2 COMPRESSIBILITYCompressibility is a measure of how much the VOLUME of matter can DECREASE under pressure.
3 COMPRESSIBILITY OF GASES According to the KMT gas particles are so SMALL in relation to the distances between them that their individual volumes are virtually insignificant. This means that there is a lot of EMPTY space between the individual gas particles. This space between the particles explains why gases can be COMPRESSED.
4 VARIABLES THAT DESCRIBE GASES Temperature (T) (KELVIN)PRESSURE (P) (atmosphere)Volume (V) (mL)Number of MOLES (N)
5 TEMPERATURE Gas laws use temperature in the SI unit of KELVIN. TEMPERATURE= measure ofthe AVERAGE kineticenergy of ParticlesConverting CELSIUS to Kelvin:Kelvin= Celsius Temperature +273EX: What is 16 degrees Celsius in Kelvin?
6 TEMPERATURE (T) Gas laws use temperature in the SI unit of KELVIN. Converting Celsius Temperatures to Kelvin:Kelvin= Celsius Temperature +273EX: What is 16 degrees Celsius in Kelvin?Kelvin= = 289
7 PRESSURE (P)Pressure is caused by MOVING molecules hitting container wallsPRESSURE can be measured in atmospheres, torr’s, mmHG, kPa, or psiThe SI unit for pressure is ATMOSPHERES.Conversion of Pressure:1 atm= 760 torr= 760mmHG= kPa= 14.7psi
8 PRESSURE CONVERSION EXAMPLES How many torrs are in 2 atm?How many HGmm are in 2 atm?How many atm are in 100 kPa?
9 PRESSURE CONVERSION ANSWERS How many torrs are in 2 atm?1520 torrHow many HGmm are in 2 atm?1520 HgmmHow many atm are in 50 kPa?.494 atm
10 VOLUME (V) Volume is measured in MILLILITERS. Volume is the amount of SPACE occupied by a substance.
11 NUMBER OF MOLES (n)The number of MOLES represents the number of PARTICLES contained in a system.6.02X particles = 1 mole22.4 L= 1 mole
12 IDEAL GAS LAW!The ideal gas law is the mathematical relationship among pressure (P), volume (V), temperature(T) , and the number of moles(n) of a gas:PV = nRT
13 WHAT IS ‘R’?“R” is the IDEAL GAS CONSTANT. The UNIT of pressure need to match up!R = L • atmmol • KR = 8.31 L • kPaR = 62.4 L • mmHg
14 Sample Problem What volume will 2.50 mol of hydrogen (H2) occupy at °C and 1.5 atm?
15 Solve…Use the Ideal Gas Law: PV = nRTSolve for V:V = nRTP
16 Substitute: V = 2.50 mol X 0.0821 L atm X 253 K 1.5 atm mol K Don’t forget to convert temperature to Kelvin!!!V = 2.50 mol X L atm X 253 K1.5 atm mol K
24 BOYLES LAW EXAMPLEConsider a 1.53-L sample of gaseous SO2 at a pressure of 5.6 x 103 Pa. If the pressure is changed to 1.5 x 104 Pa at constant temperature, what will be the new volume of the gas?
25 CHARLES LAW TEMPERATURE vs. VOLUME Charles Law states that the volume of a fixed mass of gas varies directly with temperature at a constant pressure.
26 Charles’ LawDIRECT Relationship: As temperature increases the volume increases.
27 CHARLES LAW EQUATION V= kT or k=V/T K is a CONSTANT for certain sample of gas that depends on the mass of gas and its pressure.
28 CHARLES LAW EQUATION- RATIO If we have a set of new conditions for the same sample of gas they will have the same k so we can use a ratio to solve for our variable.V1/T1 = k = V2/ T2V1/T1 = V2/ T2
29 CHARLES LAW EXAMPLEA gas with a volume of 600 mL has a temperature of 30 0C. At constant pressure the gas is heated until the gas expands to 1,200 mL. What is the new temperature of the gas if the pressure remains constant?
30 CHARLES LAW ANSWERThe first step is to convert the temperature from Celsius to Kelvin: = 303 KRearrange Charles’ law to solve for the new temperature: T2= V2 T1 / V1T2 = 1200mL X 303 K = 2 X 303K = mLT2 = 606K
31 GAY-LUSSAC’s GAS LAWLussac’s Gas Law states that if the TEMPERATURE of a gas is increased, and the volume is held constant, the pressure of the gas will also INCREASE.
32 Direct RelationshipGay-Lussac's Law is a DIRECT relationship when one increases the other also increases.PressureTEMPERATURE
33 GAY LUSSACS EQUATION P ÷ T = k If we have two sets of data for the same gas we can set the equations EQUAL to each other.We know this: P1 ÷ T1 = kAnd we know this: P2 ÷ T2 = kSince k = k, we can conclude that:P1 ÷ T1 = P2 ÷ T2
34 GAY LUSSACS EXAMPLE10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What would be the required temperature (in Celsius) to change the pressure to standard pressure( kPa)?
35 GAY LUSSACS STEPSSTEP 1: Change 25.0°C to K and remember that standard pressure in kPa isSTEP 2: Insert values into the equation.STEP 3: Solve for x
36 GAY LUSSACS ANSWERThe answer is K, but the question asks for Celsius, so you subtract 273 to get the final answer of 38.3°C.BONUS: Can you convert the temperature to KELVIN?
37 COMBINED GAS LAWBOYLES LAW, Charles law, and Gay Lussacs Law can be combined to create a COMBINED GAS LAW.The Combined gas Law equation is:P1V1 / T1 = P2V2 / T2P= PressureV= VOLUMET= Temperature.
38 ……..Now It’s time for you to practice the different kinds of Gas Laws in your Gas Laws Packet. If you don’t finish in class- It’s HOMEWORK!