10 Fisika Termal Bag 2 – Nov 12 revisi
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Transcript 10 Fisika Termal Bag 2 – Nov 12 revisi
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.
Ekspansi Linier
• Benda akan memanjang
bila temperaturnya
meningkat
• Pertambahan panjang
proporsional terhadap
perubahan temperatur
DL/L = a DT atau
DL = a L DT atau
dL
aL
dT
L
DL
DL/L = aDT
L
DL/L = aDT
DL
Contoh:
• Suatu penggaris besi dikalibrasi pada suhu 20 °C. Hitung
error dalam pengukuran jika penggaris tersebut digunakan
untuk mengukur sepanjang 500 mm pada temperatur 45 °C.
abesi= 1,2x10-5 C-1
DL/L = a DT
DL = L a DT
DL = 500 x10-3 m x 1,2 x10-5 C-1 x 25 °C
DL = 1,5x10-4 m = 0,15 mm
Ekspansi Volume
• Setiap sisi panjang berubah dari L
menjadi L+DL = L + La DT
• Volume awal = L3
• Volume baru (V+DV)
= (L +DL)3 = (L + La 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
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 DT
DV = 3a V DT
L
• 3a b → DV = b V DT
DL
Unusual Behavior of Water
Cairan umumnya akan meningkat volumenya dengan
peningkatan temperatur. Kecuali untuk air dingin seperti
ditunjukkan pada gambar.
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.
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)
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):
N
n
NA
The number of moles contained in a sample can also be
found from its mass.
m
n
M
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.
Avogadro’s Law
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:-
V n
p1V1 p2V2
constant
T1n1 T2 n2
Deriving Ideal Gas Equation
• From Boyle’s Law:
1
V
p
• From Charles’s Law:
V T
• From Avogadro’s Law:
V n
• Combining these three:
• Rewriting using the gas
constant R:
nT
V
p
nT
V R
p
Therefore:-
pV nRT
PV Diagram in Thermodynamic Process
Kinetic Theory of Gases
The pressure that a gas
exerts is caused by the
impact of its molecules
on the walls of the
container.
Consider a gas molecule colliding
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:
DP
F Dt
(
m v ( m v m v2
F
2L
v
L
The force on one of the molecule:
(
m v ( m v m v2
2L
v
L
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:
m v2
L
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:
2
m
v
N
rms
F
3 L
Vrms = root-mean-square velocity.
Pressure is force per unit area, so the pressure P acting on a
wall of area L2 is
2
F N m vrms
P 2
L 3 L3
Since the volume of the box is V = L3, the equation above
can be written as
2 1
2
PV N m vrms
3 2
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)
TERIMAKASIH