Gases

Chapter 5

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Topics:• Gases & Pressure• Gas Laws• Ideal Gas Equation• Molecular Weight/Density• Gas Stoichiometry• Gas mixtures• Kinetic molecular theory• Deviations from Ideal

Substances that exist as gasesElements that exist as gases at 250C and 1 atmosphere

5.1

5.1

• Gases assume the volume and shape of their containers.

• Gases are the most compressible state of matter.

• Gases will mix evenly and completely when confined to the same container.

• Gases have much lower densities than liquids and solids.

5.1

Physical Characteristics of Gases

Our Atmosphere:• exerts pressure on earth• more at sea level• less on mountain top• The air we breathe:• 80% N2• 20% O2

Sea level1 atm

4 miles 0.5 atm

10 miles 0.2 atm

SI Units of Pressure

1 pascal (Pa) = 1 N/m2

[N = newton = 1 kg (m/s2)]

Atmospheric pressure1 atm = 760 mmHg = 760 torr(1 torr = 1mm Hg)

1 atm = 101,325 Pa1 atm = 101 kPa 5.2Barometer

Pressure = ForceArea

(force = mass x acceleration)

Pressure of Gas

5.2

Manometers Used to Measure Gas PressuresClosed endPgas< 1atm

Open endPgas> 1atm

The Gas Laws

Pressure and Temperature• Pressure = force/area(we will use torr, mm Hg, Pa & atm)• Always use Kelvin temperature (K)K = ° C + 273

4 variables are involved:• P = pressure• V = volume• n = # of moles• T = temperature (in Kelvin)

The Pressure-Volume Relationship: Boyle’s Law

P 1/VP x V = constantP1 x V1 = P2 x V2

If temperature T and amount of gas n does not change, pressure P is inversely proportional to volume V.

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

P1 x V1 = P2 x V2

P1 = 726 mmHg

V1 = 946 mL

P2 = ?

V2 = 154 mL

P2 = P1 x V1

V2

726 mmHg x 946 mL154 mL= = 4460 mmHg

5.3

P x V = constant

As T increases V increases 5.3

Temperature-Volume Relationship: Charle’s Law

Variation of gas volume with temperatureat constant pressure.

5.3T (K) = t (0C) + 273.15

Temperature must bein Kelvin

Charles’ & Gay-Lussac’s Law

V TV = constant x TV1/T1 = V2 /T2

If n and P do NOT change, then V is proportional to T

What is the final temperature (in K), under constant-pressure condition, when a sample of hydrogen gas initially at 88oC and 9.6 L is cooled until its finial volume is 3.4 L?

V1 = 9.6 L

T1 =(88+273.15) K=361.15K

V2 = 3.4 L

T2 = ?

T2 = V2 x T1

V1

3.4 L x 361.15 K9.6 L= = 128K

5.3

V1 /T1 = V2 /T2

Group practice problem chap4-5:Q13

The Volume-amount Relationship: Avogadro’s Law

V number of moles (n)

V = constant x n

V1 / n1 = V2 / n2

5.3

If P and T do NOT change, then n is proportional to V

Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?

4NH3 + 5O2 4NO + 6H2O

1 mole NH3 1 mole NO

At constant T and P

1 volume NH3 1 volume NO

5.3

Ideal Gas Equation

5.4

Charles’ law: V T(at constant n and P)

Avogadro’s law: V n(at constant P and T)

Boyle’s law: V (at constant n and T)1P

V nTP

V = constant x = R nTP

nTP

R is the gas constant

PV = nRT

Ideal gas is a hypothetical gas whose pressure-volume-temperature behavior can be completely accounted for by the ideal gas equation.

The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).

PV = nRT

R = PVnT

=(1 atm)(22.414L)(1 mol)(273.15 K)

R = 0.082057 L • atm / (mol • K)=8.314J/(K·mol) = 8.314 L·kPa/(K·mol)

5.4

Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

Molar volume of gas• 1 mole of gas at STP = 22.4 Liters• Example:2 moles of gas at STP = 44.8 L

In calculation, the units of R must match those for P,V,T and n.

What is the pressure of the gas (in atm) when 5.0 moles of CO gas are present in a container of 20.0 L at 27 oC?

n= 5.0mole, V=20.0L, T= 27 oC=(27+273.15)K=300.15K

PV=nRT

P=nRT/V= 5.0mole*0.082 L• atm / (mol • K)*300.15K/20.0L=6.15L

Group practice problem chap4-5:Q15

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

PV = nRT

V = nRTP

T = 0 0C = 273.15 K

P = 1 atm

n = 49.8 g x 1 mol HCl36.45 g HCl

= 1.37 mol

V =1 atm

1.37 mol x 0.0821 x 273.15 KL•atmmol•K

V = 30.6 L

5.4Practice problem of chap4-5: Q14

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?

PV = nRT n, V and R are constantnRV = P

T = constant

P1

T1

P2

T2=

P1 = 1.20 atmT1 = 291 K

P2 = ?T2 = 358 K

P2 = P1 x T2

T1

= 1.20 atm x 358 K291 K

= 1.48 atm

5.4

R

P1V1

n1T1R=

The combined gas law

P2V2

n2T2

P1V1

n1T1=

P2V2

n2T2R=

If n1=n2

P2V2

T2

P1V1

T1=

Density (d) Calculations

d = mV = PM

RTm is the mass of the gas in gM is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance

dRTPM = d is the density of the gas in g/L

5.4

What is the density of HCl gas in grams per liter at 700 mmHg and 25 oC?

d = mV = PM

RT

P=700mmHg=700/760atm=0.92atmT= 25 oC=25+273.15K=298.15K

d=0.92 atm x 36.45g/mol

x 298.15 K 0.0821 L•atmmol•K

=1.37g/L

Group practice problem chap4-5:Q18

What is the molar mass (g/mol) of 7.10 grams of gas whose volume is 5.40 L at 741 torr and 40 oC?

5.4

dRTPM = d = m

V7.10 g5.40 L

= = 1.31 gL

M =1.31 g

L

0.975 atm

x 0.0821 x 313.15 KL•atmmol•K

M = 34.6 g/mol

T=313.15K P= 741torr=741/760atm=0.975atm

Group practice problem chap4-5:Q17

Gas Stoichiometry

The combustion process for methane is  CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) 

If 15.0 moles of methane are reacted, what is the volume of carbon dioxide (in L) produced at 23.0 oC and 0.985 atm?

V = nRT

P

15mol x 0.0821 x 296.15 KL•atmmol•K

0.985 atm= = 369.8 L

5.5

x 1CO2/1CH4 15 mole CH4 ---------------- 15 mole CO2

Group practice problem chap4-5:Q20

Dalton’s Law of Partial PressuresPartial pressure is the pressure of the individual gas in the mixture.

V and T are

constant

P1 P2 Ptotal = P1 + P2

5.6

Consider a case in which two gases, A and B, are in a container of volume V.

PA = nARTV

PB = nBRTV

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB XA = nA

nA + nBXB =

nB

nA + nB

PA = XA PT PB = XB PT

Pi = Xi PT

5.6

mole fraction (Xi) = ni

nT

A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?

Pi = Xi PT

Xpropane = 0.116

8.24 + 0.421 + 0.116

PT = 1.37 atm

= 0.0132

Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

5.6

Group practice problem chap4-5:Q21

Kinetic Molecular Theory of Gases1. A gas is composed of molecules that are separated from

each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.

2. Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic.

3. Gas molecules exert neither attractive nor repulsive forces on one another.

4. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy

5.7KE = ½ mu2 u2 = (u1

2 + u22 + …+ uN

2)/N KE T

Mean square speed

Application to the Gas Laws

• Compressibility of Gases (assumption 1)

• Boyle’s LawP collision rate with wallCollision rate number density (number of molecules

per unit volume)

Number density 1/VP 1/V

• Charles’ LawP collision rate with wallCollision rate average kinetic energy of gas moleculesAverage kinetic energy TP T expand volume until gas pressure is balanced

by the external constant pressure 5.7

• Avogadro’s LawP collision rate with wallCollision rate number densityNumber density nP n

• Dalton’s Law of Partial PressuresMolecules do not attract or repel one anotherP exerted by one type of molecule is unaffected by the

presence of another gasPtotal = Pi

5.7

Application to the Gas Laws

Distribution of molecular speeds

5.7

At constant temperature, the average kinetic energy and the mean square speed will remain unchanged. The motion of molecules are random.How many molecules are moving at a particular speed?

The distribution of speedsfor nitrogen gas molecules

at three different temperatures

The distribution of speedsof three different gases

at the same temperature

5.7

Maxwell speed distribution curve

Root-mean-square speed

Total KE of one mole gas = (½ mu2)NA= 3/2RT

urms = 3RTM

NAm=M

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.

5.7

NH3

17 g/molHCl

36 g/mol

NH4Cl

r1

r2

M2

M1=

Under the same conditions of T and P, rates of diffusion for gases are inversely proportional to the square roots of their molar masses. Gramham’s Law

Gas effusion is the is the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening.

5.7

r1

r2

t2

t1

M2

M1= =

Nickel forms a gaseous compound of the formula Ni(CO)x.What is the value of x given that under the same conditions methane (CH4) effuses 3.3 times faster than the compound?

r1 = 3.3 x r2

M1 = 16 g/mol

M2 = r1

r2( )2

x M1 = (3.3)2 x 16 = 174.2

58.7 + x • 28 = 174.2 x = 4.1 ~ 4Group practice problems chap4-5:Q22

Deviations from Ideal Behavior

1 mole of ideal gas

PV = nRT

n = PVRT = 1.0

5.8

Repulsive Forces

Attractive Forces

Assumption of gas laws and kinetic theory:Gas molecules do not exert any force, either attractive or repulsive;The volume of the molecules are negligible;

Effect of intermolecular forces on the pressure exerted by a gas.

5.8

The speed of the gas molecule that is moving toward the wall is reduced by the attractive force of neighbor molecules.Consequently the measured pressure is lower than the pressure the gas would exert if it behave ideally.

5.8

Van der Waals equationnonideal gas

P + (V – nb) = nRTan2

V2( )}

correctedpressure

}

correctedvolume