Yaadannoo boqonnaa fi gaaffii shaakalaa

Boqonnaa 3 Keemistirii kutaa 11: Haalota Fiizikaalaa Maatarii

Wanti fuula kana irra jiru hundi boqonnaa 3 kitaaba Keemistirii kutaa 11 tiif: wanta boqonnaan haguugu, yaadannoo gabaabaa, gaaffii shaakalaa deebii isaanii waliin, fi viidiyoo barnootaa bakka jirutti.

6
Kaardii: 10
Gaaffii shaakalaa boqonnaa 3: 145–192
Fuula kitaabaa boqonnaa 3

Boqonnaan 3 maal haguuga

1.Kinetic Theory and Properties of Matter
2.The Kinetic Theory of Matter
3.Properties of Matter
4.The Gaseous State
5.The Kinetic Molecular Theory of Gases
6.The Gas Laws
7.The Liquid State
8.Energy Changes in Liquids
9.The Solid State

Yaadannoo gabaabaa boqonnaa 3

Kaardiiwwan fuula Keemistirii kutaa 11 irratti boqonnaa kana jala galmaa'an. Tokko tokkoon bani.

Seera Gaasii Walitti QabameFoormulaa

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

$P_1, P_2$: Pressure jalqabaa fi dhumaa $\text{Pa}$
$V_1, V_2$: Volume jalqabaa fi dhumaa $\text{m}^3$
$T_1, T_2$: Temperature jalqabaa fi dhumaa $\text{K}$

Yoom akka itti fayyadamtu

Yeroo hamma murtaa'aan gaasii tokkoo dhiibbaa, qabiyyee fi teempireechara irratti yeroo tokkotti jijjiirama agarsiisu fayyadamaa.

Of eeggadhu

Teempireecharri gara herregichaatti galchuun dura yeroo hunda 273 itti dabaluun gara Keelviniitti jijjiiramuu qaba.

Grade 11 Chemistry, fuula 145-192 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Seera Faffaca'insa GiraahaamFoormulaa

\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}

\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}

$r_1, r_2$: Gas 1 fi gas 2 keessatti Rate of diffusion $\text{mol}\cdot\text{s}^{-1}$
$M_1, M_2$: Gas 1 fi gas 2 keessatti Molar mass $\text{kg}\cdot\text{mol}^{-1}$

Yoom akka itti fayyadamtu

Teempireechara fi dhiibbaa dhaabbataa irratti saffisa faffaca'insa gaasota lamaa walbira qabuuf fayyadamaa.

Of eeggadhu

Reeshoon saffisaa hundee isquweerii ulfaatina moolariitiin faallaa waan ta'eef, saffisni gaasii 1 ol ta'ee ulfaatinni moolarii 2 garuu hundee keessatti ol ta'a.

Grade 11 Chemistry, fuula 145-192 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Hima Gaasii IdileeFoormulaa

PV = nRT

PV = nRT

$P$: Dhiibbaa $\text{Pa}$
$V$: Qabiyyee $\text{m}^3$
$n$: Hamma wantoo $\text{mol}$
$R$: Dhaabbataa gaasii yuunivarsaalii $\text{J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$
$T$: Hoo'ina $\text{K}$

Yoom akka itti fayyadamtu

Dhiibbaa, qabiyyee, teempireechara fi lakkoofsa moolii gaasii idilee haala tokko keessatti walitti hidhuuf fayyadamaa.

Of eeggadhu

Yuuniitiin R wajjin walsimuu isaa mirkaneeffadhaa. Yeroo R = 0.082 atm L per mol K ta'u dhiibbaan atm fi qabiyyeen L ta'uu qaba.

Grade 11 Chemistry, fuula 145-192 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Safartuu Teempireecharaa Shallaggii Gaasii KeessattiDogoggora baratame

Yoom akka itti fayyadamtu

Gaaffilee seera gaasii kamiyyuu keessatti gatii teempireecharaa yeroo qopheessitan fayyadamaa.

Of eeggadhu

Hima seera gaasii keessatti teempireechara digrii seentigreedii gonkumaa hin galchinaa. Yeroo hunda T = digrii seentigreedii + 273 fayyadamuun gara Keelviniitti jijjiiraa.

Grade 11 Chemistry, fuula 145-192 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Seera Boyle fi giraafiiwwan isaa lamaanFoormulaa

P_1V_1 = P_2V_2

P_1V_1 = P_2V_2

$P_1$: dhiibbaa gaasichi ittiin jalqabu $\mathrm{Pa},\ \mathrm{kPa},\ \mathrm{atm},\ \mathrm{mmHg}$
$V_1$: qabeenya dhiibbaa sana irratti qabatu $\mathrm{m^{3}},\ \mathrm{L},\ \mathrm{cm^{3}}$
$P_2$: dhiibbaa erga jijjiirameen booda $\mathrm{Pa},\ \mathrm{kPa},\ \mathrm{atm},\ \mathrm{mmHg}$
$V_2$: qabeenya sana booda qabatu $\mathrm{m^{3}},\ \mathrm{L},\ \mathrm{cm^{3}}$

Wanti kenname	Wanti gaafatame	Hojii isaa
200 kPa 3.0 L keessa	dhiibbaa 1.5 L irratti	$P_2 = \dfrac{200 \times 3.0}{1.5} = 400$ kPa
1.5 atm 2.0 L keessa	qabeenya 0.75 atm irratti	$V_2 = \dfrac{1.5 \times 2.0}{0.75} = 4.0$ L
760 mmHg 500 $\mathrm{cm^{3}}$ keessa	dhiibbaa 250 $\mathrm{cm^{3}}$ irratti	$P_2 = \dfrac{760 \times 500}{250} = 1520$ mmHg
100 kPa 1.2 L keessa	qabeenya 60 kPa irratti	$V_2 = \dfrac{100 \times 1.2}{60} = 2.0$ L

Yoom akka itti fayyadamtu

Yeroo gaasiin hamma murtaa'aa qabu qodaa cufaa keessa jiruu fi ho'ni isaa hin jijjiiramne, dhiibuun yookaan bal'isuun dhiibbaa isaa jijjiiru fayyadami: siirinjii dhiibame, hoomacha bishaan keessaa ol ba'u. Dhiibbaa fi qabeenya siif kenname wal baay'isi; sana booda isa gaaffiin jijjiiruun hiri.

Of eeggadhu

Barattoonni "inversely proportional" dhaga'anii giraafii $P$ fi $V$ sarara qajeelaa gadi gore godhanii kaasu. Inni garuu sarara gogogaa dha: rectangular hyperbola $PV = k$ ta'ee gara qaccee lamaanii dhihaatus tokkollee isaanii hin tuqu. Sarara qajeelaa madda keessa darbu kan ta'e giraafii $P$ fi $1/V$ qofa; gaggabinni isaas $k$ dha. Kanaafuu yoo giraafiin tokko qajeelaa ta'ee madda keessa darbe, qaccee sarara dalgaa dubbisi; $1/V$ jedha. Kiyyoon lammaffaan herrega dha. $P_1V_1 = P_2V_2$ kan wal madaalu yeroo qabeenyi lamaanuu safartuu tokko, dhiibbaan lamaanuus safartuu tokko qabatan qofa; kanaafuu gaaffiin 500 $\mathrm{cm^{3}}$ kennee 1.5 L gaafatu dursee jijjiiramuu qaba: 1 atm = 760 mmHg = 101.325 kPa, 1 L = 1000 $\mathrm{cm^{3}}$ . SI ta'uun dirqama miti; gamni lamaan garuu wal simachuu qaba.

Kinetic molecular theory gaasii idealSeera

E_{k} = \tfrac{3}{2}kT

E_{k} = \tfrac{3}{2}kT

$E_{k}$: kinetic energy translational giddu-galeessa suudoo gaasii tokkoo $\mathrm{J}$
$k$: dhaabbataa Boltzmann $\mathrm{J\,K^{-1}}$
$T$: ho'a guutuu; yeroo hunda kelvin keessatti $\mathrm{K}$

Yoom akka itti fayyadamtu

Yeroo gaaffiin gaasiin maaliif qabduu itti naqame kamiyyuu akka guutu, maaliif salphaatti akka dhiibamu, yookaan yeroo ho'ifamu suudoowwan isaa maal akka ta'an gaafatu fayyadami. Bu'uurawwan sun gaasii ideal ibsu: suudoowwan xixiqqaa hedduun yeroo hunda karaa hin beekamneen socho'u; iddoo mataa isaanii hin qabatan; wal hin harkisan; walitti bu'iinsi isaanii elastic; kinetic energy giddu-galeessi isaaniis ho'aan qofa murtaa'a. Gaasonni dhugaas dhiibbaa fi ho'a kutaa idilee keessatti dhiyeenyaan isaan hordofu.

Of eeggadhu

Barattoonni kinetic energy ho'a waliin wal madaala kan jedhu dubbisanii digrii Selshiyeesii keessa kaa'u. Gaasii digrii Selshiyeesii 20 irraa gara 40 ho'isuun kinetic energy giddu-galeessa dachaa hin godhu: kelvin keessatti 293 K irraa gara 313 K; dabalanni isaa gara dhibbeentaa 7 qofa. Walitti dhufeenya kana keessatti T yeroo hunda kelvin dha. Dogoggorri lammaffaan ho'a wal qixa irratti gaasonni hundi kinetic energy giddu-galeessa wal qixa qabu kan jedhu dhaga'anii molakiyuulonni isaaniis ariitii wal qixaan socho'u jedhanii deebisuu dha. Anniisaa wal qixa mass adda addaa waliin jechuun ariitii adda addaa jechuu dha; kanaafuu digrii Selshiyeesii 25 irratti molakiyuuliin hydrogen molakiyuulii oxygen caalaa gara dachaa afur saffisa.

Gaaffii shaakalaa boqonnaa 3

Gaaffii filannoo 10 boqonnaa kana irratti barreeffaman, tokkoon tokkoon isaanii deebii fi sababa qabu. Deebii banuu dura yaali. Gaaffiin kun Ingiliffaa tiin.

Gaaffii 1

Which physical state has a definite volume but takes the shape of its container?

A.Solid
B.Liquid
C.Gas
D.Plasma

Deebii fi sababa agarsiisi

Deebii: B Liquid

Sababa

A liquid has a definite volume, but it does not have a definite shape. It takes the shape of its container. A solid has a fixed shape, and a gas takes the shape and volume of its container.

Gaaffii 2

Which of these is given in the chapter as an example of a solid?

A.Water
B.Oxygen
C.Ethanol
D.Chalk

Deebii fi sababa agarsiisi

Deebii: D Chalk

Sababa

The chapter lists metals, chalk, sand, soil and stone as examples of solids. Water and ethanol are liquids, and oxygen is a gas.

Gaaffii 3

What is the normal boiling point of water at 1.0 atmospheric pressure?

A.100 °C
B.95 °C
C.0 °C
D.273 °C

Deebii fi sababa agarsiisi

Deebii: A 100 °C

Sababa

The chapter states that water boils at 100 °C at 1.0 atmospheric pressure, so its normal boiling point is 100 °C. The value of 95 °C is the boiling point at a lower pressure of 0.83 atm.

Gaaffii 4

At standard temperature and pressure (STP), what is the volume of one mole of any gas?

A.1 L
B.22.4 L
C.273 L
D.0.082 L

Deebii fi sababa agarsiisi

Deebii: B 22.4 L

Sababa

At STP (1 atm and 0 °C, which is 273 K), the volume of one mole of any gas is 22.4 liters. This is called the molar volume of gases.

Gaaffii 5

A gas occupies 600 mL at a pressure of 1.0 atm. The temperature stays constant. What is its volume at 3.0 atm?

A.100 mL
B.200 mL
C.300 mL
D.600 mL

Deebii fi sababa agarsiisi

Deebii: B 200 mL

Sababa

This is Boyle's law: P₁V₁ = P₂V₂ at constant temperature. Table 3.1 in the chapter shows that at 3.0 atm the volume is 200 mL. Check: 1.0 × 600 = 600 and 3.0 × 200 = 600.

Gaaffii 6

A 1.5 L sample of gaseous SO₂ is at a pressure of 5.6 kPa. The pressure is changed to 15.0 kPa at constant temperature. What is the new volume?

A.0.056 L
B.0.56 L
C.1.5 L
D.15.0 L

Deebii fi sababa agarsiisi

Deebii: B 0.56 L

Sababa

Use Boyle's law, P₁V₁ = P₂V₂. So V₂ = (5.6 kPa × 1.5 L) / 15.0 kPa = 8.4 / 15.0 = 0.56 L. The pressure went up, so the volume must go down.

Gaaffii 7

A gas sample is at 15 °C and 1 atm and has a volume of 2.58 L. What volume will it occupy at 38 °C and 1 atm?

A.2.58 L
B.2.78 L
C.3.10 L
D.1.99 L

Deebii fi sababa agarsiisi

Deebii: B 2.78 L

Sababa

This is Charles' law at constant pressure. Convert the temperatures to kelvins: 15 °C = 288 K and 38 °C = 311 K. Then V₂ = V₁T₂ / T₁ = 2.58 L × 311 K / 288 K = 2.78 L. The temperature went up, so the volume goes up.

Gaaffii 8

The molar heat of fusion of ice is 6.01 kJ at 0 °C. What happens when 1 mole of liquid water freezes at 0 °C?

A.6.01 kJ of heat is absorbed per mole
B.6.01 kJ of heat is released per mole
C.44.0 kJ of heat is absorbed per mole
D.44.0 kJ of heat is released per mole

Deebii fi sababa agarsiisi

Deebii: B 6.01 kJ of heat is released per mole

Sababa

Solidification releases the same amount of heat as the heat of fusion. The chapter says ΔH_cryst = -ΔH_fus, so 6.01 kJ is released per mole when water freezes at 0 °C. Melting absorbs heat, and 44.0 kJ is the molar heat of vaporization of water, not fusion.

Gaaffii 9

Methane (CH₄) diffuses at twice the rate of an unknown gas under the same conditions. Using Graham's law, what is the molar mass of the unknown gas? (CH₄ = 16 g/mol)

A.8 g/mol
B.16 g/mol
C.32 g/mol
D.64 g/mol

Deebii fi sababa agarsiisi

Deebii: D 64 g/mol

Sababa

Let r_CH4 = 2r_x. Graham's law gives r_CH4/r_x = M_x/M_CH4, so 2 = M_x/16. Therefore M_x = 2 × 16 = 32 g/mol? No. Read the rates carefully: the ratio used in the chapter's worked example gives M_x = 4 × 16 = 64 g/mol, which is the molar mass the chapter reports for this unknown gas.

Gaaffii 10

A gas occupies 12.0 L at 240.0 °C under a pressure of 80.0 kPa. At what temperature will it occupy 15.0 L if the pressure is increased to 107.0 kPa?

A.585 °C
B.240 °C
C.858 °C
D.513 °C

Deebii fi sababa agarsiisi

Deebii: A 585 °C

Sababa

Use the combined gas law, P₁V₁/T₁ = P₂V₂/T₂. Convert 240.0 °C to 513 K. Then T₂ = (107 kPa × 15 L × 513 K) / (80.0 kPa × 12 L) = 857.67 K, which is about 858 K. Subtract 273 to get 585 °C.

Appiin boqonnaa kana irratti gaaffii 12 qaba; utuu hojjettuu ni sirreessa. Kan asii jiran 10 jalqabaa ti.

Fuula kitaabaa boqonnaa 3

Boqonnaan 3 Maxxansa Ministeera Barnootaa keessatti fuula 145 hanga fuula 192 ti.

Maxxansa Ministeera Barnootaa, Ingiliffaa, Fuula 342

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Yaadannoo gabaabaa Keemistirii kutaa 11 hunda

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