NORAFIZAH BINTI AHMAD SHAYUTIBSc (HONS.) OLEOCHEMISTRY, UKMMSC CHEMISTRY, UKM

CHAPTER 6GASES

PROPERTIES OF GASESGeneral properties of gasCompressibleHave low densityDiffuses quicklyFills up a container uniformlyExerts pressure uniformly on all sides of a container independently of the height or depth

GAS PRESSUREPressure of air is measured with a BAROMETER (developed by Torricelli in 1643)Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). (Just like a straw in a soft drink)P of Hg pushing down related to Hg densitycolumn height

PRESSURE CONVERSIONPressure is defined as force per unit area :Pressure = force / area

1 atm = 760 mmHg = 760 torr = 101325 Pa = 101325 Nm-2

Conversion : 475 mmHg to atm? = 475 mmHg x 1 atm 760 mmHg = What is 2 atm expressed in torr? 2 atm x 760 torr = 1 atm1520 torr

0.625 atm

BOYLES LAWStates that : the volume (V) of a sample of gas inversely proportional with the pressure (P) of the gas at constant temperature (T) and number of moles of gas (n). V 1(at constant T and n) P V = kor PV = k Pwhere k is constantThe initial condition: P1V1 = k , The final condition: P2V2 = kBecause PV, initial and final, is constant and is equal to k

P1V1 = P2V2

BOYLES LAWA bicycle pump is a good example of Boyles law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.

(Temperature is held constant)

PROBLEMA 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 cm3. At the surface the pressure is 1 atmosphere. What is the volume of each bubble when it reaches the surface?

How we work this out:We assume that the temperature is constant, so Boyles Law applies:Formula first:P1 x V1 = P2 x V2

Then numbers:= 3.0 x 2 = 1.0 x V2Now rearrange the numbers so that you have V2 on one side, and the rest of the numbers on the other side of the equals symbol.

Heres what you should have calculatedV2 = 3.0 x 2 1.0therefore volume of bubbles = 6 cm3

Note that P1 and P2 have the same unit, as will V1 and V2

PROBLEMAn ideal gas occupies a volume of 1.5 dm3 under a pressure 0f 100.5 kPa. Assume that the temperature is maintained constant, calculate its volume if the pressure is increased to 150 kPa.

CHARLESS LAWStates that : the volume (V) of a gas varies directly with the temperature (in Kelvin) if pressure (P) and number of moles of gas (n) are constant V T(at constant P and n) V = kTor V = k T where k is constant and T is the absolute temperatureInitial condition:Final condition: V1= k V2= k T1 T2Since k is constant, we may equate them, resulting in V1 = V2 T1 T2

CHARLESS LAW

Charless law only applies to the volume of a gas at Kelvin Temperature. The relationship between the Kelvin and Celcius temperature scale is K = oC + 273

ALERT!!!!

(Pressure is held constant)

PROBLEMAt constant pressure, the volume of a gas is increased from 150 dm3 to 300 dm3 by heating it. If the original temperature of the gas was 20 oC, what will its final temperature be (oC)?

T1 = 20 oC + 273 = 293 KT2 = X KV1 = 150 dm3V2 = 300 dm3 150 dm3 293 K 300 dm3 T2 T2 = 586 KoC = 586 K - 273T2 = 313 oC=

PROBLEMAn ideal gas in a balloon has a volume of 205 mL at temperature of 5 oC. Assume that the pressure is maintained constant, determine the volume of the balloon if the temperature is lowered to -17 o C.

GAY-LUSSACS LAWStates that : the pressure exerted by a gas is directly related to the Kelvin temperature P T(at constant V and n) P = kT or P = k Twhere k is constant and T is the absolute temperatureInitial condition:Final condition: P1= k P2= k T1 T2Since k is constant, we may equate them, resulting in

P1 = P2T1 T2

GAY-LUSSACS LAW(Volume is held constant)

PROBLEM A gas has a pressure at 2.0 atm at 18 C. What is the new pressure when the temperature is 62 C? (V and n constant)

P1 = 2.0 atmT1 = 18 C + 273 = 291 K T2 = 62 C + 273 = 335 KP2 = ? P1 = P2 P2 T1 T2

P2 P2 = P1 x T2 T1 = 2.0 atm x 335K 291 K =

2.3 atm

PROBLEM A gas has a pressure of 645 torr at 128 C. What is the temperature if the pressure increases to 824 torr? (n and V remain constant)

COMBINED GAS LAWIdeal gases obey Boyles law & Charless law.All pressure-volume-temperature relationship for gases that we have studied may be combined into a single relationship called the combined gas law.The combined gas law in an equation which can be derived from Boyles law and Charless law.

Boyles law = V 1 P Charless law = V T Combined gas law : V T P V = kT P hence; k = PV T

A relationship comparing initial and final conditions:

(At constant number of moles, n)

COMBINED GAS LAWP1V1 = P2V2T1 T2

AVOGADROS LAWStates that: equal volumes of any ideal gas contain the same number of moles if measured under the same temperature and pressure. V n(at constant T and P) V = kn orV = k nA relationship comparing initial and final conditions, may be derived:

V1 = V2n1 n2

Molar volume of a gas is the volume occupied by 1 mole of gas.At standard temperature and pressure(STP), the molar volume of any gas is 22.4 L (22.4 dm3).

At room temperature (25 oC or 298 K) and pressure 1 atm, the molar volume of a gas is 24 dm3.

Standard temperature , T = 273 K (or oC)Standard pressure, P = 1 atm

AVOGADROS LAW

PROBLEMCalculate the volume of the following gaseous at STP. a)0.05 mol H2b)14 g of CO2(Ar: H = 1, C = 12, O = 16)

SOLUTION (a)At STP, a mole of gas occupied 22.4dm3Hence;The V occupied by 0.05 mol H2 = 0.05 mol x 22.4dm3 1 mol = 1.12 dm3

SOLUTION (b)Convert the mass of CO2 to moleHence;No. of mol of CO2 = 14 g x 1 mol CO2 44 g = 0.318 molThe V occupied by 0.318 mol CO2 = 0.318 mol x 22.4dm3 1 mol = 7.12 dm3