GAS LAWS

ASSUMPTIONS ABOUT GASES KINETIC THEORY OF GASES The volume of gas particles is so small that they do not contribute to the volume in a container. Particles of gas do not attract or repel each other Particles of gas are in constant motion

WHAT IS PRESSURE? In your notes write down characteristics of what you think pressure is:

BOOK DEFINITION ALL GASES EXERT PRESSURE Pressure: the force per unit area on a surface In other words, it’s how hard something is pushing against some sort of surface (for example a balloon or tire)

UNITS OF PRESSURE Atmosphere (atm) Millimeters of mercury (mm Hg) Pascal (Pa) Conversions: 1atm = 760mm Hg = 101.3 kPa

YOU HAVE TO BE ABLE TO CONVERT Example: If you have 560 mm Hg, how many atmospheres is this? Use dimensional analysis Conversion factor 1 atm = 760 mm Hg 560mm Hg | 1 atm___ = 0.734atm | 760mm Hg

TRY THESE If you start with 57.3 kPa, how many atm is this? If you have 6.5 atm, how many mm Hg is this? If you start with 435 mm Hg, how many kPa is this?

ANSWER 0.566 atm 4.9 x 103 mm Hg 58.0 kPa

DALTON’S LAW OF PARTIAL PRESSURES Dalton’s Law of Partial Pressures: The total pressure of a system is equal to the sum of all the partial pressures of the gases in the system. PTotal = P1 + P2 + P3 + . . .

EXAMPLE You have three gases in a container (N2, H2, and CO2). If they have the following pressures that they exert: N2 = 360 mm Hg H2 = 225 mm Hg CO2 = 86 mm Hg What is the total pressure in the container?

ANSWER PTotal = PN2 + PH2 + PCO2 = 671 mm Hg PTotal = 360mm Hg + 225mm Hg + 86mm Hg = 671 mm Hg

TRY THIS You have a container that has the following gases (O2 + Cl2 + NO2). If the total pressure is 6.99atm and the PO2 is 2.25atm and the PCl2 is 4.04atm, what is the P NO2?

ANSWER PT = PO2 + PCl2 + PNO2 6.99atm = 2.25atm + 4.04atm + PNO2

BOYLE’S LAW THOUGHT QUESTION: If you squeeze a balloon (so that the volume is less), does the pressure inside the balloon increase or decrease?

BOYLE’S LAW As we all just saw, the pressure increases. If we were to increase the space of the balloon (without adding anymore air), would the pressure increase or decrease.

BOYLE’S LAW The pressure would decrease Therefore pressure and volume are inversely proportional When pressure goes up, volume goes down and vice versa The individual that noticed this relationship was Robert Boyle. Pressure (P) = 1/Volume (V)

THE RESULT Mathematical relationship of an inverse proportion: PV = k (k is a constant that depends on the environment . . . Temperature, number or air particles, etc . . .) ASSUMPTIONS: Temperature remains constant and the number of air molecules does not change

BOYLE’S LAW If have the same conditions of temperature and number of air molecules, then different pressures and volumes will still equal the same k: P1V1 = k P2V2 = k

BOYLE’S LAW The end result: You can predict pressure or volume based on how you change the system: P1V1 = P2V2 This equation is known as Boyle’s Law

EXAMPLE If you start with a balloon that has a volume of 3.5L and a pressure of 10 atm, what will be the new pressure if you squeeze the volume down to 1700mL?

STEP 1 Write down the information you know P1 = 10atm V1 = 3.5L P2 = ? V2 = 1700mL

STEP 2 Make sure that the units for volume and pressure are the same. If not, then make them the same P1 = 10atm V1 = 3.5L P2 = ? V2 = 1700mL = 1.7L L and mL are not the same, so we convert mL to L to make them the same unit

STEP 3 Put the values in the formula P1V1 = P2V2 (10atm)(3.5L) = P2 (1.7L)

STEP 4 Solve the equation so that the units cancel out (10atm)(3.5L) = P2 (1.7L) P2 = (10atm)(3.5L) 1.7L P2 = 21 atm

TRY THESE If you have a container that has a volume of 25.0L and a pressure of 760mmHg, what is the new volume if you increase the pressure to 2.40atm? If you start with a volume of 250mL and a pressure of 67.3kPa and change the volume to 1.00L, what is the new pressure?

SOLUTION 10.4L 16.8 kPa

What happens to air as it is heated? Think of a hot air balloon CHARLE’S LAW Thought question: What happens to air as it is heated? Think of a hot air balloon

CHARLE’S LAW Answer: As you heat air, or any gas, it causes the volume to increase In addition, if you cool air, the volume of the gas will decrease

CHARLE’S LAW Therefore, temperature and volume are directly proportional As temperature increases, volume increases As temperature decreases, volume decreases The individual who noticed this relationship was Jacques Charles Volume (V) = Temperature (T)

CHARLE’S LAW Mathematical relationship of a direct relationship: V/T = k (k is the same constant as before) ASSUMPTIONS: That the system has a constant pressure and the same number of gas particles.

CHARLE’S LAW If have the same conditions of pressure and number of air molecules, then different temperatures and volumes will still equal the same k: V1/T1 = k V2/T2 = k

CHARLE’S LAW The end result: You can predict temperature or volume based on how you change the system: V1/T1 = V2/T2 This equation is known as Charle’s Law

REVIEW OF KELVIN TEMPERATURE Before we go any further, we need to review a special unit of temperature: Kelvin (K) A temperature scale based on absolute 0 (the coldest possible temperature) The Kelvin temperature converts to Celsius: K = °C + 273

KELVIN TEMPERATURE In any of the gas laws, you MUST first convert all temperatures into K before you can use the formulas

EXAMPLE If you have a container that has a temperature of 25 °C and a volume of 3.4L, what would be the new temperature if you increased the volume to 5.0L?

STEP 1 Write down your information T1 = 25°C V1 = 3.4L T2 = ?

STEP 2 Make sure that the units for volume are the same and temperature is in kelvins (K). If not, then convert. T1 = 25°C + 273 = 298K V1 = 3.4L T2 = ? V2 = 5.0L Converted Celsius to kelvin and the volume has the same unit already.

STEP 3 Put the values in the formula V1/T1 = V2/T2 (3.5L)/(298K) = (5.0L)/T2 NOTE: This is where you cross multiply

STEP 4 Solve the equation so that the units cancel out (3.5L)/(298K) = (5.0L)/T2 T2 = (5.0L)(298K)/(3.5L) T2 = 426K

TRY THESE A container has a temperature of 30°C and a volume of 335mL. If you heat the container to 50 °C, what is the new volume? A balloon has a volume of 25.0L and a temperature of 0 °C, what is the new volume if you decrease the temperature to -25 °C?

SOLUTION 357 mL 22.7L

TRY THIS A box has a temperature of -25°C and the dimensions of 25cm x 10cm x 0.5cm. If you heat the container to 280 K, what is the new volume?

SOLUTION Find the following: V1= 25cm x 10cm x 0.5cm = 125mL T1= -25°C + 273 = 248K V2= ? T2= 280K V2 = 141mL

COMBINED GAS LAW There is a way of looking at all three conditions of: Temperature Volume Pressure at the same time

COMBINED GAS LAW Let’s look at Boyle’s Law PV = k Let’s also look at Charles’ Law V/T = k THOUGHT QUESTION: How do you think you could combine P, V, and T to equal k in one equation?

COMBINED GAS LAW Notice: If you put them together: P and V are multiplied together (PV) V is divided by T (V/T) If you put them together: PV/T = k ASSUMPTIONS: You have the same number of air molecules.

COMBINED GAS LAW If have the same number of air molecules, then different temperatures, volumes and pressures will still equal the same k: P1V1/T1 = k P2V2/T2 = k

COMBINED GAS LAW The end result: You can predict temperature, volume or pressure based on how you change the system: P1V1/T1 = P2V2/T2 This equation is called the combined gas law Temperature (T) must always be in Kelvin (K)

EXAMPLE A container has a starting pressure of 2.50atm, a starting temperature of 33°C and a starting volume of 1250 mL. If you change the temperature to 10 °C and the volume to 1.0L, what is the new pressure?

STEP 1 Write down your information P1 = 2.50atm V1 = 1250mL T1 = 33 °C V2 = 1.0L T2 = 10 °C

STEP 2 Make sure that the units for volume and pressure are the same and temperature is in kelvins (K). If not, then convert. P1 = 2.50atm V1 = 1250mL = 1.250L T1 = 33 °C + 273 = 306K P2 = ? V2 = 1.0L T2 = 10 °C + 273 = 283K

STEP 3 P1V1/T1 = P2V2/T2 Put the values in the formula (2.50atm)(1.250L)/306K = P2(1.0L)/283K

STEP 4 Solve the equation so that the units cancel out (2.50atm)(1.250L)/303K = P2(1.0L)/283K P2 = (2.50atm)(1.250L)(283K)/(303K)(1.0L) P2 = 2.9atm

TRY THIS A container has a volume of 2500mL, a pressure of 700 mmHg and a temperature of 15 °C. If you change the pressure to 2.5atm and you change the temperature to 300K, what is the new volume?

SOLUTION 9.6 x 102 mL or 0.96L

IDEAL GAS LAW How does an ideal gas behave? The volume of gas particles is so small that they do not contribute to the volume in a container. Particles of gas do not attract or repel each other Particles of gas are in constant motion

IDEAL GAS LAW When we make these assumptions, we can create a universal IDEAL GAS LAW PV = nRT

VARIABLES P = pressure (must be in the units of atm) V = volume (must be in the units of L) n = moles T = temperature (must be in the units of K) R = gas constant R = 0.0821 L*atm/mol*K only works with ideal gases

EXAMPLE A gas is enclosed in a box that is 5.4L in volume. If you have 22.4 moles at a temperature of 22°C, what is the pressure exerted on the box?

STEP 1 Write down your information P = ? V = 5.4L n = 22.4 moles R = 0.0821 L*atm/mol*K T = 22°C

STEP 2 Make sure that each measurement has the correct units. P(atm), V(L), n(moles) and T (K). P = ? V = 5.4L n = 22.4 moles R = 0.0821 L*atm/mol*K T = 22°C = 295K

STEP 3 PV = nRT Put the values in the formula (P)(5.4L) = (22.4moles)(0.0821 L*atm/mol*K)(295K)

STEP 4 Solve the equation so that the units cancel out (P)(5.4L) = (22.4moles)(0.0821 L*atm/mol*K)(295K) (P)(5.4L) = 542 L*atm 5.4L 5.4L P = 1.0 x 102 or 100 atm

TRY THIS A container has a volume of 2500mL, a pressure of 3.5 atm and a temperature of 25 °C. If this is an ideal gas, how many moles do you have?

SOLUTION 0.36 moles

MORE ON ASSUMPTIONS For each of the gas laws to work we have to follow certain assumptions . . . just like the Ideal Gas Law With these assumptions we can convert the ideal gas law into the other gas laws

ASSUMPTIONS Boyle’s Law Charles’ Law Combined Gas Law Temperature remains constant and the number of moles does not change That the system has a constant pressure and the number of moles does not change. You have the same number moles.

CONVERTING THE IDEAL GAS LAW We start with the ideal gas law PV = nRT We look at the assumptions of each law to see what must remain constant If the condition remains constant, we can remove it from the ideal gas law We set the equation equal to the gas constant (R) to get the other gas laws EXAMPLES ON THE BOARD

STOICHIOMETRY REVISITED With gases, we sometimes work at a state called STANDARD TEMPERATURE AND PRESSURE (STP). The values of STP are: Pressure = 1 atm Temperature = 273K (0°C)

VOLUME If you have one mole of a gas at STP, what is the volume? TRY IT OUT

VOLUME AT STP Given: Pressure = 1 atm Volume = ? n = 1 mole R = 0.0821 L*atm/mol*K T = 273 K

ANSWER 1 mole of gas = 22.4L at STP

NOW WE CAN USE THIS IN DIMENSIONAL STOICHIOMETRY Try the following: You begin with 9.66x1033 molecules of nitrogen gas. What is the volume at STP?

SOLUTION Just like you did previously, you convert molecules to moles Since you know 1 mole = 22.4L at STP, that is the second step 9.66x1033 mq | 1 mole | 22.4L | 6.02x1023 mq | 1 mole

ANSWER 3.85x1011 L Notice: It doesn’t matter what gas you have, they are all 22.4 L/mole Therefore, you don’t need the density to solve for the volume of a gas

TRY THIS You have 336mL of O2 gas at STP. How many oxygen molecules do you have?

ANSWER 9.03X1021 molecules of O2

STOICHIOMETRY Since we do not need density to solve for volume, we can also solve normal stoichiometry problems with gases IF THEY ARE AT STP

TRY THIS Translate and balance: Hydrogen gas is combined with nitrogen gas to form ammonia (NH3) gas. If you start with 25g of nitrogen gas, what volume of ammonia gas do you create?

= 40L 3H2 + N2  2NH3 | 28 g N2 | 1 mole N2 | 1 mole NH3 25g N2 | 1 mole N2 | 2 moles NH3 | 22.4 L NH3__ | 28 g N2 | 1 mole N2 | 1 mole NH3 = 40L

WHAT DO YOU USE? A sample of nitrogen gas has a mass of 55.4g. If the gas is at STP, what is the volume that the gas occupies?

MORE CHALLENGING A sample has a volume of 250mL at a temperature of 25°C. If this sample is composed of 0.50g of H2, what is the pressure of the sample?

WHAT DO YOU USE? A container has a volume of 254mL and a temperature of 25°C. If you increase the temperature to 373K, what is the new volume?

WHAT DO YOU USE? A system has a volume of 550mL and a pressure of 225kPa. If you have 32.5g of chlorine gas, what is the temperature?

WHAT DO YOU USE? A sample of gas has a pressure of 1250 mmHg in a volume of 3.55L. If you change the pressure to 295kPa, what is the new volume?

WHAT DO YOU USE? A container contains a gas that has a pressure of 2.95atm, a volume of 455mL and a temperature of 15°C. If you change the pressure to 700mm Hg and the temperature to 35°C, what will be the new volume?