FISIKA TERMAL

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FISIKA TERMAL BAGIAN 2

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FISIKA TERMAL. BAGIAN 2. Ekspansi termal dari benda padat dan cair. Fenomena terjadinya peningkatan volume dari suatu materi karena peningkatan temperatur disebut dengan ekspansi termal. Ekspansi termal adalah konsekuensi dari perubahan average separation antar atom dalam suatu materi. - PowerPoint PPT Presentation

Transcript of FISIKA TERMAL

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FISIKA TERMALBAGIAN 2

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Ekspansi termal dari benda padat dan cair

Fenomena terjadinya peningkatan volume dari suatu materi karena peningkatan temperatur disebut dengan ekspansi termal.

Ekspansi termal adalah konsekuensi dari perubahan average separation antar atom dalam suatu materi.

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• Benda akan memanjang bila temperaturnya meningkat

• Pertambahan panjang proporsional terhadap perubahan temperaturDL/L = a DT atauDL = a L DT atau

dL

dTL

L DL

L DL

DL/L = aDT

DL/L = aDT

Ekspansi Linier

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Contoh:

• Suatu penggaris besi dikalibrasi pada suhu 20 °C. Hitung error pengukuran jika penggaris tersebut digunakan untuk mengukur sepanjang 500 mm pada temperatur 45 °C. abesi= 1,2x10-5 C-1

DL/L = a DTDL = L a DTDL = 500 mm x 1,2 x10-5 C-1 x 25 °C DL = 0,15 mm

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The initial gap between the ends of the bolts is 5.0 m at 27°C. At what temperature will the bolts touch? Assume that the distance between the walls of the device is not affected by the temperature change.

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Ekspansi Volume

• Setiap sisi panjang berubah dari L menjadi L+DL = L + L a DT

• Volume awal = L3

• Volume baru (V+DV)= (L +DL)3 = (L + L a DT)3

= (L (1+ a DT))3

= L3 (1+ a DT)3

= V [1+ 3a DT+ 3(a DT) 2 +(a DT)3]

L DL

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Ekspansi Volume (lanjutan)

• V+DV = V [1+ 3a DT+ 3(a DT) 2 +(a DT)3]DV / V = [3a DT+ 3(a DT) 2 +(a DT)3]

• Karena a DT < 1 untuk nilai DT < 100 °C maka nilai 3(a DT) 2 dan (a DT )3 dapat diabaikan. Sehingga:DV / V = 3a DTDV = 3a V DT

• 3 a = b → DV = b V DT

L DL

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Unusual Behavior of Water

Cairan umumnya akan meningkat volumenya dengan peningkatan temperatur. Kecuali untuk air dingin seperti ditunjukkan pada gambar.

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Ideal Gas

An ideal gas is an idealized model for real gases that have sufficiently low densities.

The condition of low density means that the molecules of the gas are so far apart that they do not interact (except during collisions that are effectively elastic).

The ideal gas law expresses the relationship between the absolute pressure (P), the Kelvin temperature (T), the volume (V), and the number of moles (n) of the gas.

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Misal gas ideal didalam wadah silinder yang volumenya dapat divariasikan dengan piston yang dapat bergerak. Jika diasumsikan sistem tertutup, maka massa (atau jumlah mol) gas tetap konstan. Pada sistem tersebut dapat diperoleh informasi:

Pada temperatur konstan, tekanan berbanding terbalik dengan volume (Hukum Boyle)

Pada tekanan konstan, volume berbanding lurus dengan temperatur (Hukum Charles)

Pada volume konstan, tekanan berbanding lurus dengan temperatur (Hukum Gay-Lussac)

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Hukum Charles menyatakan bahwa pada tekanan tetap, volume gas ideal bermassa tertentu berbanding lurus terhadap temperaturnya (dalam Kelvin).Secara matematis, hukum Charles dapat ditulis sebagai:

denganV: volume gas (m3),T: temperatur gas (K), dank: konstanta.

Hukum Boyle menyatakan bahwa "dalam suhu tetap" untuk massa yang sama, tekanan absolut dan volume udara terbalik secara proporsional.

P V = kdimana:

P = tekananV = volumek = konstanta.

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Gay-Lussac menemukan bahwaTekanan dari sejumlah tetap gas pada volum yang tetap berbanding lurus dengan temperaturnya

atau

where:

P adalah tekanan gas.T adalah temperatur gas (dalam Kelvin).k adalah sebuah konstanta.

Hukum ini dapat dibuktikan melalui teori kinetik gas, karena temperatur adalah ukuran rata-rata energi kinetik, dimana jika energi kinetik gas meningkat, maka partikel-partikel gas akan bertumbukan dengan dinding/wadah lebih cepat, sehingga meningkatkan tekanan.

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Amount of Gas

• Better to describe gas in terms of number of moles.

• The number of moles n contained in any sample is the number of particles N in the sample divided by the number of particles per mole NA (Avogadro's number):

AN

Nn

The number of moles contained in a sample can also be found from its mass.

M

mn

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Avogadro’s Constant

• One mole of any gas contains the same number of particles. This number is called Avogadro’s constant and has the symbol NA. The value of NA is 6.02 × 1023 particles per mole.

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Avogadro’s Law

nV

The most significant consequence of Avogadro's law is that the ideal gas constant has the same value for all gases. This means that the constant is given by:-

constant22

22

11

11 nT

Vp

nT

Vp

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A spray can containing a propellant gas at twice atmospheric pressure (202 kPa) and having a volume of 125.00 cm3 is at 22°C. It is then tossed into an open fire. When the temperature of the gas in the can reaches 195°C, what is the pressure inside the can? Assume any change in the volume of the can is negligible.

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If the gas is compressed, dV is negative and the work done on the gas is positive.If the gas expands, dV is positive and the work done on the gas is negative. If thevolume remains constant, the work done on the gas is zero.

dW = - P dV

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(a) the volume of the gas is first reduced from Vi to Vf at constant pressure Pi and the pressure of the gas then increases from Pi to Pf by heating at constant volume Vf . The work done on the gas along this path is Pi(Vf -Vi).

(b) the pressure of the gas is increased from Pi to Pf at constant volume Vi and then the volume of the gas is reduced from Vi to Vf at constant pressure Pf . The work done on the gas is Pf (Vf - Vi). This value is greater than that for the process (a) because the piston is moved through the same displacement by a larger force.

(c) Both P and V change continuously, the work done on the gas has some value between the values obtained in the first two processes. To evaluate the work in this case, the function P(V ) must be known so that we can evaluate the Integral.

W = Pf (Vf – Vi)W = Pi (Vf – Vi)

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Cyclic process, that is, a process that starts and ends at the same state.

Characterize the paths in Figure 20.9 as isobaric, isovolumetric, isothermal, or adiabatic. For path B, Q = 0.

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Isolated system no energy transfer by heat takes place (Q = 0) internal energy remains constant (Eint = 0; therefore, Eint,i = Eint,f )

An energy reservoir is a source of energy that is considered to be so great that a finite transfer of energy to or from the reservoir does not change its temperature.

Q = 0

Q = W = 0Adiabatic Free expansion

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PV Diagram for an isothermal expansion

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SOAL:A 1.0-mol sample of an ideal gas is kept at 0.0°C during an expansion from 3.0 L to 10.0 L. (R=8.31 J / mol K)

(A)How much work is done on the gas during the expansion?

(B) How much energy transfer by heat occurs between the gas and its surroundings in this process?

(C) If the gas is returned to the original volume by means of an isobaric process, how much work is done on the gas?

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Kinetic Theory of Gases

The pressure that a gas exerts is caused by the impact of its molecules on the walls of the container.

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Consider a gas molecule collising elastically with the right wall of the container and rebounding from it.

The force on the molecule is obtained using Newton’s second law as follows:

t

PF

L

mv

vL

mvmvF

2

2

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The force on one of the molecule:

According to Newton's law of action–reaction, the force on the wall is equal in magnitude to this value, but oppositely directed.

The force exerted on the wall by one molecule: L

mv 2

L

mv

vL

mvmv 2

2

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If N is the total number of molecules, since these particles move randomly in three dimensions, one-third of them on the average strike the right wall. Therefore, the total force is:

Vrms = root-mean-square velocity.

L

vmNF rms

2

3

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Pressure is force per unit area, so the pressure P acting on a wall of area L2 is

Since the volume of the box is V = L3, the equation above can be written as

3

2

2 3 L

vmN

L

FP rms

2

2

1

3

2rmsvmNPV

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Contoh soal:

1. Berapa banyaknya massa Cl2 (gram) yang dapat disimpan dalam suatu wadah (kontainer) dengan volume 10 L pada suhu 30 °C dan tekanan 1000 kPa?

2. Pada suhu 150 °C dan tekanan 100 kPa, sebuah senyawa dengan massa 2,506 gram memiliki volume 1 L. Hitung massa molar senyawa tersebut.

3. Berapa vrms dari atom helium yang mengisi sebuah balon dengan diameter 30 cm pada suhu 20 °C dan tekanan 1 atm? (diketahui jumlah atom helium dalam balon sebanyak 3,54 x 1023 dan massa atom helium 6,64 x 10-24 g)

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