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GAS LAWS!

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COMPRESSIBILITY Compressibility is a measure of how much the VOLUME of matter can DECREASE under pressure.

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**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.

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**VARIABLES THAT DESCRIBE GASES**

Temperature (T) (KELVIN) PRESSURE (P) (atmosphere) Volume (V) (mL) Number of MOLES (N)

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**TEMPERATURE Gas laws use temperature in the SI unit of KELVIN.**

TEMPERATURE= measure of the AVERAGE kinetic energy of Particles Converting CELSIUS to Kelvin: Kelvin= Celsius Temperature +273 EX: What is 16 degrees Celsius in Kelvin?

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**TEMPERATURE (T) Gas laws use temperature in the SI unit of KELVIN.**

Converting Celsius Temperatures to Kelvin: Kelvin= Celsius Temperature +273 EX: What is 16 degrees Celsius in Kelvin? Kelvin= = 289

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PRESSURE (P) Pressure is caused by MOVING molecules hitting container walls PRESSURE can be measured in atmospheres, torr’s, mmHG, kPa, or psi The SI unit for pressure is ATMOSPHERES. Conversion of Pressure: 1 atm= 760 torr= 760mmHG= kPa= 14.7psi

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**PRESSURE CONVERSION EXAMPLES**

How many torrs are in 2 atm? How many HGmm are in 2 atm? How many atm are in 100 kPa?

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**PRESSURE CONVERSION ANSWERS**

How many torrs are in 2 atm? 1520 torr How many HGmm are in 2 atm? 1520 Hgmm How many atm are in 50 kPa? .494 atm

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**VOLUME (V) Volume is measured in MILLILITERS.**

Volume is the amount of SPACE occupied by a substance.

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NUMBER OF MOLES (n) The number of MOLES represents the number of PARTICLES contained in a system. 6.02X particles = 1 mole 22.4 L= 1 mole

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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

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WHAT IS ‘R’? “R” is the IDEAL GAS CONSTANT. The UNIT of pressure need to match up! R = L • atm mol • K R = 8.31 L • kPa R = 62.4 L • mmHg

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Sample Problem What volume will 2.50 mol of hydrogen (H2) occupy at °C and 1.5 atm?

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Solve… Use the Ideal Gas Law: PV = nRT Solve for V: V = nRT P

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**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 K 1.5 atm mol K

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Final Answer V = 34.6 L

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PART 2!

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**BOYLES LAW PRESSURE vs. VOLUME**

As one increase the other decreases. This is an INVERSE relationship.

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BOYLES LAW GRAPH As the Volume decreases the pressure increases.

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**BOYLES LAW EQUATION k = VxP**

k is a CONSTANT for a certain sample of gas that depends on the mass of the gas and the temperature.

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**BOYLES 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 set up a RATIO.

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BOYLES LAW EXAMPLE

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BOYLES LAW EXAMPLE Consider 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?

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**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.

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Charles’ Law DIRECT Relationship: As temperature increases the volume increases.

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**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.

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**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/ T2 V1/T1 = V2/ T2

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CHARLES LAW EXAMPLE A 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?

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CHARLES LAW ANSWER The first step is to convert the temperature from Celsius to Kelvin: = 303 K Rearrange Charles’ law to solve for the new temperature: T2= V2 T1 / V1 T2 = 1200mL X 303 K = 2 X 303K = mL T2 = 606K

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GAY-LUSSAC’s GAS LAW Lussac’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.

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Direct Relationship Gay-Lussac's Law is a DIRECT relationship when one increases the other also increases. Pressure TEMPERATURE

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**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 = k And we know this: P2 ÷ T2 = k Since k = k, we can conclude that: P1 ÷ T1 = P2 ÷ T2

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GAY LUSSACS EXAMPLE 10.0 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)?

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GAY LUSSACS STEPS STEP 1: Change 25.0°C to K and remember that standard pressure in kPa is STEP 2: Insert values into the equation. STEP 3: Solve for x

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GAY LUSSACS ANSWER The 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?

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COMBINED GAS LAW BOYLES 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 / T2 P= Pressure V= VOLUME T= Temperature.

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……..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!

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Gas Laws Lesson 2.

Gas Laws Lesson 2.

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