G A S _KIMIA INDUSTRI_ DEWI HARDININGTYAS, ST, MT, MBA WIDHA KUSUMA NINGDYAH, ST, MT AGUSTINA EUNIKE, ST, MT, MBA

Transkripsi

1 G A S _KIMIA INDUSTRI_ DEWI HARDININGTYAS, ST, MT, MBA WIDHA KUSUMA NINGDYAH, ST, MT AGUSTINA EUNIKE, ST, MT, MBA

2 Elemen Berwujud Gas pada 25 0 C dan 1 atm

3 Karakteristik Fisika dari Gas Gas diasumsikan mempunyai volume dan bentuk sesuai tempatnya. Gas adalah wujud materi yang (paling) dapat terkompresi (mendapat variasi tekanan) untuk mampat (atau memuai). Gas akan bercampur jika tergabung dalam satu tempat. Gas mempunyai kerapatan dan berat jenis lebih ringan dibandingkan wujud cair atau padat. Terpengaruh tekanan pada lingkungannya.

4 Perubahan Tekanan

5 Satuan Tekanan Tekanan = Gaya Area Satuan Tekanan 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mmhg = 760 torr 1 atm = Pa

6 Tekanan Udara 6 10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm

7 Manometer

8 Barometer

9 Hukum Boyle (Boyle s Law) Dikemukakan pada 1660 oleh Robert Boyle Jika temperatur tetap konstan, volume suatu gas dengan massa tertentu, berbanding terbalik dengan tekanan V 1 / P P.V = konstan V1 / V2 = P2 / P1

10 Hukum Boyle (Boyle s Law)

11 Hukum Boyle (Boyle s Law) As P (h) increases V decreases

12 Hukum Boyle (Boyle s Law) P 1/V P x V = constant P 1 x V 1 = P 2 x V 2 Constant temperature Constant amount of gas

13 Hukum Boyle (Boyle s Law) Pressure x Volume = Constant (T = constant) P 1 V 1 = P 2 V 2 (T = constant) V 1/P (T = constant) ( * Holds precisely only at very low pressures.)

14 Tekanan ditambah, volume SO 2 menurun

15 Contoh Perhitungan A sample of chlorine gas occupies a volume of 946 ml at a pressure of 726 mmhg. What is the pressure of the gas (in mmhg) if the volume is reduced at constant temperature to 154 ml? P 1 = 726 mmhg P 1 x V 1 = P 2 x V 2 P 2 =? V 1 = 946 ml V 2 = 154 ml P 2 = P 1 x V 1 V mmhg x 946 ml = = 4460 mmhg 154 ml

16 Hukum Charles (Charles s Law) Dikemukakan pada 1787 oleh Jacques Charles dan dirumuskan pada 1802 oleh Joseph L. Gay Lussac Jika tekanan tak berubah, volume gas dengan massa tertentu, berbanding lurus dengan temperatur V T

17 Hukum Charles (Charles s Law) As T increases V increases

18 Hukum Charles (Charles s Law)

19 Hukum Charles (Charles s Law) The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin. V = bt (P = constant) b = a proportionality constant

20 Hukum Charles (Charles s Law) V 1 V 2 ( P constant) T 1 T 2

21 Contoh Perhitungan A sample of carbon monoxide gas occupies 3.20 L at C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 /T 1 = V 2 /T 2 V 1 = 3.20 L T 1 = K V 2 = 1.54 L T 2 =? T 2 = V 2 x T 1 V L x K = = 192 K 3.20 L

22 Hukum Gay Lussac Dikemukakan pada 1703 oleh Joseph L. Gay Lussac dan Guillaume Amontons Tekanan suatu gas dengan massa tertentu berbanding lurus dengan temperatur P T

23 Hukum Avogadro Dikemukakan pada 1811 oleh Amadeo Avogadro Molekul yang sama banyak terdapat dalam gas-gas berlainan yang volumenya sama, jika tekanan dan temperaturnya sama V n

24 Hukum Avogadro V number of moles (n) V = constant x n V 1 /n 1 = V 2 /n 2 Constant temperature Constant pressure

25 Efek penambahan mol partikel gas pada temperatur dan tekanan konstan.

26 Persamaan Gas Ideal Boyle s law : V 1 (at constant n and T) P Charles law : V T (at constant n and P) Gay Lussac law : P T (at constant n and V) Avogadro s law : V n (at constant P and T) V nt P V = constant x nt P = R nt P R is the gas constant PV = nrt

27 Hukum Gas Ideal PV = nrt P = pressure in atm V = volume in liters n = moles R = proportionality constant = L atm K -1 mol -1 T = temperature in Kelvins Holds closely at P < 1 atm

28 Tetapan R Gas Rumus Volume Tetapan R Ideal 22,414 0, Hydrogen H 2 22,428 0, Helium He 22,426 0, Neon Ne 22,425 0, Nitrogen N 2 22,404 0, Carbon Monoxide CO 22,403 0, Oxygen O 2 22,394 0, Argon Ar 22,393 0, Gas Rumus Volume Tetapan R Nitrogen Oxyde NO 22,389 0, Methane CH 4 22,360 0, Carbon Dioxide Hydrogen Chloride CO 2 22,256 0, HCl 22,249 0, Ethilene C 2 H 4 22,241 0, Asetilene C 2 H 2 22,190 0, Ammonia NH 3 22,094 0, Chloride Cl 2 22,063 0,080760

29 Tetapan R The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies L. PV = nrt R = PV nt = (1 atm)(22.42l) (1 mol)( K) R = L atm / (mol K)

30 Contoh Perhitungan What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0 C = K PV = nrt V = nrt P V = P = 1 atm n = 49.8 g x 1 atm L atm mol K 1 mol HCl g HCl 1.37 mol x x K V = 30.6 L = 1.37 mol

31 Contoh Perhitungan Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? nr V P 1 T 1 PV = nrt = = P T P 2 T 2 P 2 = P 1 x n, V and R are constant = constant T 2 T 1 = 1.20 atm x P 1 = 1.20 atm T 1 = 291 K 358 K 291 K P 2 =? T 2 = 358 K = 1.48 atm

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