TUGAS TERMODINAMIKA

KELOMPOK IV

Nama Anggota Kelompok :

1) Arya Dharma

2) Aulia

3) Enda Hutabarat

4) Michael Joy

5) Yos Pawer Ambarita

SOAL DAN PENYELESAIAN

1.1) What is the value of g, and what are its units in a system in which the second,

the foot, and the pound mass are defined as in Sec. 1.2, and the unit of force is

the poundal, defined as the force required to give l (lbm) an acceleration of l(ft)

(s)-2

Penyelesaian :

F =m.ag c

1 N =1 kg. m/s 2g c

14,448 lbf =

10,454

lbm. 10,304

ft/s2

g c

0,225 lbf =2,2 03 lbm. 3,2 8 9 ft/s 2g c

gc = 32,2 lbm.ftlbf . s2

1.2) Electric current is the fundamental electrical dimension in SI; its unit is the

ampere (A). Determine units for the following quantities, as combinations

fundamental SI units.

(a) Electric power; (b) Electric charge; (c) Electric potential difference;

(d) Electric resistance; (e) Electric capacitance.

Penyelesaian :

(a) Daya Listrik

Daya listrik didefinisikan sebagai laju hantaran energi listrik dalam sirkuit

listrik. Satuan SI daya listrik adalah watt yang menyatakan banyaknya

tenaga listrik yang mengalir per satuan waktu (Joule/detik).

Daya listrik : P =Wt

= I2.R

(b) Muatan Listrik

Coulomb, dilambangkan dengan C, adalah satuan SI untuk muatan listrik,

dan didefinisikan dalam ampere: 1 coulomb adalah banyaknya muatan

listrik yang dibawa oleh arus sebesar 1 ampere mengalir selama 1 detik.

Muatan listrik : Q = I.t

(c) Beda Potensial

Volt (V) adalah satuan turunan di dalam Standar Internasional (SI) untuk

mengukur perbedaan tegangan listrik atau beda potensial. 1 Volt berarti

beda tegangan yang diperlukan untuk membuat arus tepat sebesar

1 ampere di dalam suatu rangkaian dengan resistensi 1 ohm.

Beda potensial : V = I.R

(d) Hambatan Listrik

Ohm (lambang: Ω) adalah satuan SI dari impedansi listrik, atau dalam

kasus arus searah, hambatan listrik. Nama satuan ini berasal dari

ilmuwan Georg Ohm. Satu ohm (yang diukur oleh alat ohm-meter) adalah

hambatan listrik pembawa arus yang menghasilkan perbedaan tegangan

satuvolt ketika arus satu ampere melewatinya.

Hambatan listrik: R =VI

(e) Kapasitansi Listrik

Kapasitansi atau kapasitans adalah ukuran jumlah muatan listrik yang

disimpan (atau dipisahkan) untuk sebuah potensial listrik yang telah

ditentukan. Bentuk paling umum dari piranti penyimpanan muatan adalah

sebuah kapasitor dua lempeng/pelat/keping. Jika muatan di

lempeng/pelat/keping adalah +Q dan –Q, dan V adalah tegangan listrik

antar lempeng/pelat/keping.

Kapasitansi listrik : C =QV

=I.tI.R

1.3) Liquidlvapor saturation pressure Psat is often represented as a function of

temperature by an equation of the form:

log10Psat/torr = a - bt/ oC + c

Here, parameters a, b, and c are substance-specific constants. Suppose it is

required torepresent Psat by the equivalent equation:

log10Psat/kPa = a - bt/K + c

Show how the parameters in the two equations are related.

1.4) At what absolute temperature do the Celsius and Fahrenheit temperature scales

give the same numerical value? What is the value?

Penyelesaian :

Celcius : Fahrenheit = 5 : 9 (+32)

Celcius = Fahrenheit

5 = 9 (+32) (kedua ruas dikalikan dengan x)

5x = 9x + 32 ....(1)

-4x = 32

x = -8

substitusi nilai x=-8 ke persamaan (1)

5(-8) = 9(-8) + 32

-40 = -40

Jadi Celcius dan Fahrenheit menunjukkan angka yang sama pada temperatur

absolut -40.

1.5) Pressures up to 3000 bar are measured with a dead-weight gauge. The piston

diameter is 4 rnm. What is the approximate mass in kg of the weights required?

Penyelesaian :

Dik : P = 3000 bar

D = 4 mm

Dit : m = .....?

Jawab :

P = 3000 bar x 1 x 10 5 N/m21 bar

= 3000 x 105 N/m2

D = 4 mm x 1 m1000 mm

= 4 x 10-3 m

A = 14

πD2

= 14

(3,14) (4 x 10-3 m)2

= 12,56 x 10-6 m2

P = FA

↔ F = P . A

= (3000 x 105 N/m2) (12,56 x 10-6 m2)

= 3768 N

F = m.g (g = 9,8 m/s2)

m = Fg

= 3768 N9,8 m/s2

= 384,49 kg

1.6) Pressures up to 3000 atm are measured with a dead-weight gauge. The piston

diameter is 0.17 (in). What is the approximate mass in (lb,) of the weights

required?

Penyelesaian :

Dik : P = 3000 atm

D = 0,17 inch

Dit : m (dalam lbm) = ....?

Jawab :

1 inch = 2,54 cm

D = 0,17 inch x 2,54 cm1 inch

x 1 m100 cm

= 0,4318 x 10-2 m

P = 3000 atm x 1 bar0,986923 atm

x 10 5 Pa1 bar

= 3039,75 x 105 Pa

A = 14

πD2

= 14

(3,14) (0,4318 x 10-2 m)2

= 0,146 x 10-4 m2

P = FA

↔ F = P . A

= (3039,75 x 105 N/m2) (0,146 x 10-4 m2)

= 4438,035 N

F = m.g (g = 9,8 m/s2)

m = Fg

= 4438,035 N9,8 m/s2

= 452,86 kg

m = 452,86 kg x 2,20462 lbm1 kg

= 998,38 lbm

1.7) The reading on a mercury manometer at 298.15 K (25°C) (open to the

atmosphere at one end) is 56.38 cm. The local acceleration of gravity is 9.832

m.s-2. Atmospheric pressure is 101,78 kPa. What is the absolute pressure in kPa

being measured? The density of mercury at 298.15 K (25°C) is 13.534 g.cm-3.

Penyelesaian :

Dik : h = 56,38 cm

g = 9,832 m/s2

Patm = 101,78 kPa

ρ = 13,543 gr/cm3

Dit : Pabs (dalam kPa) = ....?

Jawab :

ρ = 13,543 gr/cm3 x 10 6 cm31 m 3

x 1 kg1000 gr

= 13,543 x 103 kg/m3

h = 56,38 cm x 1 m100 cm

= 56,38 x 10-2 m

Pgauge = ρ.g.h

= (13,543 x 103 kg/m3) (9,832 m/s2) (56,38 x 10-2 m)

= 75072,66 Pa

Pgauge = 75072,66 Pa x 1 kPa1000 Pa

= 75,07266 kPa

Pabs = Pgauge + Patm

= 75,07266 kPa + 101,78 kPa = 176,85266 kPa

1.8) The reading on a mercury manometer at 70 (oF) (open to the atmosphere at one

end) is 25,62 (in). The local acceleration of gravity is 32,243 ft.s-2. Atmospheric

pressure is 29,86 (inHg). What is the absolute pressure in Psia being measured?

The density of mercury at 70 (°F) is 13.534 g.cm-3.

Penyelesaian :

Dik : h = 25,62 in

g = 32,243 ft/s2

Patm = 28,86 inHg

ρ = 13,543 g/cm3

Dit : Pabs (dalam Psia) = ....?

Jawab :

g = 32,243 ft/s2 x 0,304 m1 ft

= 9,802 m/s2

h = 25,62 in x 2,54 cm1 in

x 1 m100 cm

= 65,07 x 10-2 m

ρ = 13,543 g/cm3 x 10 6 cm31 m 3

x 1 kg1000 gr

= 13,543 x 103 kg/m3

Pgauge = ρ.g.h

= (13,543 x 103 kg/m3) (9,802 m/s2) (65,07 x 10-2 m)

= 86379,4 Pa

Patm = 29,86 inHg x 1 Psia2,0360 inHg

= 14,666 Psia

Pgauge = 86361,8 Pa x 1 Psia6894,76 Pa

= 12,5257 Psia

Pabs = Patm + Pgauge

= (14,666 + 12,5257) Psia

= 27,19 Psia

1.9) Liquids that boil at relatively low temperatures are often stored as liquids under

their vapor pressures, which at ambient temperature can be quite large. Thus, n-

butane stored as a liquid/vapor system is at a pressure of 2.581 bar for a

temperature of 300 K. Largescale storage (>50 m3) of this kind is sometimes

done in spherical tanks. Suggest two reasons why.

Penyelesaian :

Tangki bola (spherical tanks) banyak digunakan untuk menyimpan cairan yang

mudah mendidih pada suhu yang relatif rendah, karena :

1) Tekanan dalam tangki bola didesain berada di bawah tekanan uap cairan

pengisi.

2) Tangki bola menawarkan volume maksimum untuk luas permukaan yang

kecil dan memiliki ketebalan 1½ kali ketebalan tangki silinder dengan

diameter yang sama

1.10) The first accurate measurements of the properties of high-pressure gases were

made by E. H. Amagat in France between 1869 and 1893. Before developing

the dead-weight gauge, he worked in a mine shaft, and used a mercury

manometer for measurements of pressure to more than 400 bar. Estimate the

height of manometer required.

Dik : P = 400 bar

g = 9,8 m/s2

ρ = 13,5 gr/cm3

Dit : h = ....?

Jawab :

P = 400 bar x 10 5 Pa1 bar

= 4 x 107 Pa

ρ = 13,5 gr/cm3 x 10 6 cm31 m 3

x 1 kg1000 gr

= 13,5 x 103 kg/m3

P = ρ.g.h

h = Pρ.g

= 4 x 10 7 Pa(13,5 x 103 kg/m 3) (9,8 m/s 2)

= 0,03023 x 104 m

= 302,3 m

1.11) An instrument to measure the acceleration of gravity on Mars is constructed of

a spring from which is suspended a mass of 0.40 kg. At a place on earth where

the local acceleration of gravity is 9.81 m.s-2, the spring extends 1.08 cm. When

the instrument package is landed on Mars, it radios the information that the

spring is extended 0.40 cm. What is the Martian acceleration of gravity?

Penyelesaian :

Dik : m = 0,4 kg

gbumi = 9,8 m/s2

xbumi = 1,08 cm

xmars = 0,4 cm

Dit : gmars = ....?

Jawab :

Pada Bumi → xbumi = 1,08 cm x 1 m100 cm

= 1,08 x 10-2 m

F = m.gbumi = K. xbumi

(0,4 kg)( 9,8 m/s2) = K. 1,08 x 10-2 m

K = 3,6296 x 102 kg/s2

Pada Mars → xbumi = 0,4 cm x 1 m100 cm

= 0,4 x 10-2 m

F = m. gmars = K. xmars

(0,4 kg) gmars = (3,6296 x 102 kg/s2) (0,4 x 10-2 m)

gmars = 3,629 m/s2

1.12) The variation of fluid pressure with height is described by the differential

equation:

Here, ρ is specific density and g is the local acceleration of gravity. For an ideal

gas, ρ = MP/RT, where M is molar mass and R is the universal gas constant.

Modeling the atmosphere as an isothermal column of ideal gas at 283.15 K (l0

°C), estimate the ambient pressure in Denver, where z = l (mile) relative to sea

level. For air, take M = 29 g mol-1; values of R are given in App. A.

Penyelesaian :

Dik : z = 1 mile

T = 10 oC

M = 29 g/mol

R = 82,06 cm 3 .atmmol.K

g = 9,8 m/s2

Dit : PDenver = ....?

Jawab :

T = 10 oC + 273,15 = 283,15 K

z = 1 mile x 5280 ft1 mile

x 1 m3,2808 ft

x 100 cm1 m

= 1609,36 x 102 cm

g = 9,8 m/s2 x 100 cm1 m

= 980 cm/s2

dPdz

= - ρg .....(1)

ρ = M.PR.T

.....(2)

substitusi pers (2) ke pers (1) :

dPdz

= - M.PR.T

g

∫1

P1P

dP = - M.gR.T

∫0

z

dz

ln P 1P = -

M.gR.T

z

ln P1

= - M.gR.T

z

ln P = - M.gR.T

z

P = e-M.gR.T

z ..... (1)

- M.gR.T z =

-29 g cm /mol.980 cm/s 2

82,06 cm 3 .atmmol.K

.283,15 K 1609,36 x 102 cm

= - 0,196 .....

(2)

Substitusi pers (2) ke pers (1) :

P = e-M.gR.T

z

P = e−0,196 = 0,822 atm

1.13) A group of engineer has landed on the moon, and they wish to determine the

mass of some rocks. They have a spring scale calibrated to read pounds mass of

at a location where the acceleration of gravity is 32,186 (ft)(s)-2. One of the

moon rocks gives a reading of 18,76 on this scale. What is its mass? What is its

weight on the moon? Take g (moon) = 5,32 (ft)(s)-2.

Penyelesaian :

Dik : gbumi = 32,186 (ft)(s)-2

gbulan = 5,32 (ft)(s)-2

Δlbulan = 18,76

Dit : m = ....?

wbulan = ....?

Jawab :

mbulan = mbumi

k. Δl bulang bulan

=k . Δl bumig bumi

Δlbumi =Δl bulang bulan

gbumi

=18,765,32 (ft)(s) -2

32,186 (ft)(s)-2

= 113,498

m = Δlbumi lbm

= 113,498 lbm

Wdibulan = m . gbulan

= 113,498 lbm . 5,32 ft s-2

= 603,809 lbf

1.14) A 70 W outdoor security light burns, on average, 10 hours a day. A new bulb

costs $5.00, and the lifetime is about 1000 hours. If electricity costs $0.10 per

kWh, what is the yearly price of "security," per light?

Penyelesaian :

Dik : P = 70 W

t = 10 jam/hari

costbulb = 5 dollar/1000 jam

costelec = 0,1 dollar/kWh

Dit : costtotal per tahun = ....?

Jawab :

costbulb per tahun = costbulb . t . 365 hari

= (5 dollar/1000 jam)(10 jam/hari)(365 hari)

= 18,25 dollar

costelec per tahun = costelec . P . t . 365 hari

= (0,1 dollar/kW.jam)(0,07 kW)(10 jam/hari)

(365 hari)

= 25,55 dollar

costtotal per tahun = costbulb per tahun + costelec per tahun

= (18,25 + 25,55) dollar

= 43,8 dollar

1.15) A gas is confined in a 1,25(ft)-diameter cylinder by a piston, on which rests a

weight. The mass of the piston and weight together is 250 (lbm). The local

acceleration of gravity is 32,169 (ft)(s-2), and atmospheric pressure is 30,12 (in

Hg).

(a) What is the force in lbf exerted on the gas by the atmosphere, the piston,

and the weight, assuming no friction between the piston and cylinder?

(b) What is the pressure of the gas in kPa?

(c) If the gas in the cylinder is heated, it expands, pushing the piston and

weight upward. If the piston and weight are raised 1,7 ft, what is the work

done by the gas in (ft lbf)? What is the change in potential energy of the

piston and weight?

Penyelesaian :

Dik : D = 1,25 ft

m = 250 lbm

g = 32,169 ft.s-2

Patm= 30,12 in Hg

Dit : a) F = ..?

b) P = ..? dalam kPa

c) w = ...? dan Ep = ..? jika ∆x = 1,7 ft

Jawab :

A = 14

πD2

= 14

(3,14) (1,25 ft)2

= 1,227 m2

a) F = patm . A + m.g

patm = 30,12 inHg x

1 psia2,0360 inHg

x 6,89476. 104 dyn cm-2

1 psia x

2,0886 10-3 lbf ft-2

1 dyn cm-2

= 2130,35 lbf ft-2

F = 2130,35 lbf ft-2 . 1,226 ft2 + 250 lbm . 32,169 ft s-2

F = 10,1726 x 103 lbf

b) P = FA

= 10 ,1726 .103

1,2265 ft2

= 8,294. 103 lbf ft-2

P = 8,294. 103 lbf ft-2 x 1 dyn cm-2

2 ,0883 . 10-3 lbf ft-2 x 1 psia6,89476.104 dyn cm-2

P = 57,595 psia

c) w = F. ∆l

= 10,1726 x 103 lbf . 1,7 ft

= 17,2934 lbf ft

Ep = m.g.∆l

= 250 lbm . 32,169 ft.s-2 . 1,7 ft

= 13,671. 103 lbf ft

1.16) A gas is confined in a 0.47-m-diameter cylinder by a piston, on which rests a

weight. The mass of the piston and weight together is 150 kg. The local

acceleration of gravity is 9.813 m sP2, and atmospheric pressure is 101.57 kPa.

(a) What is the force in newtons exerted on the gas by the atmosphere, the

piston, and the weight, assuming no friction between the piston and

cylinder?

(b) What is the pressure of the gas in kPa?

(c) If the gas in the cylinder is heated, it expands, pushing the piston and

weight upward. If the piston and weight are raised 0.83 m, what is the work

done by the gas in kJ? What is the change in potential energy of the piston

and weight?

Dik : D = 0,47 m A = 0,1734 m2

m = 150 kg

g = 9,813 ms-2

patm = 101,57 kPa

Dit : a) F = ..?

b) P = ..? dalam kPa

c) w = ...? dan Ep = ..? jika ∆x = 0,83 m

Jawab : A = 14

πD2

= 14

(3,14) (0,47 m)2

= 0,1734 m2

a) F = Patm . A + m.g

F = 101,57. 103 Pa . 0,1734 m2 + 150 kg . 9,813 ms-2

F = 19,08395. 103 N

b) P = FA

= 19 ,08395 . 103N0,1734 m2 = 110,057 kPa

c) w = F. ∆x

= 19,08395. 103 N. 0,83 m

= 15,8396 kJ

Ep = m.g.∆x = 150 kg . 9,813 ms-2 . 0,83 m = 1,2217 kJ

1.17) Verify that the SI unit of kinetic and potential energy is the joule.

Penyelesaian :

Ek = EP

½ mv2 = m.g.h

½ (kg)(m/s)2 = kg.m/s2.m

½ kg.m2/s2 = kg.m2/s2

Joule = Joule

1.18) An automobile having a mass of 1250 kg is traveling at 40 m.s-1. What is its

kinetic energy in kJ? How much work must be done to bring it to a stop?

Penyelesaian :

Dik : m = 1250 kg

v = 40 m.s-1

Dit : Ek = ....?

W = ....?

Jawab :

Ek = ½ mv2 = ½ (1250 kg)( 40 m/s)2

= 1000000 J = 1000 kJ

W = ΔEk = 1000 kJ