• Therefore: m/V = PM/RT = d = density

• Therefore: M = dRT/P = molar mass

Dalton’s Law of Partial Pressure

• “the total pressure of a mixture of gases is the sum of the pressures that would exist if each gas were present by itself”

• Consider a mixtuer of gases A, B, and C–PA=nART/V, PB=nBRT/V, and PC=nCRT/V

–PTotal=PA+PB+PC

=nART/V + nBRT/V + nCRT/V

= (nA+nB+nC)(RT/V)

= nTotal (RT/V)

Dalton’s Law of Partial Pressure

• A gaseous mixture made from 6.00 g O2 and 9.00 g CH4 is placed in a 15 L vessel at 0oC. What is the particle pressure of each gas and the total pressure in the vessel?

Dalton’s Law of Partial Pressure

Dalton’s Law of Partial Pressure

• Example: If 50mL of gas are collected at a pressure of 750.0 torr at 25oC, how many grams of Zinc Reacted?

Zn(s) + HCl(aq) H2(g) + Zn2+(aq) + Cl-(aq)

Dalton’s Law of Partial Pressure

Kinetic Molecular Theory of Gases• Assumptions

• Distance between particles is very large compared to the size of individual particles

• Gas molecules undergo constant, random, rapid motion and experience frequent collisions with other particles

• Gas particles do not experience any attractive or repulsive forces w.r.t. other molecules

• The average kinetic energy per particle is: KE=1/2 mu2 where m is mass and u2 is the mean square speed– Any gas at the same temperature has the same KE– KE a T or 1/2mu2 a T

Kinetic Molecular Theory of Gases

• Results–Gas pressures result from collisions between

particles and the container walls–Absolute temperature of a gas is really a

measure of the average KE of the particles• High T: more KE, more frequent collisions• Low T: less KE, less frequent collisions

–Many observations can be explained by the Kinetic Molecular Theory of Gases• EX: gasses are easily compressed, P a T

Molecular Effusion and Diffusion

• Distribution of Molecular Speeds

Molecular Effusion and Diffusion

• All gases at the same temperature have the same KE–KE=1/2mu2

• A lighter gas (small m) must have a greater speed, u

–u= = because M is in the denominator, a less massive gas will be faster

Note: T = K, M = kg/mol, R= 8.314 J/mol∙K

Molecular Effusion and Diffusion

• Graham’s Law of Effusion• Effusion= the process by which a gas under

pressure escapes through a pinhole to an area a lower pressure

• The effusion rate of a gas is inversely proportional to the square root of the M of the gas

• Assuming we have two gases at the same T and P in identical containers,

How 235U and 238U are separated

Molecular Effusion and Diffusion

• Example: An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is only 0.355 times that of O2 at the same T. What is the molar mass of the unknown gas and identify it.

Molecular Effusion and Diffusion

• Graham’s Law of Diffusion• Diffusion= the gradual mixing of molecules of

different gases by virtue of their kinetic properties

• Diffusion is a slow process• It turns out that the rate of diffusion is also

inversely proportional to the molar mass so the same equation applies

Deviations of Gases from Ideal Behavior

• All gases deviate from ideal behavior in one way or another–Real gases deviate from ideal behavior at

high pressures, and low temperatures• The van der Waals Equation corrects the

Ideal Gas Equation to account for this non-ideal behavior *****a and b are van der

Waals constants determined experimentally for

each gas, found in table 5.4

Deviations from Ideal Gas Behavior

• Example: Estimate the real pressure exerted by 1.000 mol of Cl2(g) in 22.4 L at 0.0oC. Compare this to the idealized pressure.