Ratio Method of Solving Basic Gas Laws. Boyle’s Law Don’t hate me just because I’m beautiful Robert Boyle.

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Presentation transcript:

Ratio Method of Solving Basic Gas Laws

Boyle’s Law Don’t hate me just because I’m beautiful Robert Boyle

Boyle’s Law Demonstrations that illustrate Boyle’s Law Marshmallow or balloon in the bell jar (marshmallow or balloon gets big when pressure gets low)

Boyle’s Law Volume is inversely related to Pressure If volume decreases, pressure increases If volume increases, pressure decreases (they move in opposite directions) You can see this relationship when temperature and moles of gas are held constant

Boyle’s Law Mathematically, PV = k Where “k” is a constant

Boyle’s Law Another way to write Boyle’s law is P 1 V 1 = P 2 V 2 Temperature is constant!! If it is not, then you can not use Boyle’s Law

Boyle’s Law Problem #1 A 0.030L marshmallow undergoes a drop in pressure from 1.1 atm to 0.20 atm. What will be the new volume? Start with a “data table” to organize what you know and what you don’t know V 1 = 0.030L V 2 = ? P 1 = 1.1 atm P 2 = 0.20 atm

Boyle’s Law Problem #1 A 0.030L marshmallow undergoes a drop in pressure from 1.1 atm to 0.20 atm. What will be the new volume? V 1 = 0.030L V 2 = ? P 1 = 1.1 atm P 2 = 0.20 atm V 2 =0.030L x 1.1 atm 0.20 atm = 0.17 L Bigger number on top makes Volume get larger Pressure got smaller so Volume must get larger

Boyle’s Law Problem #2 A 0.025L marshmallow at 0.60 atm experiences a pressure increase to 0.95 atm. What will be the new volume? Start with your data table V 1 = 0.025L V 2 = ? P 1 = 0.60 atm P 2 = 0.95 atm

Boyle’s Law Problem #2 A 0.025L marshmallow at 0.60 atm experiences a pressure increase to 0.95 atm. What will be the new volume? V 1 = 0.025L V 2 = ? P 1 = 0.60 atm P 2 = 0.95 atm V 2 =0.025L x 0.60 atm 0.95 atm = L Smaller number on top Makes volume get smaller Pressure got larger, so Volume must get smaller

Charles’ Law Hey Ladies!!!

Charles’ Law Demonstrations that illustrate Charles’ Law Balloon on flask (heat the flask then the balloon expands)

Charles’ Law Volume is directly related to Temperature If temperature increases, volume increases If temperature decreases, volume decreases (they move in the same direction) You can see this relationship when pressure and moles of gas are held constant

Charles’ Law V = kT Where “k” is a constant of proportionality Note that V does NOT have to equal T, but they are directly proportional to each other

Charles’ Law Another way to write this is: (V 1 / T 1 ) = (V 2 /T 2 ) Or V 1 T 2 = V 2 T 1 Pressure must remain constant or you can not use this law!

Charles’ Law Problem #1 A 2.0L sample of air is collected at 298K then cooled to 278K. What will be the new volume? Start with a data table to organize what you know and what you don’t know V 1 = 2.0 L V 2 = ? T 1 = 298 K T 2 = 278 K

Charles’ Law Problem #1 A 2.0L sample of air is collected at 298K then cooled to 278K. What will be the new volume? V 1 = 2.0 L V 2 = ? T 1 = 298 K T 2 = 278 K V 2 =2.0 L x 278 K 298 K = 1.9 L Smaller number on top makes volume get smaller Temperature decreased so volume must decrease

Charles’ Law Problem #2 A 3.25L balloon at 298K changes volume to 2.50L. What temperature would cause this to happen? Start with your data table V 1 = 3.25 L V 2 = 2.50 L T 1 = 298 K T 2 = ?

Charles’ Law Problem #2 A 3.25L balloon at 298K changes volume to 2.50L. What temperature would cause this to happen? V 1 = 3.25 L V 2 = 2.50 L T 1 = 298 K T 2 = ? T 2 =298 K x 2.50 L 3.25 L = 229 K Smaller number on top makes temperature smaller Volume got smaller so temperature must get smaller

Gay-Lussac Law O.K. So I have a funny name. Go ahead get it out of your system so we may continue.

Gay-Lussac Law Demonstrations that illustrate Gay-Lussac’s Law Aerosol can in the campfire (As temperature increases, pressure increases until the can explodes)

Gay-Lussac Law Pressure is directly related to Temperature If temperature increases, pressure increases If temperature decreases, pressure decreases (they move in the same direction) You can see this relationship when volume and moles of gas are held constant

Gay-Lussac Law P = kT Where “k” is a constant of proportionality

Gay-Lussac Law Another way to write it: (P 1 / T 1 ) = (P 2 / T 2 ) Or T 1 P 2 = P 1 T 2 Volume stays constant!

Gay-Lussac Law Problem #1 A can of hairspray contains a gaseous propellant at 1.50 atm and 298K. What is the new pressure if the can is heated to 500.K? Start with a data table T 1 = 298K T 2 = 500.K P 1 = 1.50 atm P 2 = ?

Gay-Lussac Law Problem #1 A can of hairspray contains a gaseous propellant at 1.50 atm and 298K. What is the new pressure if the can is heated to 500.K? T 1 = 298K T 2 = 500.K P 1 = 1.50 atm P 2 = ? P 2 =1.50 atm x 500 K 298 K = 2.52 atm Big number on top makes pressure get bigger Temperature increased so pressure must increase

Gay-Lussac Law Problem #2 A can of spraypaint at 298K experiences an increase in pressure from kPa to 425 kPa. What temperature would cause such a change? Start with a data table T 1 = 298 K T 2 = ? P 1 = kPa P 2 = 425 kPa

Gay-Lussac Law Problem #2 A can of spraypaint at 298K experiences an increase in pressure from kPa to 425 kPa. What temperature would cause such a change? T 1 = 298 K T 2 = ? P 1 = kPa P 2 = 425 kPa T 2 =298 K x 425 kPa kPa = 1250 K Big number on top makes Temperature get bigger Pressure increased so temperature must increase

Avogadro’s Law I’m dead sexy and you know it!! Text me. You got my number.

Avogadro’s Law Demonstration Blowing up a balloon As you add moles of gas, the volume increases

Avogadro’s Law Volume is directly related to Moles If moles of gas increases, volume increases If moles of gas decreases, volume decreases (they move in the same direction) You can see this relationship when temperature and pressure of gas are held constant

Avogadro’s Law V = kn Where “k” is a constant of proportionality

Avogadro’s Law Another way to write this: V 1 /n 1 = V 2 /n 2 Or V 1 n 2 = V 2 n 1

Avogadro’s Law Problem #1 A 12.2 L sample of gas contains mol of O 2 at 1.00 atm and 25.0 o C. If 3.00 mol of O 2 are added, what will be the new volume? Start with a data table V 1 = 12.2 L V 2 = ? n 1 = mol n 2 = mol

Avogadro’s Law Problem #1 A 12.2 L sample of gas contains mol of O 2 at 1.00 atm and 25.0 o C. If 3.00 mol of O 2 are added, what will be the new volume? V 1 = 12.2 L V 2 = ? n 1 = mol n 2 = mol V 2 =12.2 L x mol mol = 85.4 L Bigger number on top Makes volume get bigger Moles increased so volume must increase

Avogadro’s Law Problem #2 A 2.25 L balloon contains mol of helium. To increase the volume of the balloon to 6.50 L, how many moles of helium must it contain? Start with a data table V 1 = 2.25 L V 2 = 6.50 L n 1 = mol n 2 = ?

Avogadro’s Law Problem #2 A 2.25 L balloon contains mol of helium. To increase the volume of the balloon to 6.50 L, how many moles of helium must it contain? V 1 = 2.25 L V 2 = 6.50 L n 1 = mol n 2 = ? n 2 =0.475 mol x 6.50 L 2.25 L = 1.37 mol Bigger number on top makes Moles get bigger Volume increased so moles must increase