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Physical Chemistry I(TKK-2246)13/14 Semester 2Instructor: Rama OktavianEmail: rama.oktavian86@gmail.comOffice Hr.: M.13-15, Tu. 13-15, W. 13-15, Th. 13-15, F. 09-11

Outlines 1. Gas: Properties 2. Gas laws: Boyle and Charles law 3. Ideal gas law 4. Equation of stateGas stateState of matterThere are 4 states of matter

Gas stateState of matterProperties of solid

Microscopic view of a solidretains a fixed volume and shape rigid - particles locked into placenot easily compressible little free space between particlesdoes not flow easily rigid - particles cannot move/slide past one anotherGas stateState of matterProperties of liquidMicroscopic view of a solidassumes the shape of the part of the container which it occupies particles can move/slide past one anothernot easily compressible little free space between particlesflows easily particles can move/slide past one another

Gas stateState of matterProperties of gasMicroscopic view of a solidassumes the shape and volume of its container particles can move past one anothercompressible lots of free space between particlesflows easily particles can move past one another

Microscopic propertiesGas stateState of matterProperties of gasMicroscopic view of a solidProperties that can be observed and measured

Macroscopic properties Properties of bulk gases Observable Pressure, volume, mass, temperature..How to make relation between those macroscopic properties of gas??The general form of an equation of state isp=f(T,V,n)Gas state

Gas propertiesGases Exert Pressure: What is Pressure?Pressure is defined as the force exerted divided by the area it acts overPressure = Force/AreaThe SI unit of pressure, the pascal(Pa), is defined as 1 newton per metre-squared:1 Pa =1 N m2

1 Pa =1 kg m1s21 atm =1.013 25 105Pa exactly 1 bar =105PaGas propertiesGases Exert Pressure: What is Pressure?Pressure is defined as the force exerted divided by the area it acts overPressure = Force/AreaSelf-test 1.1 Calculate the pressure (in pascals and atmospheres) exerted by a mass of 1.0 kg pressing through the point of a pin of area 1.0 102 mm2 at the surface of the Earth. Gas propertiesPressure measurementBarometer device that measures atmospheric pressureInvented by Evangelista Torricelli in 1643

the height of the mercury column is proportional to the external pressureGas propertiesPressure measurement

Derive an equation for the pressure at the base of a column of liquid of mass density (rho) and height h at the surface of the Earth.p=F/AF = mgm = VV = Ahm = AhF = mg = Ahg

the pressure is independent of the shape and cross-sectional area of the column. Gas propertiesPressure measurement

Gas propertiesPressure measurementA manometer measures the pressure of a gas in a containerGas pressure is the force exerted by the collisions of gas particles with a surface

Gas propertiesPressure measurementAtmospheric pressureBy definition the average pressure at sea level will support a column of 760 mm of mercury. (760 torr)

g = 9.81 m.s-2 h = 0.76 m = 13.6 g.cm-3 = 13.6 kg.L-1 = 13.6x103kg.m-3P = 9.81x0.76x13.6x103 = 1.013x105 Pa (N.m-2)Gas propertiesPressure measurementAtmospheric pressure problemIf we made a barometer out of water, what would be the height of the water column if the pressure is 745 torr?Gas propertiesPressure measurementVariation in atmospheric pressureChanging weather conditions

Gas propertiesPressure measurementVariation in atmospheric pressureChanging altitude

Gas lawsBoyles law Boyles Law is one of the laws in physics that concern the behaviour of gases When a gas is under pressure it takes up less space: The higher the pressure, the smaller the volume Boyles Law tells us about the relationship between the volume of a gas and its pressure at a constant temperature The law states that pressure is inversely proportional to the volume

Gas lawsBoyles lawPressure-volume relationshippressure-volume behavior of gases were made by Robert Boyle in 1662P a 1/VP x V = constantP1 x V1 = P2 x V2

Gas lawsPressure and Volume: Boyles LawVolume and pressure are inversely proportional. If one increases the other decreases.

Gas lawsPressure and Volume: Boyles LawBoyles Law and Breathing: InhalationDuring inhalation,the lungs expandthe pressure in the lungs decreasesair flows towards the lower pressure in the lungs

Gas lawsPressure and Volume: Boyles LawBoyles Law and Breathing: InhalationDuring exhalationlung volume decreasespressure within the lungs increasesair flows from the higher pressure in the lungs to the outside

Gas lawsPressure and Volume: Boyles LawA 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?P1 x V1 = P2 x V2P1 = 726 mmHgV1 = 946 mLP2 = ?V2 = 154 mLP2 = P1 x V1V2726 mmHg x 946 mL154 mL== 4460 mmHgGas lawsPressure and Volume: Boyles LawGuide to Calculations with Gas Laws

Gas lawsPressure and Volume: Boyles LawProblemA deep sea diver is working at a depth where the pressure is 3.0 atmospheres. He is breathing out air bubbles. The volume of each air bubble is 2 cm2. At the surface the pressure is 1 atmosphere. What is the volume of each bubble when it reaches the surface?

Gas lawsCharless law French chemist Jacques Charles discovered that the volume of a gas at constant pressure changes with temperature. As the temperature of the gas increases, so does its volume, and as its temperature decreases, so does its volume. The law says that at constant pressure, the volume of a fixed number of particles of gas is directly proportional to the absolute (Kelvin) temperatureGas lawsCharless lawVolume-temperature relationshipVariation of gas volume with temperatureat constant pressureV a TV = constant x TV1/T1 = V2/T2

Gas lawsCharless lawVolume-temperature relationshipFor two conditions, Charless law is writtenV1 = V2 (P and n constant) T1 T2

Rearranging Charless law to solve for V2 givesT2 x V1 = V2 x T2 T1 T2

V2= V1 x T2 T1Gas lawsCharless lawExample problemA balloon has a volume of 785 mL at 21 C. If thetemperature drops to 0 C, what is the new volume ofthe balloon (P constant)?

STEP 1 Set up data table:

Conditions 1 Conditions 2 Know PredictV1 = 785 mL V2 = ? V decreasesT1 = 21 C T2 = 0 C = 294 K = 273 K T decreasesBe sure to use the Kelvin (K) temperature in gascalculations.Gas lawsCharless law STEP 2 Solve Charless law for V2: V1 = V2 T1 T2

V2 = V1 x T2 T1Temperature factor decreases T

STEP 3 Set up calculation with data: V2 = 785 mL x 273 K = 729 mL 294 KGas lawsCharless lawA sample of oxygen gas has a volume of 420 mL at a temperature of 18 C. At what temperature (in C) will the volume of the oxygen be 640 mL (P and n constant)?1) 443 C2) 170 C 3) 82 CGas lawsAvogadros lawAvogadros law states thatthe volume of a gas is directly related to the number of moles (n) of gas T and P are constant V1 = V2 n1 n2

Ideal Gas law

The combination of those laws gives

Usually written as:

R is gas constantIdeal Gas lawR is known as universal gas constant Using STP conditions

Ideal Gas lawR is known as universal gas constant Learning checkWhat is the value of R when the STP value for P is 760 mmHg?Ideal Gas lawProblemLearning checkA 5.0 L cylinder contains oxygen gas at 20.0C and 735 mm Hg. How many grams of oxygen are in the cylinder?

Ideal Gas lawProblemLearning checkIn an industrial process, nitrogen is heated to 500 K in a vessel of constant volume. If it enters the vessel at 100 atm and 300 K, what pressure would it exert at the working temperature if it behaved as a perfect gas?Equation of stateEquation of stateThe general form of an equation of state isp=f(T,V,n)

Ideal gas equation is equation of stateEquation of stateEquation of state

P, V, n, T are propertiesIntensive properties independent on the quantity of materialP, T Extensive properties dependent on the quantity of materialn, V

Intensive properties The ratio of any two extensive variables is always an intensive variableEquation of stateEquation of state

Expressed in intensive properties gives

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