Gases Entry Task: Oct 25 th Block 2 Question: What is the relationship between pressure and...
-
Upload
caitlin-miles -
Category
Documents
-
view
227 -
download
0
Transcript of Gases Entry Task: Oct 25 th Block 2 Question: What is the relationship between pressure and...
Gases
Entry Task: Oct 25th Block 2
Question:
What is the relationship between pressure and temperature?
You have 5 minutes!!
Gases
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!)
Gases
BREAK OUT AP EQUATION SHEET
These formulas are rarely or not at all on the AP Exam
Ideal gas law
Van der Waals equation
Daltons Partial pressure
Moles= molar mass/molarity
Kelvin/Celsius
Combination gas law
Gases
I can…
• Describe the characteristics of gases• Interconvert the various SI units for
pressure• Manipulate mathematically the 4
variables that pertain to gases.
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.
Gases
• Pressure is the amount of force applied to an area.
Pressure
• Atmospheric pressure is the weight of air per unit of area.
P =FA
Gases
Units of Pressure• Pascals
1 Pa = 1 N/m2
• Bar 1 bar = 105 Pa = 100 kPa
• mm Hg or torrThese units are literally the difference in the
heights measured in mm (h) of two connected columns of mercury.
• Atmosphere1.00 atm = 760 torr
Derivation of Pressure / barometer height relationship
Gases
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
Gases
Standard Pressure
• Normal atmospheric pressure at sea level.
• It is equal to1.00 atm760 torr (760 mm Hg)101.325 kPa
Gases
The Gas Laws
• The 4 variables needed to define the physical conditions of gas are:Temperature (T) in kelvinPressure (P)Volume (V)Number of moles (n)
Gases
Boyle’s Law
The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.
Gases
Boyle’s Law at Work…
Doubling the pressure reduces the volume by half. Conversely, when the volume doubles, the pressure decreases by half.
Gases
As P and V areinversely proportional
A plot of V versus P results in a curve.
Since
V = k (1/P)This means a plot of V versus 1/P will be a straight line.
PV = k
Gases
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)
Gases
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 x106 atm
V1 = 0.050 L
P2 = 1 atm
V2 = X
(4.0 x106atm) (0.050L) = (1 atm) (X)
Gases
Finishing the algebra
(4.0 x106 atm) (0.05 L) = (1 atm) (X)(1 atm)(1 atm)
(200000 L) = (X)(1)
X =200000 L or 2.0 x 105 L
Gases
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) (3.7 dm3) = (X) (6.0 dm3)
(121 kPa)(3.7 dm3)= (X)
6.0 dm3
75 kPa
Gases
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.
• i.e., VT
= k
Gases
Charles’ Law at Work…
As the temperature increases, the volume increases. Conversely, when the temperature decreases, volume decreases.
Gases
WHY must you convert Celsius to Kelvin?
For the math to work AND to show a proportional relationship, the absolute temperature (Kelvin) is needed.
Gases
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 ˚
Gases
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
-173
27
298
373
243
130
-273
Gases
V1 / T1 = V2 / T2
• If a 1.0 L balloon is heated from 22°C to 100°C, what will its new volume be?
Need to convert C° to KV1 = 1.0 L
T1 = 22°C + 273 = 295 K
V2 = X
T2 = 100°C + 273 = 373 K
1.0L / 295K = X / 373K
Gases
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)(295 K)
(X L)
(0.5 L)(277 K)= (X)
295 K
0.469 L
277 K=
Gases
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)(293 K)
(X L)
(0.4 L)(523 K)= (X)
293 K
0.714 L
523 K=
Gases
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.
Gases
Avogadro’s Law
• The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas.
• Mathematically, this means V = kn
Gases
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
Gases
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
Gases
Now the ALGEBRA!!!
• Set it up- Algebraically
3.20 atm / 295 K = X / 333 K
Cross multiply (295 K) X =
3.20 atm = X333 K295 K
(3.20 atm)(333 K)
Gases
Finishing the algebra
(295 K) (X) = (3.20 atm) (333 K)(295K)(295 K)
X =3.61 atm
X = (3.20 atm) 333
295
(295 K) X = (3.20 atm)(333 K)
Gases
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?
Need to convert C° to K
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
Gases
Now the ALGEBRA!!!
• Set it up- Algebraically
1.50 atm / 294 K = X atm / 394 K
Cross multiply (294 K) X atm =
1.50 atm = X atm394 K294 K
(1.50 atm)(394 K)
Gases
Finishing the algebra
(294K) (X) = (1.50 atm) (394 K)(294K)(294K)
X =2.01 atm
X = (1.50 atm) 394
294
(294 K) X atm = (1.50 atm)(394 K)
Gases
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
60903 = (X) (125 kPa)
125 kPa303 K
487 – 273 = 214 ˚C
201 kPaX=
60903 K = X125
Gases
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
583 = (X) (298K)
1.88 atm298 K
1.95 atm
X310 K=
583 atm = X298
Gases
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
Gases
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.
Gases
Explain V= constant x n
• If you double the number of moles of gas (n) then the volume will double- proportionally.