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Entry Task: Oct 25 th 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.

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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
• 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