Entry Task: Oct 25 th Block 2

Entry Task: Oct 25th Block 2 Question: What is the relationship between pressure and temperature? You have 5 minutes!!

Agenda: • QUICKLY discuss Ch. 10 sec 1-3 • Gas Inquiry Lab • HW: B, C, G-L and Combo gas law ws (review from last year!)

BREAK OUT AP EQUATION SHEET Ideal gas law Van der Waals equation Daltons Partial pressure Moles= molar mass/molarity Kelvin/Celsius Combination gas law These formulas are rarely or not at all on the AP Exam

BREAK OUT AP EQUATION SHEET These formulas are rarely or not at all on the AP Exam

Chapter 10Gases

I can… • Describe the characteristics of gases • Interconvert the various SI units for pressure • Manipulate mathematically the 4 variables that pertain to gases.

Characteristics of Gases • Unlike liquids and solids, they • Expand to fill their containers • Are highly compressible. • Create homogeneous mixtures • Molecules of gases are very far apart thus having extremely low densities.

F A P = Pressure • Pressure is the amount of force applied to an area. • Atmospheric pressure is the weight of air per unit of area.

Units of Pressure • Pascals • 1 Pa = 1 N/m2 • Bar • 1 bar = 105 Pa = 100 kPa • mm Hg or torr • These units are literally the difference in the heights measured in mm (h) of two connected columns of mercury. • Atmosphere • 1.00 atm = 760 torr Derivation of Pressure / barometer height relationship

Manometer Used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel. Manometer at <, =, > P of 1 atm

Standard Pressure • Normal atmospheric pressure at sea level. • It is equal to • 1.00 atm • 760 torr (760 mm Hg) • 101.325 kPa

The Gas Laws • The 4 variables needed to define the physical conditions of gas are: • Temperature (T) in kelvin • Pressure (P) • Volume (V) • Number of moles (n)

Boyle’s Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

Boyle’s Law at Work… Doubling the pressure reduces the volume by half. Conversely, when the volume doubles, the pressure decreases by half.

PV = k • Since • V = k (1/P) • This means a plot of V versus 1/P will be a straight line. As P and V areinversely proportional A plot of V versus P results in a curve.

Application of Boyle’s Law A gas has a volume of 3.0 L at 2 atm. What will its volume be at 4 atm? P1 = 2 atm V1 = 3.0 L P2 = 4 atm V2 = X (2 atm) (3.0L) = (4 atm) (X)

Finishing the algebra (2 atm) (3.0 L) = (4 atm) (X) (4 atm) (4 atm) (6 L) = (X) (4) X =1.5 L

P1V1 = P2 V2 In a thermonuclear device, the pressure of 0.050 liters of gas within the bomb casing reaches 4.0 x 106 atm. When the bomb casing is destroyed by the explosion, the gas is released into the atmosphere where it reaches a pressure of 1.00 atm. What is the volume of the gas after the explosion? P1 = 4.0 x106atm V1 = 0.050 L P2 = 1 atm V2 = X (4.0 x106atm) (0.050L) = (1 atm) (X)

Finishing the algebra (4.0 x106atm) (0.05 L) = (1 atm) (X) (1 atm) (1 atm) (200000 L) = (X) (1) X =200000 L or 2.0 x 105 L

If some neon gas at 121 kPa were allowed to expand from 3.7 dm3 to 6.0 dm3 without changing the temperature, what pressure would the neon gas exert under these new conditions? (121 kPa) (X) (6.0 dm3) (3.7 dm3) = (121 kPa)(3.7 dm3)= (X) 6.0 dm3 75 kPa

V T = k • i.e., Charles’s Law • The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. A plot of V versus T will be a straight line.

Charles’ Law at Work… As the temperature increases, the volume increases. Conversely, when the temperature decreases, volume decreases.

WHY must you convert Celsius to Kelvin? For the math to work AND to show a proportional relationship, the absolute temperature (Kelvin) is needed.

Converting K to C˚ and C˚ to K • Kelvin temp = C ˚ + 273 Its 32 C ˚, what is this temp in Kelvin? 32 + 273 = 305 K • Celsius temp = K - 273 Its 584 K, what is the temp in C ˚? 584 - 273 = 311C ˚

Kelvin Practice 0°C = _______ K 100°C = _______ K 100 K = _______ °C –30°C= _______ K 300 K = _______ °C 403 K = _______ °C 25°C = _______ K 0 K = _______ °C 273 373 -173 243 27 130 298 -273

V1 / T1 = V2 / T2 • If a 1.0 L balloon is heated from 22°C to 100°C, what will its new volume be? V1 = 1.0 L T1 = 22°C + 273 = 295 K V2 = X T2 = 100°C + 273 = 373 K 1.0L / 295K = X / 373K Need to convert C° to K

Finishing the algebra (295K) (X) = (1.0 L) (373 K) (295K) (295K) X = (1.0 L) 373 295 X =1.26 L

The temperature inside my refrigerator is about 40 Celsius. If I place a balloon in my fridge that initially has a temperature of 220 C and a volume of 0.5 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator? (0.5 L) (X L) = (295 K) 277 K (0.5 L)(277 K)= (X) 295 K 0.469 L

A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 0C, what will the volume of the balloon be after he heats it to a temperature of 250 0C? (0.4 L) (X L) = (293 K) 523 K (0.4 L)(523 K)= (X) 293 K 0.714 L

Charles’ Law: Summary • Volume / Temperature = Constant • V1 / T1 = V2 / T2 • With constant pressure and amount of gas, you can use these relationships to predict changes in temperature and volume.

V = kn • Mathematically, this means Avogadro’s Law • The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas.

Gay-Lussac’s Law combines pressure and temperature. This is a review from last year.When volume remains constant, pressure and temperature have a direct relationship. P1/T1 = P2/T2

Gay-Lussac’s Law • The pressure of gas in a tank is 3.20 atm at 22.0 C°. If the temperature rises to 60.0 C°, What will be the gas pressure in the tank? P1 = 3.20 atm T1 = 22 + 273 = 295 K P2 = X T2 = 60 + 273 = 333 K 3.20 atm / 295 K = X / 333 K Need to convert C° to K

Now the ALGEBRA!!! 3.20 atm / 295 K = X / 333 K • Set it up- Algebraically 3.20 atm = X 295 K 333 K • Cross multiply • (295 K) X = (3.20 atm)(333 K)

Finishing the algebra • (295 K) X = (3.20 atm)(333 K) (295 K) (X) = (3.20 atm) (333 K) (295 K) (295K) X = (3.20 atm) 333 295 X =3.61 atm

Gay-Lussac’s Law • A rigid container has an initial pressure of 1.50 atm at 21oC. What will the pressure be if the temperature is increased to 121oC? P1 = 1.50 atm T1 = 21 + 273 = 294 K P2 = X T2 = 121+ 273 = 394 K 1.50 atm / 294 K = X atm / 394 K Need to convert C° to K

Now the ALGEBRA!!! 1.50 atm / 294 K = X atm / 394 K • Set it up- Algebraically 1.50 atm = X atm 294 K 394 K • Cross multiply • (294 K) X atm = (1.50 atm)(394 K)

(1.50 atm)(394 K) Finishing the algebra • (294 K) X atm = (294K) (X) = (1.50 atm) (394 K) (294K) (294K) X = (1.50 atm) 394 294 X =2.01 atm

A gas in a sealed container has a pressure of 125 kPa at a temperature of 30.0 ˚C. If the pressure in the container is increased to 201 kPa, what is the new temperature- in Celsius? P1 = 125 kPa T1 = 30 + 273 = 303K P2 = 201 kPa T2 = X 125 kPa 201 kPa = 303 K X 60903 = (X) (125 kPa) 60903 K = X 125 487 – 273 = 214 ˚C

The pressure in an automobile tire is 1.88 atm at 25 ˚C. What will be the pressure if the temperature warms up to 37.0 ˚C? P1 = 1.88 atm T1 = 25 + 273 = 298K P2 = X T2 = 37 + 273 = 310K 1.88 atm X = 298 K 310 K 583 = (X) (298K) 583 atm = X 298 1.95 atm

Explain Gay-Lussac’s law of combining volumes. • At a given pressure and temperature the volumes of gases that react with one another are in ratio of small whole numbers

Explain the difference between Avogadro’s hypothesis and Avogadro’s Law • Avogadro’s hypothesis states that there are at equal volumes with P and T constant, they have the same number of gas molecules • Avogadro’s Law takes it one step further, the equal volumes means equal number of moles.

Explain V= constant x n • If you double the number of moles of gas (n) then the volume will double- proportionally.

Gas Law Inquiry Lab • HW: B, C, G-L and combination gas law ws.

Entry Task: Oct 25 th Block 2