FI1101 Fisika Dasar IA Program studi Fisika, FMIPA ITB

FI1101 Fisika Dasar IA

Program studi Fisika, FMIPA ITB

§ Mempelajari dan penerapan energy panas (energy dalam) system. Salah satu konsep utama

termodinamika adalah suhu/temperatur

§ Temperatur merupakan dasar kuantitas SI yang terkait dengan pengindraan panas dan dingin.

§ Alat ukurnya adalah termometer, yg mana

mengandung suatu zat yang bekerja dengan sifat yang terukur, seperti panjang atau tekanan, yang berubah secara regular jika terjadi berubahan menjadi lebih panas atau lebih dingin

§ Fisikawan mengukur temperatur dalam skala Kelvin.

§ Dua buah benda berada dalam kesetimbangan termal jika keduanya berada pada temperatur yang sama sehingga tidak ada panas yang akan mengalir dari satu benda ke benda lainnya

§ Gambar disamping menunjukan kondisi dari triple point water, yaitu suatu kondisi dimana padatan es, cairan air, dan uap air, ketiganya ada dalam suatu wadah dengan kesetimbangan termal

§ Hal ini tidak akan terjadi pada tekanan atmosfir yang normal

§ Temperatur dari gabungan ketiganya adalah 273.16 K

§ Dari gambar terlihat bahwa ditengah-tengah wadah tersebut disisipkan tabung termometer gas dengan volum konstan

§ Gambar disamping menunjukan termometer gas dgn volume konstan. Terdiri dari bola gas yang dihubungkan dengan manometer raksa. Dengan menaikan dan menurunkan reservoir R, level raksa disisi kiri tabung U dapat selalu mencapai nol

untuk menjaga volume gas tetap konstan.

§ Rumusan untuk mengukur suhu dengan termometer gas adalah,

§ Dimana, p adalah tekanan yang diamati dan p₃ adalah tekanan saat kondisi triple point water

§ Skala temperature Celcius:

§ T dalam kelvin

§ Skala temperature Fahrenheit:

§ Ekspansi linier

§ Semua benda mengalami perubahan bentuk seiring dengan perubahan

temperatur. Untuk perubahan temperatur ΔT, perubahan panjang ΔL benda memenuhi persamaan:

§ dimana, 𝛂 adalah koeefisien ekspansi linier

§ Ekspansi Volum

§ Jika temperatur padatan atau cairan yang volume nya V bertambah sebesar ΔT, penambahan volume nya adalah,

§ dimana, 𝛽 adalah koeefisien ekspansi volum. Hubungan antara koefisien ekspansi linier dan volum adalah,

§ Kalor Q adalah energi yang ditransfer

antara sistem dan lingkungannya karena adanya perbedaan temperatur diantara keduanya

§ Dapat diukur dalam Joule (J), kalori (kal), kilokalori (kkal) atau satuan Britis

thermal (Btu),

§ Penyerapan kalor oleh padatan dan cairan

§ Kapasitas kalor C benda sebanding dengan Kalor Q yang diserap/dilepas benda dan menghasilkan perubahan temperatur ΔT benda,

§ Jika benda memiliki massa m, maka

§ Dimana, c merupakan kalor jenis

§ Kalor lebur/uap

§ Kalor dan Kerja/Usaha

§ Sebuah gas dapat mentransfer energi pada

lingkungan melalui kerja/usaha. Besarnya usaha W yang dilakukan gas (mengembang/termanpatkan) adalah

§ Prinsip kekekalan energi untuk proses termodinamika diberikan pada hk. 1 termodinamika, yaitu:

§ Atau,

492 CHAPTER 18 TEMPERATURE, HEAT, AND THE FIRST LAW OF THERMODYNAMICS

HALLIDAY REVISED

18-11 Some Special Cases of the First Law of Thermodynamics

Here are four thermodynamic processes as summarized in Table 18-5.

1. Adiabatic processes. An adiabatic process is one that occurs so rapidly or occurs in a system that is so well insulated that no transfer of energy as heat occurs between the system and its environment.Putting Q ! 0 in the first law (Eq.18-26) yields

"E_int ! #W (adiabatic process). (18-28) This tells us that if work is done by the system (that is, if W is positive), the internal energy of the system decreases by the amount of work. Conversely, if work is done on the system (that is, if W is negative), the internal energy of the system increases by that amount.

Figure 18-15 shows an idealized adiabatic process. Heat cannot enter or leave the system because of the insulation. Thus, the only way energy can be transferred between the system and its environment is by work. If we remove shot from the piston and allow the gas to expand, the work done by the system (the gas) is positive and the internal energy of the gas decreases. If, instead, we add shot and compress the gas, the work done by the system is negative and the internal energy of the gas increases.

2. Constant-volume processes. If the volume of a system (such as a gas) is held con- stant, that system can do no work. Putting W ! 0 in the first law (Eq. 18-26) yields

"E_int ! Q (constant-volume process). (18-29) Thus, if heat is absorbed by a system (that is, if Q is positive), the internal energy of the system increases. Conversely, if heat is lost during the process (that is, if Q is negative), the internal energy of the system must decrease.

3. Cyclical processes. There are processes in which, after certain interchanges of heat and work, the system is restored to its initial state. In that case, no intrinsic property of the system—including its internal energy—can possibly change.

Putting "E_int ! 0 in the first law (Eq. 18-26) yields

Q ! W (cyclical process). (18-30) Thus, the net work done during the process must exactly equal the net amount of energy transferred as heat; the store of internal energy of the system remains unchanged. Cyclical processes form a closed loop on a p-V plot, as shown in Fig. 18-14f. We discuss such processes in detail in Chapter 20.

4. Free expansions. These are adiabatic processes in which no transfer of heat occurs between the system and its environment and no work is done on or by the system.Thus, Q ! W ! 0, and the first law requires that

"E_int ! 0 (free expansion). (18-31) Figure 18-16 shows how such an expansion can be carried out. A gas, which is

The First Law of Thermodynamics: Four Special Cases The Law: "E_int!Q # W(Eq. 18-26)

Process Restriction Consequence

Adiabatic Q ! 0 "E_int ! #W

Constant volume W ! 0 "E_int! Q

Closed cycle "E_int!0 Q ! W

Free expansion Q ! W !0 "E_int !0 Fig. 18-15 An adiabatic expansion can

be carried out by slowly removing lead shot from the top of the piston.Adding lead shot reverses the process at any stage.

Lead shot

W

Insulation

We slowly remove lead shot, allowing an expansion without any heat transfer.

Table 18-5 Fig. 18-16 The initial stage of a

free-expansion process.After the stopcock is opened, the gas fills both chambers and eventually reaches an equilibrium state.

Vacuum

Insulation

Stopcock

halliday_c18_476-506v2.qxd 22-10-2009 12:03 Page 492

492 CHAPTER 18 TEMPERATURE, HEAT, AND THE FIRST LAW OF THERMODYNAMICS

HALLIDAY REVISED

18-11 Some Special Cases of the First Law of Thermodynamics

Here are four thermodynamic processes as summarized in Table 18-5.

1. Adiabatic processes. An adiabatic process is one that occurs so rapidly or occurs in a system that is so well insulated that no transfer of energy as heat occurs between the system and its environment.Putting Q ! 0 in the first law (Eq.18-26) yields

"E_int! #W (adiabatic process). (18-28) This tells us that if work is done by the system (that is, if W is positive), the internal energy of the system decreases by the amount of work. Conversely, if work is done on the system (that is, if W is negative), the internal energy of the system increases by that amount.

Figure 18-15 shows an idealized adiabatic process. Heat cannot enter or leave the system because of the insulation. Thus, the only way energy can be transferred between the system and its environment is by work. If we remove shot from the piston and allow the gas to expand, the work done by the system (the gas) is positive and the internal energy of the gas decreases. If, instead, we add shot and compress the gas, the work done by the system is negative and the internal energy of the gas increases.

2. Constant-volume processes. If the volume of a system (such as a gas) is held con- stant, that system can do no work. Putting W ! 0 in the first law (Eq. 18-26) yields

"E_int!Q (constant-volume process). (18-29) Thus, if heat is absorbed by a system (that is, if Q is positive), the internal energy of the system increases. Conversely, if heat is lost during the process (that is, if Q is negative), the internal energy of the system must decrease.

3. Cyclical processes. There are processes in which, after certain interchanges of heat and work, the system is restored to its initial state. In that case, no intrinsic property of the system—including its internal energy—can possibly change.

Putting "Eint!0 in the first law (Eq. 18-26) yields

Q ! W (cyclical process). (18-30) Thus, the net work done during the process must exactly equal the net amount of energy transferred as heat; the store of internal energy of the system remains unchanged. Cyclical processes form a closed loop on a p-V plot, as shown in Fig. 18-14f. We discuss such processes in detail in Chapter 20.

4. Free expansions. These are adiabatic processes in which no transfer of heat occurs between the system and its environment and no work is done on or by the system.Thus, Q ! W ! 0, and the first law requires that

"E_int!0 (free expansion). (18-31) Figure 18-16 shows how such an expansion can be carried out. A gas, which is

The First Law of Thermodynamics: Four Special Cases The Law: "E_int!Q # W(Eq. 18-26)

Process Restriction Consequence

Adiabatic Q !0 "E_int! #W

Constant volume W !0 "E_int!Q

Closed cycle "E_int!0 Q ! W

Free expansion Q ! W !0 "Eint!0 Fig. 18-15 An adiabatic expansion can

be carried out by slowly removing lead shot from the top of the piston.Adding lead shot reverses the process at any stage.

Lead shot

W

Insulation We slowly remove lead shot, allowing an expansion without any heat transfer.

Table 18-5 Fig. 18-16 The initial stage of a

free-expansion process.After the stopcock is opened, the gas fills both chambers and eventually reaches an equilibrium state.

Vacuum

Insulation

Stopcock

halliday_c18_476-506v2.qxd 22-10-2009 12:03 Page 492

§ Konduksi panas

§ Dimana, k merupakan konduktivitas termal material

§ Radiasi

§ Melalui emisi energi EM

§ Dimana, σ (= 5.6704×10^-8 W/m².K⁴) adalah konstanta the Stefan– Boltzmann. 𝜀adalah Emisivitas permukaan benda. Temperatur dalam Kelvin

§ Suatu mesin berisi gas ideal monoatomik dengan volum dan tekanan seperti pada gambar. Gas mulai dari keadaan titik A dengan nilai volum V = 5 L, tekanan P = 1 atm, dan suhunya T_a =300 K, lalu mengalami proses isokhorik menuju keadaan B yang suhunya T_b. Diketahui perubahan energi internal pada proses AB adalah 1,52 x 10³ J. dar keadaan B menuju C gas mengalami proses isothermal, sedangkan dari keadaan C menuju A adalah proses

isobarik.

§ A. hitunglah berapa jumlah mol, dan suhu pada keadaan B

§ B. hitunglah usaha yang dilakukan gas untuk masing2 proses

§ C. hitunglah total kalor yang di terima sistem dan total kalor yang dilepas sistem.

5. Satu mol gas ideal dengan 𝛾 = 1,3 mengalami proses siklus seperti pada gambar di bawah ini. Proses 1-2 dan 3-4

adalah proses isokhorik sedangkan proses 2-3 dan 4-1 adalah proses adiabatik. Siklus tersebut menggambarkan sebuah mesin kalor berbahan bakar bensin. Diketahui volume 𝑉₃ = 4𝑉₁. Tentukan

a. Perbandingan temperatur 𝑇₂⁄𝑇₁ dan 𝑇₃⁄ 𝑇₁

b. Perbandingan antara laju rms molekul gas di keadaan 4 dan keadaan 1.

c. Efisiensi dari mesin tersebut SOLUSI:

a. Tekanan pada keadaan 2 adalah 𝑝₂ = 3𝑝₁. Karena proses 1-2 adalah isokhorik, maka

𝑇₂

𝑇₁ = 𝑝₂ 𝑝₁ = 3

1 Proses 2-3 adalah proses adiabatik, sehingga berlaku hubungan

𝑇₂𝑉₂^𝛾−1 = 𝑇₃𝑉₃^𝛾−1 Dengan menggunakan hasil sebelumnya, didapatkan,

𝑇₃

𝑇₁ = 1,98 b. Karena proses 4-1 adalah proses adiabatik, jadi didapatkan

𝑇₄

𝑇₁ = 0,66 Sehingga

(𝑣_rms)₄

(𝑣_rms)₁ = √𝑇₄

𝑇₁ = 0,812

c. Kalor masuk dan keluar hanya pada proses isokhorik (1-2 dan 3-4).

𝑄₁₂ = 𝑛𝐶_𝑉(𝑇₂ − 𝑇₁) = 𝑛𝐶_𝑉𝑇₁(𝑇₂

𝑇₁ − 1) = 2𝑛𝐶_𝑉𝑇₁ dan

𝑄₃₄ = 𝑛𝐶_𝑉(𝑇₄ − 𝑇₃) = 𝑛𝐶_𝑉𝑇₁ (𝑇₄

𝑇₁ −𝑇₃

𝑇₁) = −1,32 𝑛𝐶_𝑉𝑇₁ Jadi kalor masuk pada proses 1-2 dan kalor keluar pada proses 3-4.

Efisiensi mesin kalor adalah

𝜂 = 1 −|𝑄_𝑜𝑢𝑡|

𝑄_𝑖𝑛 = 0,34

Skor 5

Skor 5

Skor 10

c. Hitunglah jumlah kalor yang masuk dan keluar