Chapter 5The Gaseous

State

GAS PROPERTIESالغازات خصائص

5 | 2

Gases differ from liquids and solids:: يلي بما والصلب السوائل عن يختلف الغاز

They are compressible. االنضغاط او للضغط القابلية

The variables pressure, volume, temperature, and amount are related.

جميعها هي والكميات الحرارة، الحجم، الضغط، ناحية من التنوععالقة ذات

PRESSUREبالضغط Pressure, Pلنبدأ

The force exerted per unit area بالرمز له ويرمز من Pالضغط وحدة عل المؤثرة القوة هو

المساحة

It can be given by two equations: توضيحها يمكن بمعادلتين

The SI unit for pressure is the pascal, Pa.

5 | 3(pascal)Pasm

kgm

s

m

m

kg223

dgh P A

FP

(pascal)Pasm

kg

ms

mkg

22

2

PRESSURE

Other Pressure Unitsatmosphere, atmmmHgtorrBar

العادي الجوي زئبق 760الضغط ملم هو الضغط هذا عل الغاز مل 100حجم

5 | 4

PRESSURE

A barometer is a device for measuring the pressure of the atmosphere.

الجوي الضغط لقياس جهازA manometer is a device for measuring the pressure of a gas or liquid in a vessel. االنابيب في السائل او الغاز ضغط لقياس جهاز

5 | 5

EXAMPLE 1

The water column would be higher because its density is less by a factor equal to the density of mercury to the density of water.

بعامل اكثر وسترتفع اقل الماء كثافة طبعاكثافة بنسبو مضروب الزئبق لكثافة مساوي

معادلة حسب الماء كثافة الى الزئبقالضغط

5 | 6

OHOHHgHg 22hgdhgd

gdhP

OH

HgHgOH

2

2

d

dhh

اكثر سيرتفع العمودين اي بالماء وواحد بالزئبق مليء واحد الجوي الضغط لقياس بارميتر جهازين لديك ان لنفترض

GAS LAWSالغاز قانون

5 | 7

Empirical Gas Laws التجريبي الغازات قانون

All gases behave quite simply with respect to temperature, pressure, volume, and molar amount. By holding two of these physical properties constant, it becomes possible to show a simple relationship between the other two properties.

والحجم الضغط الحرارة لدرجة تعرضها بحسب الغازات تتصرف . الخواص هذه من اثنان تثبيت تم اذا الموالت كمية و فيه الموضوعة

المتبقيتين الخاصيتين بين العالقة اظهار فباالمكان للغازات الفيزيائيةThe studies leading to the empirical gas laws occurred from the mid-17th century to the mid-19th century.

من موجودة التجريبية الغازات نظرية على الموجودة الدراسات. عشر التاسع القرن وحتى عشر السابع القرن منتصف

BOYLE’S LAW

5 | 8

Boyle’s Law بويل قانون

The volume of a sample of gas at constant temperature varies inversely with the applied pressure.

الذي الضغط مع عكسيا تختلف ثابتة حرارة درجات في الغاز عينة حجمله تتعرض

The mathematical relationship:

In equation form:

وعليه معين غاز حجم هناك كان فلو مترافقين دائما والضغط الحجم اذنفي اختالف او زيادة اي مع متناسبة الحجم محصلة تبقى معين ضغط

الضغط

PV

1

ffii

constant

VPVP

PV

BOYLE’S LAW

5 | 9

Figure A shows the plot of V versus P for 1.000 g O2 at 0°C. This plot is nonlinear.

Figure B shows the plot of (1/V) versus P for 1.000 g O2 at 0°C. This plot is linear, illustrating the inverse relationship.

BOYLE’S LAW

5 | 10

At one atmosphere the volume of the gas is 100 mL. When pressure is doubled, the volume is halved to 50 mL. When pressure is tripled, the volume decreases to one-third, 33 mL.

BOYLE’S LAW

5 | 11

When a 1.00-g sample of O2 gas at 0C is placed in a containerat a pressure of 0.50 atm, it occupies a volume of 1.40 L.

When the pressure on the O2 is doubled to 1.0 atm, the volume is reduced to 0.70 L, half the original volume.

EXAMPLE 2

5 | 12

A volume of oxygen gas occupies 38.7 mL at 751 mmHg and 21°C. What is the volume if the pressure changes to 359 mmHg while the temperature remains constant?

Vi = 38.7 mLPi = 751 mmHgTi = 21°C

Vf = ?Pf = 359 mmHgTf = 21°C

f

iif P

VPV

f

iif P

VPV

5 | 13

Vi = 38.7 mLPi = 751 mmHgTi = 21°C

Vf = ?Pf = 359 mmHgTf = 21°C

mmHg)(359

mmHg)mL)(751(38.7f V

= 81.0 mL(3 significant figures)

Example 2 (Cont)

VOLUME - TEMPERATURE

5 | 14A graph of V versus T is linear. Note that all lines cross

zero volume at the same temperature, -273.15°C.

ABSOLUTE ZEROالمطلق الصفر ظارة

5 | 15

The temperature -273.15°C is called absolute zero. It is the temperature at which the volume of a gas is hypothetically zero.

- حرارة درجة درجة 273.15تسمى وهي المطلق الصفر بدرجة. صفر نظريا الغازات احجام تكون عندها التي الحرارة

This is the basis of the absolute temperature scale, the Kelvin scale (K). لتدريج االساسية الدرجة بالمناسبة وهي

الحراري كالفن

CHARLES’ LAWتشارل قانون

5 | 16

Charles’ LawThe volume of a sample of gas at constant pressure is directly proportional to the absolute temperature (K).

مع مباشرة يتناسب ثابت ضغط درجة على الغاز من عينة حجم. بالكالفن المطلقة الحرارة قيمة

The mathematical relationship:

In equation form:

TV

f

f

i

i

constant

T

V

T

VT

V

CHARLES’ LAW

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A balloon was immersed in liquid nitrogen (black container) and is shown immediately after being removed. It shrank because air inside contracts in volume.

As the air inside warms, the balloon expands to its orginial size.

CHARLES’ LAW

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A 1.0-g sample of O2 at a temperature of 100 K and a pressure of 1.0 atm occupies a volume of 0.26 L.

When the absolute temperature of the sample is raised to 200 K, the volume of the O2 is doubled to 0.52 L.

EXAMPLE 3

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You prepared carbon dioxide by adding HCl(aq) to marble chips, CaCO3. According to your calculations, you should obtain 79.4 mL of CO2 at 0°C and 760 mmHg. How many milliliters of gas would you obtain at 27°C?

Vi = 79.4 mLPi = 760 mmHgTi = 0°C = 273 K

Vf = ?Pf = 760 mmHgTf = 27°C = 300. K

i

iff T

VTV

i

iff T

VTV

5 | 20

Vi = 79.4 mLPi = 760 mmHgTi = 0°C = 273 K

Vf = ?Pf = 760 mmHgTf = 27°C = 300. K

K)(273

mL)K)(79.4(300.f V

= 87.3 mL(3 significant figures)

Example 3 (Cont)

GAY-LUSSAC’S LAWلوساك جاي قانون

5 | 21

Gay-Lussac’s Law

There is an analogous relationship to Charles’ Law that relates pressure and temperature. This relationship is called Gay-Lussac’s Law, and is described as follows:

قيم منخاللها ترتبط والتي شارل لقانون متناظرة عالقة هناكبالحرارة الضعط

The mathematical relationship:

In equation form:

P T

P

Tconstant

Pi

Ti

Pf

Tf

GAY-LUSSAC’S LAW

5 | 22

Interestingly, a plot of pressure versus temperature can also be extrapolated to estimate absolute zero, and values from Charles’ Law and Gay-Lussac’s Law give equivalent estimates.

نقطة تقدير الممكن من والحرارة الضغط بين العالقة رسمومنها والحجم الحرارة بين تشارلز وقيم فيها المطلق الصفر

غاي قانون اشراك يمكن ايضاProblems using Gas-Lussac’s Law are done in the

same fashion as Charles’ Law, and though we will not do any here, you will be accountable for solving them as well.

COMBINED GAS LAWالمشترك الغازات قانون الى ينقلنا هذا

القوانين هذه من

5 | 23

Combined Gas LawThe volume of a sample of gas at constant pressure is inversely proportional to the pressure and directly proportional to the absolute temperature.

ثابتة ضغط قيمة عند الغاز عينة حجم بان القانون هذا وينصبالكالفن المطلقة الحرارة مع وطرديا الضغط مع عكسيا تتناسب

The mathematical relationship:

In equation form:

P

TV

f

ff

i

ii

constant

T

VP

T

VPT

PV

EXAMPLE 4

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Divers working from a North Sea drilling platform experience pressure of 5.0 × 101 atm at a depth of 5.0 × 102 m. If a balloon is inflated to a volume of 5.0 L (the volume of the lung) at that depth at a water temperature of 4°C, what would the volume of the balloon be on the surface (1.0 atm pressure) at a temperature of 11°C?

Vi = 5.0 LPi = 5.0 × 101 atmTi = 4°C = 277 K

Vf = ?Pf = 1.0 atmTf = 11°C = 284 K

EXAMPLE 4 (CONT)

5 | 25

Vi = 5.0 LPi = 5.0 × 101 atmTi = 4°C = 277 K

Vf = ?Pf = 1.0 atmTf = 11°C = 284. K

fi

iiff PT

VPTV

atm)K)(1.0(277

L)atm)(5.010xK)(5.0(284 1

f V

= 2.6 x 102 L(2 significant figures)

EXAMPLE 5

5 | 26

EXAMPLE 5 (CONT)

a. Decreasing the temperature at a constant pressure results in a decrease in volume. Subsequently increasing the volume at a constant temperature results in a decrease in pressure.

b. Increasing the temperature at a constant pressure results in an increase in volume. Subsequently decreasing the volume at a constant temperature results in an increase in pressure.

5 | 27

AVOGADRO’S LAWافوجادرو قانون

5 | 28

Avogadro’s LawEqual volumes of any two gases at the same temperature and pressure contain the same number of molecules. We could also say,

الحرارة درجة نفس على مختلفين غازين من المتسوية االحجامنقول ان ونستطيع الجزيئات، من العدد نفس على تحتوي والضغط

ايضا:

V n and V

n = K

STANDARD TEMPERATURE AND PRESSURE والضغط الحرارة ظروف المثالية

5 | 29

Standard Temperature and Pressure (STP)The reference condition for gases, chosen by convention to be exactly 0°C and 1 atm pressure.

مئوي صفر بحرارة تكون ان على اتفق للغازات المرجعية الظروف هياتموسفير وواحد

The molar volume, Vm, of a gas at STP is 22.4 L/mol. هو المثالية الظروف في للغاز المولي مول 22.4الجم لكل لتر

The volume of the yellow box is 22.4 L. To its left is a basketball.

IDEAL GAS CONSTANT

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Ideal Gas Law

المثالي الغاز قانونThe ideal gas law is given by the

equation

PV=nRT

The molar gas constant, R, is the constant of proportionality that relates the molar volume of a gas to T/P.

التناسب ثابت هو المولي الغاز ثابتللغاز المولي الحجم بين يربط الذي

والضغط الحرارة T/Pمع

constant. is whereRT

V

RT

VP n

nRT PV

EXAMPLE 6

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You put varying amounts of a gas into a given container at a given temperature. Use the ideal gas law to show that the amount (moles) of gas is proportional to the pressure at constant temperature and volume.

. معينة حرارة بدرجة وعاء في الغاز من مختلفة كميات وضعتتتناب بالموالت الغاز كمية بأن الظهار المثالي الغاز قانون اتخدم

ثابتين وحجم حرارة درجات في الضغط مع

constant. is whereRT

V

RT

VP n

nRT PV

EXAMPLE 7

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A 50.0-L cylinder of nitrogen, N2, has a pressure of 17.1 atm at 23°C. What is the mass of nitrogen in the cylinder?

50 ضغطها النيترجين على تحتوي اسطوانة على 17.1لتر اتموسفيرحرارة . 23درجة النيتروجين؟ كتلة ماهي V = 50.0 Lمئوية

P = 17.1 atmT = 23°C = 296 K

RT

PVn

K)(296Kmol

atmL0.08206

L)atm)(50.0(17.1

n

mass = 986 g(3 significant figures)mol

g28.01mol35.20mass

MOLAR MASS AND DENSITYوالكثافة المولية الكتلة

5 | 33

P

dRTM

RT

PMd m

m or

Gas Density and Molar Mass

Using the ideal gas law, it is possible to calculate the moles in 1 L at a given temperature and pressure. The number of moles can then be converted to grams (per liter).

الموالت حساب الممكن من المثالي الغاز قانون باستخدامنحول وبعدها معينين، وضغط حرارة درجة على لتر في

غرامات الى الموالت

To find molar mass, find the moles of gas, and then find the ratio of mass to moles.

In equation form:

EXAMPLE 8

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What is the density of methane gas (natural gas), CH4, at 125°C and 3.50 atm?

Mm = 16.04 g/molP = 3.50 atmT = 125°C = 398 K

RT

PMd m

K)(398Kmol

atmL0.08206

atm))(3.50mol

g(16.04

dfigures)tsignifican(3L

g1.72d

EXAMPLE 9

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A 500.0-mL flask containing a sample of octane (a component of gasoline) is placed in a boiling water bath in Denver, where the atmospheric pressure is 634 mmHg and water boils at 95.0°C. The mass of the vapor required to fill the flask is 1.57 g. What is the molar mass of octane? (Note: The empirical formula of octane is C4H9.) What is the molecular formula of octane?

EXAMPLE 9 (CONT)

5 | 36

P = 634 mmHg = 0.8342 atm P

dRTM m

atm)(0.8342

K368.2Kmol

atmL0.08206

L

g3.140

m

M

figures)tsignifican(3mol

g114mM

d = 1.57 g/0.5000 L = 3.140 g/L

T = 95.0°C = 368.2 K

EXAMPLE 9 (CONT)

2

mol

g57

mol

g114

n

5 | 37

Molecular formula: C8H18

Molar mass = 114 g/molEmpirical formula: C4H9

Empirical formula molar mass = 57 g/mol

EXAMPLE 10

5 | 38

EXAMPLE 10 (CONT)

Assume the flasks are closed.

a. All flasks contain the same number of atoms.

b. The gas with the highest molar mass, Xe, has the greatest density.

c. The flask at the highest temperature (the one containing He) has the highest pressure.

d. The number of atoms is unchanged.

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STOICHIOMETRY AND THE GAS LAWS

5 | 40

Stoichiometry and Gas VolumesUse the ideal gas law to find moles from a given volume, pressure, and temperature, and vice versa.

EXAMPLE 11

5 | 41

When a 2.0-L bottle of concentrated HCl was spilled, 1.2 kg of CaCO3 was required to neutralize the spill. What volume of CO2 was released by the neutralization at 735 mmHg and 20.°C?

EXAMPLE 11 (CONT)

1.2 x103g CaCO3 1mol CaCO3

100.09 g CaCO3

1mol CO2

1mol CaCO3

5 | 42

First, write the balanced chemical equation:

CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

Moles of CO2 produced = 11.99 mol

Second, calculate the moles of CO2 produced:

Molar mass of CaCO3 = 100.09 g/mol

EXAMPLE 11 (CONT)

5 | 43

n = 12.0 molP = 735 mmHg = 0.967 atmT = 20°C = 293 K

P

nRTV

atm)(0.967

K)(293Kmol

atmL0.08206mol12.0

V

= 3.0 × 102 L(2 significant figures)

DALTON’S LAW

5 | 44

Gas MixturesDalton found that in a mixture of unreactive gases each gas acts as if it were the only the only gas in the mixture as far as pressure is concerned.

DALTON’S LAW

5 | 45

Originally (left), flask A contains He at 152 mmHg and flask B contains O2 at 608 mmHg. Flask A is then filled with oil forcing the He into flask B (right). The new pressure in flask B is 760 mmHg

DALTON’S LAW

Partial PressureThe pressure exerted by a particular gas in a mixture

Dalton’s Law of Partial PressuresThe sum of the partial pressures of all the different gases in a mixture is equal to the total pressure of the mixture:

P = PA + PB + PC + . . .

5 | 46

EXAMPLE 12

5 | 47

A 100.0-mL sample of air exhaled from the lungs is analyzed and found to contain 0.0830 g N2, 0.0194 g O2, 0.00640 g CO2, and 0.00441 g water vapor at 35°C. What is the partial pressure of each component and the total pressure of the sample?

EXAMPLE 12 (CONT)

5 | 48

mL10

L1mL100.0

K308Kmol

atmL0.08206

Ng28.01

Nmol1Ng0.0830

3

2

22

N2P

mL10

L1mL100.0

K308Kmol

atmL0.08206

Og32.00

Omol1Og0.0194

3

2

22

O2P

mL10

L1mL100.0

K308Kmol

atmL0.08206

COg44.01

COmol1COg0.00640

3

2

22

CO2P

mL10

L1mL100.0

K308Kmol

atmL0.08206

OHg18.01

OHmol1OHg0.00441

3

2

22

OH2P

atm0.749

atm0.153

atm0.0368

atm0.0619

EXAMPLE 12 (CONT)

OHCOON 2222PPPPP

5 | 49

atm0.7492N P

atm0.1532O P

atm0.03682CO P

atm0.0619OH2P

P = 1.00 atm

EXAMPLE 13

5 | 50

The partial pressure of air in the alveoli (the air sacs in the lungs) is as follows: nitrogen, 570.0 mmHg; oxygen, 103.0 mmHg; carbon dioxide, 40.0 mmHg; and water vapor, 47.0 mmHg. What is the mole fraction of each component of the alveolar air?

mmHg40.02CO P

Hg103.02O P

mmHg570.02N P

mmHg47.0OH2P

EXAMPLE 13 (CONT)

5 | 51

OHCOON 2222PPPPP

570.0 mmHg103.0 mmHg

40.0 mmHg47.0 mmHg

P = 760.0 mmHg

EXAMPLE 13 (CONT)

5 | 52

Mole fraction of N2

Mole fraction of H2OMole fraction of CO2

Mole fraction of O2

mmHg760.0

mmHg47.0

mmHg760.0

mmHg40.0

mmHg760.0

mmHg103.0

mmHg760.0

mmHg570.0

Mole fraction N2 = 0.7500

Mole fraction O2 = 0.1355

Mole fraction CO2 = 0.0526

Mole fraction O2 = 0.0618

EXAMPLE 14

5 | 53

a. Nothing happens to the pressure of H2.b. The pressures are equal because the moles are

equal.c. The total pressure is the sum of the pressures of the

two gases. Because the pressures are equal, the total pressure is double the individual pressures.

DALTON’S LAW AND WATER VAPOR

5 | 54

Collecting Gas Over WaterGases are often collected over water. The result is a mixture of the gas and water vapor. The total pressure is equal to the sum of the gas pressure and the vapor pressure of water. The partial pressure of water depends only on temperature and is known (Table 5.6).The pressure of the gas can then be found using Dalton’s law of partial pressures

COLLECTING GAS OVER WATER

5 | 55

The reaction of Zn(s) with HCl(aq) produces hydrogen gas according to the following reaction:

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

The next slide illustrates the apparatus used to collect the hydrogen. The result is a mixture of hydrogen and water vapor.

APPARATUS FOR TRAPPING GAS

5 | 56

WATER VAPOR

5 | 57

5.6)Table(See

mmHg16.5PC,19At

mmHg769

OH2

P

mmHg16.5mmHg7692

22

22

H

OHH

OHH

P

PPP

PPP

mmHg752.52H P

places)decimal(no

mmHg7532H P

EXAMPLE 15

5 | 58

You prepare nitrogen gas by heating ammonium nitrite:NH4NO2(s) N2(g) + 2H2O(l)

If you collected the nitrogen over water at 23°C and 727 mmHg, how many liters of gas would you obtain from 5.68 g NH4NO2?

Molar mass NH4NO2

= 64.05 g/mol

P = 727 mmHgPvapor = 21.1 mmHgPgas = 706 mmHgT = 23°C = 296 K

3

2

3

33 CaCOmol1

COmol1

CaCOg64.04

CaComol1CaCOg5.68

5 | 59

P = 727 mmHgPvapor = 21.1 mmHgPgas = 706 mmHgT = 23°C = 296 K P

nRTV

Molar mass NH4NO2

= 64.04 g/mol

= 0.8869 mol CO2 gas

Example 15 (Cont)

5 | 60

P = 727 mmHgPvapor = 21.1 mmHgPgas = 706 mmHgT = 23°C = 296 Kn = 0.8869 mol

P

nRTV

mmHg760atm1

mmHg706

K)(296Kmol

atmL0.08206mol0.0887

V

= 2.32 L of CO2

(3 significant figures)

Example 15 (Cont)