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Gay-Lussac’s Law: P and T

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Gay-Lussac’s Law: P and T. the pressure exerted by a gas is directly related to the Kelvin temperature . V and n are constant. P 1 = P 2 T 1 T 2. Calculation with Gay-Lussac’s Law. A gas has a pressure of 2.0 atm at 18°C. What is the new pressure when the temperature is - PowerPoint PPT Presentation
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Gay-Lussac’s Law: P and T the pressure exerted by a gas is directly related to the Kelvin temperature . V and n are constant. P 1 = P 2 T 1 T 2 1
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Page 1: Gay-Lussac’s Law: P and T

Gay-Lussac’s Law: P and T

• the pressure exerted by a gas is directly related to the Kelvin temperature.

• V and n are constant.

P1 = P2

T1 T2

1

Page 2: Gay-Lussac’s Law: P and T

A gas has a pressure of 2.0 atm at 18°C. Whatis the new pressure when the temperature is 62°C? (V and n constant)

1. Set up a data table; Conditions 1 Conditions 2

P1 = 2.0 atm P2 =

T1 = 18°C + 273 T2 = 62°C + 273

= 291 K = 335 K

Calculation with Gay-Lussac’s Law

?

2

Page 3: Gay-Lussac’s Law: P and T

Calculation with Gay-Lussac’s Law (continued)

2. Solve Gay-Lussac’s Law for P2:

P1 = P2

T1 T2

P2 = P1 x T2

T1

P2 = 2.0 atm x 335 K = 2.3 atm 291 KTemperature ratioincreases pressure

3

Page 4: Gay-Lussac’s Law: P and T

Try it

A gas has a pressure of 645 torr at 128°C. What is the

temperature in Celsius if the pressure increases to

1.50 atm (n and V remain constant)?

1. Set up a data table:

Conditions 1 Conditions 2

P1 = 645 torr P2 = 1.50 atm x 760 torr = 1140 torr

1 atm

T1 = 128°C + 273 T2 = ?K K – 273 = ?°C

= 401 K4

Page 5: Gay-Lussac’s Law: P and T

Solution

2. Solve Gay-Lussac’s Law for T2:

P1 = P2

T1 T2

T2 = T1 x P2

P1

T2 = 401 K x 1140 torr = 709 K - 273 = 436°C 645 torr

Pressure ratioincreases temperature

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Page 6: Gay-Lussac’s Law: P and T

Avogadro's Law: Volume and Moles

• the volume of a gas is directly related to the number of moles (n) of gas.

• T and P are constant.

V1 = V2

n1 n2

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Page 7: Gay-Lussac’s Law: P and T

Learning Check

If 0.75 mole helium gas occupies a volume of 1.5 L, what volume will 1.2 moles helium occupy at the same temperature and pressure?

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Page 8: Gay-Lussac’s Law: P and T

Solution

STEP 1 Conditions 1 Conditions 2 V1 = 1.5 L V2 = ??? n1 = 0.75 mole He n2 = 1.2 moles He

STEP 2 Solve for unknown V2

V2 = V1 x n2

n1

STEP 3 Substitute values and solve for V2.V2 = 1.5 L x 1.2 moles He = 2.4 L

0.75 mole He

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Page 9: Gay-Lussac’s Law: P and T

STP and Molar VolumeSTP = standard temperature (273 K) and pressure (1 atm)

At STP: 1 mol of a gas (any gas) has a volume of 22.4 L

22.4 L 1 mole

Can be used as a conversion factor :

a) What is the volume of 2.50 moles of N2 at STP?

b) What is the volume in liters of 6.40 g of O2 at STP?

= molar volume

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Page 10: Gay-Lussac’s Law: P and T

Partial Pressure: Dalton’s Law

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Many gases are mixtures of gases, e.g. air.

The partial pressure of a gas

• is the pressure of each gas in a mixture.

• is the pressure that gas would exert if it were by itself in the container.

Page 11: Gay-Lussac’s Law: P and T

Dalton’s Law of Partial Pressures indicates that

• pressure depends on the total number of gas particles, not on the types of particles.

• the total pressure exerted by gases in a mixture is the sum of the partial pressures of those gases.

PT = P1 + P2 + P3 +.....

So: if total pressure (PT) is known and all but one partial

pressure…..can determine partial pressure of that one.

Dalton’s Law of Partial Pressures

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Page 12: Gay-Lussac’s Law: P and T

Learning Check

A scuba tank contains O2 with a pressure of 0.450 atm and He at 855 mm Hg. What is the total pressure in mm Hg in the tank?

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Page 13: Gay-Lussac’s Law: P and T

1. Convert the pressure in atm to mm Hg 0.450 atm x 760 mm Hg = 342 mm Hg = P(O2) 1 atm

2. Calculate the sum of the partial pressures.

Ptotal = P(O2) + P(He)

Ptotal = 342 mm Hg + 855 mm Hg

= 1197 mm Hg

Solution

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Page 14: Gay-Lussac’s Law: P and T

For a deep dive, a scuba diver uses a mixture of helium and oxygen with a total pressure of 8.00 atm. If the oxygen has a partial pressure of 1280 mm Hg, what is the partial pressure of the helium?

Learning Check

14

PTotal = 8.00 atm x 760 mm Hg = 6080 mm Hg 1 atm

PTotal = PO + PHe 2

PHe = PTotal - PO2

PHe = 6080 mm Hg - 1280 mm Hg = 4800 mm Hg

Page 15: Gay-Lussac’s Law: P and T

Gases We Breathe

The air we breathe • is a gas mixture.

• contains mostly N2 and O2 and small amounts of other gases.

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Page 16: Gay-Lussac’s Law: P and T

A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the

air?

1) 35.6 2) 156 3) 760

B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?

1) 557 2) 9.14 3) 0.109

Learning Check

Hint: air is 21% oxygen

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Page 17: Gay-Lussac’s Law: P and T

Blood Gases

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Lungs

Page 18: Gay-Lussac’s Law: P and T

Blood GasesIn the lungs/blood/body:

• O2 flows (combined with hemoglobin) into the tissues because the partial pressure of O2 is higher in arterial blood, and lower in the tissues.

• CO2 flows out of the tissues because the partial pressure of CO2 is higher in the tissues, and lower in the blood.

Partial Pressures (mmHg) in Blood and Tissue

Oxygenated Deoxygenated

(arterial) (venous)

Gas Blood Blood Tissues

O2 100 40 30 or less

CO2 40 46 50 or greater

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