Chapter 5 - Gas Laws-Student

8/17/2019 Chapter 5 - Gas Laws-Student

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Chapter 5: Gas Laws5.1 Objective5.2 Introduction5.3 Particles Arrangement and Gas Properties5.4 Boyle’s Law 5.5 Charles’s Law 5.6 Combined Gas Law5.7 Avogadro’s Law 5.8 Ideal Gas Law5.9 Molar Mass and Gas Density5.10 Dalton’s Law 5.11 Graham’s Law 5.12 Exercise Questions

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5.1 Objectives

At the end of lecture on this chapter, students willbe able

to explain the arrangement of

particles/molecules and properties of gases,

to explain Boyle, Charles, Avogadro, CombinedGas, Ideal Gas, Dalton and Graham Laws,

to solve numerical problems on gas laws.

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5.2 Introduction

Elements (light blue) that exist as gases at 25 oCand 1 atmosphere

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Some Substances Found as Gasesat 1 atm and 25 oC

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In the gas phase, particles/molecules arerelatively far apart and move randomly compare

to liquid and solid phases.

This makes gases very compressible andexpandable to fill the volume of its container.

Gases exert a pressure on the walls of theircontainer.

5.3 Particles Arrangement

and Properties of Gas

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Collisions causePressure

The pressure of a gas is caused by thecollision of molecules against the sides ofthe container. The force of the collisionagainst the container can be calculatedby Newton’s Second Law of Motion:

F = ma

where F = force, m = mass in kg and a isthe acceleration in m/s2.

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Pressure

Pressure is defined as force, F per unit area, A :

Pressure units:

1 pascal (Pa) = 1 N/m2

1 atm = 760 mmHg = 760 torr

1 atm = 101,325 Pa = 101.325 kPa

F P=

A

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Atmospheric Pressure The atmospheric pressure can be measured using a barometer.

The standard atmosphere is defined as the pressure exerted by

a mercury column of exactly 76 cm in height when the

density of mercury equals 13.6 g/cm3.

Units of Standard Atmospheric Pressure

1.00 atm

76 cm Hg = 760 mm Hg = 760 torr101.325 kPa

1.01325 bar

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Problem 1

If the barometer reads 753.3 mm Hg, what is

the atmospheric pressure in (i) atm and (ii) kPa?

=100.4 kPa

753.3 mm Hg

0.9912 atm

X

Solution:

X

(i)

(ii)

= 0.9912 atmHgmm760atm1

1 atm

101.325 kPa

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The gas laws are experimental relationships among

pressure (P ), volume (V ), temperature (T ), andmoles (n ) for individual gas.

All gases behave similarly. The gas laws assume

ideal behavior that one exhibits simple linearrelationships among volume, pressure, temperatureand gas quantity .

Gases only approach ideal behavior at low pressure(< 1 atm) and high temperature

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Units You Have To Be FamiliarWhen Using Gas Laws

Pressure: atm, mmHg, torr, Pa1 atm =760 mmHg =760 torrs =101,325 Pa =101.325 kPa

Volume: L = dm3, mL = cc = cm3, m3

1 L = 1 dm3 = 1000 cc =1000 cm3 =1000 mL

Temperature : Kelvin, K where T(K) =T( oC)+273

Gas quantity : Mole,n where

=mass (g)

molar mass (g/mole) n

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Lord Kelvin proposed an absolute temperature scaledefined by:

T(K) = T(

O

C) + 273

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Standard Temperature Pressure (STP)

Known as standard state.

The values are P = 1 atm, T = 273 K. Applied to all gases at ideal state.

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5.4 Boyle’s Law

P V = P V 1 1 2 2

V 1/P ; when P V and if P V So, PV = k (k is a constant)

For initial (1) and final (2) states:

In 1662, Robert Boyle discovered that volume is inversely

proportional to pressure (V 1/P )

Boyle’s Law states that at constant temperature (T ) thevolume (V ) of a fixed amount of gas (n ) is inversely

proportional to its pressure (P ).

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Graphs of:

(b) 1/V vs P (a) V vs P

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A sample of chlorine gas occupies a volume of 946 mL at apressure of 726 mmHg. What is the pressure of the gas (in

mmHg) if the volume is reduced at constant temperature to154 mL?

P 1V

1= P

2V

2 P 1 = 726 mmHg P 2= ?

V 1 = 946 mL V 2= 154 mL

P 2= P 1x V 1

V 2

726 mmHg x 946 mL154 mL

= = 4460 mmHg

PROBLEM 3

SOLUTION:

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Chemistry in Action:

Scuba Diving and the Gas Laws

P V

Depth (ft) Pressure(atm)

0 1

33 2

66 3

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5.5 Charles’s Law

At constant pressure and fixed amount of gas:

V T or V = kTwhere k is a proportionality constant.

For initial (1) and final (2) states

V V k

T T 1 2

1 2

= =

Charles discovered that volume is directlyproportional to temperature.

1 2

1 2

=V V

T T or

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Graph of Volume vs Temperature

A linear relationship between Vand T .

This relationship betweenvolume and temperaturedescribes a direct relationship.This means when temperature

increases, so does the volume.T , V

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The volume of a gas extrapolates to zero at

-273 C. This must be the lowest temperature

possible.

Absolute zero

(0 K)

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Variation of Gas Volume withTemperature at Constant Pressure.

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A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if

the pressure remains constant?

PROBLEM 4

SOLUTION:V 1 = 3.20 L V 2 = 1.54 L

T 1 =125 + 273=398 K T 2 = ?

T 2 = 1.54 L x 398 K3.20 L= = 192 KV 2 x T 1

V 1

V V

T T

1 2

1 2

=

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PROBLEM 5

A sample of a gas at 15 °C and 1 atm has a volume

of 2.58 L. What volume will the gas occupy at 38 °Cand 1 atm?

SOLUTION:V 1 = 2.58 L V 2 = ?

T 1 =15 + 273=288 K T 2 = 38 + 273 = 311 K

V 2 =V 1 x T 2

T 1

2.58 L x 311 K288 K

= = 2.79 L

V V

T T

1 2

1 2

=

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5.6 The Combined Gas Law

The combined gas law was derived from Boyle’s andCharles’s Laws at constant quantity of gas.

A direct relationship was observed.

As temperature increased, volume increased. As volumeincreased pressure decreased.

This resulted in a combined formula to calculate changesobserved in a gas due to changes in either temperature,

pressure or volume.

PV PV n T T

1 1 2 2

1 2

= ( constant) where 1 and 2 are initial andfinal states

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P 1 = 2.0 atm P 2= 1 atm

V 1 = 2 mL V 2 = ?

2.0 atm x 2 mL x 311 K1 atm x 288 K

= = 4.32 mL

SOLUTION:

PROBLEM 6

A sample of a gas at 15°C and 2.0 atm has a volume of 2 mL.What volume will the gas occupy at 38°C and 1 atm?

T 1 = (15+273) K =288 K T 2 = (38+273) K = 311 K

1 1 2

22 1

PV T V

P T

=1 1 2 2

1 2

=PV PV

T T

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5.7 Avogadro’s Law

Gases are usually measured by volume so a relationship

between volume and number of moles is needed.

In 1811, Avogadro proposed that:

“ Equal volumes of gases at the same temperature

and pressure contain equal numbers of molecules ”.

It follows that the volume of a gas at constanttemperature and pressure is proportional to the numberof moles.

V V

n n 1 2

1 2

=

V n (at constant T and P )

For initial (1 ) and final (2 ) state,

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Avogadro’s Law

“Equal volumes of gases at the same temperatureand pressure contain equal numbers of molecules”

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5.8 Ideal Gas Equation

Charles’s 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 )1

P

V = constant x = RnT

P

nT

P

R is the gas constant So, PV = nRT

Combining: or

This is the ideal gas equation

nT

V P

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PV = nRT R = PV

nT

= (1 atm)(22.4 L)(1 mol)(273 K)

= 0.0821 L.atm/mol.K

Ideal Gas Constant, R

Experiments show that at STP, 1 mole of an ideal gasoccupies 22.4 L in volume

R has other values for other sets of units .

R = 82.05 mL.atm/mol.K= 8.314 J/mol.K

= 1.987 cal/mol.K

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How many moles of N2 are in a 750 mL vessel at

26°C and 625 mm Hg?

V = 750 mLmL1000

L1 = 0.750 L

P = 625 mm HgHgmm760

atm1 = 0.822 atm

T = 26 + 273 = 299 K

=PV n RT

=0.822 atm 0.750 L

0.0821L.atm

mol.Kx 299 K

= 0.0251 mol

SOLUTION:

PROBLEM 7

PV

PV = nRT n = RT

x

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T = 0

0

C = 273 K

P = 1 atm

n = 49.8 g x1 mol HCl

36.5 g HCl= 1.36 mol

V = 1 atm

1.36 molx0.0821 x 273 KL.atmmol.K = 30.48 L

SOLUTION: nRT PV = nRT V = P

PROBLEM 8

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

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5.9 Density and Molar Mass of Gas

Different gases have the same volume at STP, but they

have different masses. From ideal gas equation:

Molar mass of gas, dRT M P

=PV nRT = or =m m PM

RT

M V RT

, Density of gasm PM

d V RT m : mass of gas in gram (g)

d : density in gram/liter (g/L)

M: molar mass in gram/mol(g/mol)

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PROBLEM 9

What are the densities of N2 and He at STP ?

SOLUTION:

=MP

d

RT

=atm.L0.0821mol.K x 273 K

g

mol28

x 1 atm= 1.250 g/L

He: = atm.L0.0821mol.K x 273 K

g4mol x 1 atm

= 0.1786 g/L

N2:

=

MP

d RT

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PROBLEM 10

A 2.10 L vessel contains 4.65 g of a gas at 1.00 atm and27.0 0C. What is the molar mass of the gas?

SOLUTION:dRT

PM =

d = mV 4.65 g2.10 L= = 2.21 gL

M = 2.21

g

L

1 atm

x 0.0821 x 300 KL•atm mol•K

= 54.6 g/mol

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Use of Ideal Gas Law in Gas

Stoichiometry

PROBLEM 11

What is the volume of CO2 (in liter) liberated at 3700C and 1.00

atm when 5.60 g of glucose are used up in the reaction

C6H12O6(s) + 6O2 (g ) 6CO2(g ) + 6H2O(l )

SOLUTION:(i) Check the balancing of the chemical equation:

: OK

If not, firstly you have to balance chemical equation.(ii) Write the stoichiometry:

1 mol C6H12O6 + 6 mol O2 6 mol CO2+ 6 mol H2O

C6H12O6(s) + 6O2(g ) 6CO2(g ) + 6H2O(g )

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5.60 g C6H12O6 1 mol C6H12O6180 g C6H12O6

x

6 mol CO2

1 mol C6H12O6x = 0.187 mol CO2

V =nRT

P

0.187 mol x 0.0821 x 643 KL•atm

mol•K

1.00 atm= = 9.85 L

(iii) Use the schematic to convert the given data into mole/gram

Amount of C6H12O6 used = 5.60 g

Mole of C6H12O6 used = = 0.0311 mol

From the stoichiometry 1 mol of C6H12O6 producing 6 mol CO2 .Therefore 0.0311 mol producing

0.0311 mol C6H12O6

(iv) Use the ideal gas law to calculate the volume of CO2

in liter

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5.10 Dalton’s Law of Partial

Pressure• Gases mix together uniformly.

• Total pressure depend on the individual/partial pressure of the

gas components.

• Dalton’s Law states that the total pressure of a mixture of

gases is the sum of the partial pressures of the components ofthe mixture.

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Gas Mixtures

Consider a mixture of gas 1, gas 2 and gas 3 with n 1

, n 2

and n 3 are the number of mole respectively.

Partial pressure is the pressure of the individual gaswould exert if it were the only gas present.

P 1 for gas 1 P 2 for gas 2 and P 3 for gas 3

If the gases in a mixture behave ideally then,

,

1 1

= RT

P n V

RT P n

V 2 2=

RT

P n V 3 3and =

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Since P t = P 1 + P 2 + P 3, then

t

RT RT RT P n n n

V V V 1 2 3

= + +

RT

n + n + n V 1 2 3= ( )

=t

RT n

V

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PROBLEM 12

What are the partial and total pressures of 2.00 g H2

and 8.00 g N2 in a 10.0 L vessel at 273 K?SOLUTION:

2.00 g H2 = 1.00 mol H2

8.00 g N2 = 0.286 mol N2

mol H2× 2 g H2

mol N2×

28 g N2

H2:

N2:

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PH2

1.00 mol x

273 K

10.0 L

= 2.241 atm

PN2

0.286 mol x = 0.641 atm

atm.L0.0821mol.K

x

273 K

10.0 L

atm.L0.0821mol.K

x

Ptotal = 2 2H NP + P = 2.242 + 0.641 = 2.882 atm

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Mole Fractions

The ratio of partial pressure to total pressure can

be expressed as:

PPt

1

RTnV

1

RTnV

t

nn t

1

= X 1 mole fraction of gas1

The mole fractions in a mixture must sum to 1:

X X1 2+ + . . . nn + + n nt t

1 2 ... + + ( . . . )n nn t

1 2 = 1

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PROBLEM 13

On a 25°C day with 100% humidity, the mole fractionof H

2

O vapor is 0.031. What is the partial pressure ofH2O vapor? What is the mole fraction of H2O if therelative humidity is 60%?

SOLUTION:

= 0.031 = 24 mmHg´ 760 mmHgatmOHOH P XP 22

24 mmHg 0.60 = 14 mmHg

14 mmHg

760 mmHg= 0.018

OHP 2 =

OH X 2 =

(i)

(ii)

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Measuring Gases

To measure the amount of gas produced in a reaction,

it is often collected over water.

Reaction of magnesium with HCl:

Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g )

OHHgas 22 PPP +

= Patm

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PROBLEM 13

The reaction of a sample containing Mg with

excess HCl at 22C and 753.2 mmHg yielded1207 mL of gas. What was the mass of Mg?

SOLUTION:

T = 22°C:

= 753.2 mmHg - 19.83 mmHg

´ 1760

atmmmHg

= 0.965 atm

=19.83 mmHg

= 733.4 mmHg

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The mass in gram?

T(K) = 22°C + 273 = 295 K

V(L)=1207 mL ´ 1

1000L

mL= 1.207 L

n PVRT

= 0.965 atm x 1.207 L

0 0821.atm Lmol K

× ×

x 295 K

= 0.0481 mol H2 24´

1.3 g Mgmol Mg= 0.0481 mol Mg

= 1.17 g Mg47

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PROBLEM 14

A student generates oxygen gas and collects it over water. If

the volume of the gas is 245 mL and the barometric pressure

is 758 torr at 25oC, what is the volume of the “dry” oxygen gas

at STP?

SOLUTION:

Pwater = 23.8 torr at 25oC; PO2 = Pbar - Pwater = (758 - 23.8)

= 734.2 torrP1= PO2 = 734.2 torr; P2= SP = 760 torr

T1= 298 K ; T2= 273 K ; V1= 245 mL ; V2= ?

(V1

P1

/T1

) = (V2

P2

/T2

)

V2= (V1P1T2)/(T1P2) = (245 mL)(734.2 torr)(273 K)

(298 K)(760 torr) = 217 mL

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Chapter 5 - Gas Laws-Student