Loading...

PPT – Gas Laws PowerPoint presentation | free to download - id: 78614c-MTJhM

The Adobe Flash plugin is needed to view this content

Gas Laws

- What to do when conditions are ideal

Boyles Law

- What was the relationship between pressure and

volume? - When P Then V
- When P Then V
- Algebraically this is written as Pk/V
- When you solved for k PVk
- Therefore P1V1P2V2

Gas Laws Boyles Law

- Constant Temperature
- Relationship Pressure is inversely proportional

to volume - Pressure a 1/volume
- Written As P1V1 P2V2
- Pressure is typically in atm or torr

Charles Law

- What was the relationship between Temperature and

Volume? - When T Then V
- When T Then V
- Algebraically this is written as VkT
- When you solved for k V/Tk
- Therefore V1/T1V2/T2

Gas Laws Charles Law

- Constant Pressure
- Relationship Temperature is directly

proportional to volume - Temp a Volume
- Written As V1/T1 V2/T2

Charles Law

- What unit of measure is needed for these

calculations? C or K? - Temperature is in K (K 273 C)

Gas Laws Gay-Lussacs Law

- Constant Volume
- Relationship Pressure is directly proportional

to temperature - Pressure a Temperature
- Written As P1/T1 P2/T2

The IDEAL GAS Law this is what we will use

- When we put all three laws together
- PV a nT (n number of moles)
- PV nRT (R ideal gas law constant)
- R62.4 L torr/K mol or .08206 L atm/K mol

Ideal Gases

- Behave as described by the ideal gas equation no

real gas is actually ideal - Within a few , ideal gas equation describes most

real gases at room temperature and pressures of 1

atm or less - In real gases, particles attract each other

reducing the pressure - Real gases behave more like ideal gases as

pressure approaches zero.

PV nRT

- R is known as the universal gas constant
- Using STP conditions
- P V
- R PV (1.00 atm)(22.4 L) nT

(1mol) (273K) - n T
- 0.0821 L-atm
- mol-K

Combined Gas Law

- P1V1 P2V2
- T1 T2
- Isolate V2
- P1V1T2 P2V2T1
- V2 P1V1T2
- P2T1

Learning Check G15

- What is the value of R when the STP value for P

is 760 mmHg?

Solution G15

- What is the value of R when the STP value for P

is 760 mmHg? - R PV (760 mm Hg) (22.4 L)
- nT (1mol) (273K)
- 62.4 L-mm Hg

mol-K

Learning Check G16

- Dinitrogen monoxide (N2O), laughing gas, is used

by dentists as an anesthetic. If 2.86 mol of gas

occupies a 20.0 L tank at 23C, what is the

pressure (mmHg) in the tank in the dentist office?

Solution G16

- Set up data for 3 of the 4 gas variables
- Adjust to match the units of R
- V 20.0 L 20.0 L
- T 23C 273 296 K
- n 2.86 mol 2.86 mol
- P ? ?

- Rearrange ideal gas law for unknown P
- P nRT
- V
- Substitute values of n, R, T and V and solve for

P - P (2.86 mol)(62.4L-mmHg)(296 K)
- (20.0 L) (K-mol)
- 2.64 x 103 mm Hg

Learning Check G17

- A 5.0 L cylinder contains oxygen gas at 20.0C

and 735 mm Hg. How many grams of oxygen are in

the cylinder?

Solution G17

- Solve ideal gas equation for n (moles)
- n PV
- RT
- (735 mmHg)(5.0 L)(mol K)
- (62.4 mmHg L)(293 K)
- 0. 20 mol O2 x 32.0 g O2 6.4 g O2
- 1 mol O2

Learning Check C1

- Solve the combined gas laws for T2.

Solution C1

- Solve the combined gas law for T2.
- (Hint cross-multiply first.)
- P1V1 P2V2
- T1 T2
- P1V1T2 P2V2T1
- T2 P2V2T1
- P1V1

Combined Gas Law Problem

- A sample of helium gas has a volume of 0.180 L,

a pressure of 0.800 atm and a temperature of

29C. What is the new temperature(C) of the

gas at a volume of 90.0 mL and a pressure of 3.20

atm?

Data Table

- Set up Data Table
- P1 0.800 atm V1 0.180 L T1

302 K - P2 3.20 atm V2 90.0 mL T2

??

??

Solution

- Solve for T2
- Enter data
- T2 302 K x atm x mL

K - atm mL
- T2 K - 273 C

Calculation

- Solve for T2
- T2 302 K x 3.20 atm x 90.0 mL 604 K
- 0.800 atm 180.0 mL
- T2 604 K - 273 331 C

Learning Check C2

- A gas has a volume of 675 mL at 35C and 0.850

atm pressure. What is the temperature in C when

the gas has a volume of 0.315 L and a pressure of

802 mm Hg?

Solution G9

- T1 308 K T2 ?
- V1 675 mL V2 0.315 L 315 mL
- P1 0.850 atm P2 802 mm Hg
- 646 mm Hg
- T2 308 K x 802 mm Hg x 315 mL
- 646 mm Hg

675 mL - P inc, T inc V dec,

T dec - 178 K - 273 - 95C

Volume and Moles

- How does adding more molecules of a gas change

the volume of the air in a tire? - If a tire has a leak, how does the loss of air

(gas) molecules change the volume?

Learning Check C3

- True (1) or False(2)
- 1.___The P exerted by a gas at constant V is not

affected by the T of the gas. - 2.___ At constant P, the V of a gas is directly

proportional to the absolute T - 3.___ At constant T, doubling the P will cause

the V of the gas sample to decrease to one-half

its original V.

Solution C3

- True (1) or False(2)
- 1. (2)The P exerted by a gas at constant V is not

affected by the T of the gas. - 2. (1) At constant P, the V of a gas is directly

proportional to the absolute T - 3. (1) At constant T, doubling the P will cause

the V of the gas sample to decrease to one-half

its original V.

Avogadros Law

- When a gas is at constant T and P, the V is

directly proportional to the number of moles (n)

of gas - V1 V2
- n1 n2
- initial final

STP

- The volumes of gases can be compared when they

have the same temperature and pressure (STP). - Standard temperature 0C or 273 K
- Standard pressure 1 atm (760 mm Hg)

Learning Check C4

- A sample of neon gas used in a neon sign has a

volume of 15 L at STP. What is the volume (L) of

the neon gas at 2.0 atm and 25C? - P1 V1 T1 K
- P2 V2 ?? T2 K
- V2 15 L x atm x K

6.8 L - atm K

Solution C4

- P1 1.0 atm V1 15 L T1 273 K

- P2 2.0 atm V2 ?? T2 248 K
- V2 15 L x 1.0 atm x 248 K

6.8 L - 2.0 atm 273 K

Molar Volume

- At STP
- 4.0 g He 16.0 g CH4 44.0 g CO2
- 1 mole 1 mole 1mole
- (STP) (STP) (STP)
- V 22.4 L V 22.4 L V

22.4 L

Molar Volume Factor

- 1 mole of a gas at STP 22.4 L
- 22.4 L and 1 mole
- 1 mole 22.4 L

Learning Check C5

- A.What is the volume at STP of 4.00 g of CH4?
- 1) 5.60 L 2) 11.2 L 3) 44.8 L
- B. How many grams of He are present in 8.0 L of

gas at STP? - 1) 25.6 g 2) 0.357 g 3) 1.43 g

Solution C5

- A.What is the volume at STP of 4.00 g of CH4?
- 4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) 5.60 L
- 16.0 g CH4 1 mole

CH4 - B. How many grams of He are present in 8.0 L of

gas at STP? - 8.00 L x 1 mole He x 4.00 g He 1.43

g He - 22.4 He 1 mole

He

Daltons Law of Partial Pressures

- Partial Pressure
- Pressure each gas in a mixture would exert if it

were the only gas in the container - Dalton's Law of Partial Pressures
- The total pressure exerted by a gas mixture is

the sum of the partial pressures of the gases in

that mixture. - PT P1 P2 P3 .....

Gases in the Air

- The of gases in air Partial pressure (STP)

- 78.08 N2 593.4 mmHg
- 20.95 O2 159.2 mmHg
- 0.94 Ar 7.1 mmHg
- 0.03 CO2 0.2 mmHg
- PAIR PN PO PAr PCO 760 mmHg
- 2 2

2 - Total Pressure 760 mm Hg

Learning Check C6

- 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

Solution C6

- A.If the atmospheric pressure today is 745 mm Hg,

what is the partial pressure (mm Hg) of O2 in the

air? - 2) 156
- B. At an atmospheric pressure of 714, what is the

partial pressure (mm Hg) N2 in the air? - 1) 557

Partial Pressures

- The total pressure of a gas mixture depends
- on the total number of gas particles, not on
- the types of particles.
- P 1.00 atm P 1.00

atm

1 mole H2

0.5 mole O2 0.3 mole He 0.2 mole Ar

Health Note

- When a scuba diver is several hundred feet
- under water, the high pressures cause N2 from

the tank air to dissolve in the blood. If the

diver rises too fast, the dissolved N2 will form

bubbles in the blood, a dangerous and painful

condition called "the bends". Helium, which is

inert, less dense, and does not dissolve in the

blood, is mixed with O2 in scuba tanks used for

deep descents.

Learning Check C7

- A 5.00 L scuba tank contains 1.05 mole of O2

and 0.418 mole He at 25C. What is the partial

pressure of each gas, and what is the total

pressure in the tank?

Solution C7

- P nRT PT PO PHe
- V

2 - PT 1.47 mol x 0.0821 L-atm x 298 K
- 5.00 L (K mol)
- 7.19 atm

Molar Mass of a gas

- What is the molar mass of a gas if 0.250 g of

the gas occupy 215 mL at 0.813 atm and 30.0C? - n PV (0.813 atm) (0.215 L) 0.00703

mol - RT (0.0821 L-atm/molK) (303K)
- Molar mass g 0.250 g 35.6

g/mol - mol 0.00703 mol

Density of a Gas

- Calculate the density in g/L of O2 gas at STP.

From STP, we know the P and T. - P 1.00 atm T 273 K
- Rearrange the ideal gas equation for moles/L
- PV nRT PV nRT P n
- RTV RTV

RT V

- Substitute
- (1.00 atm ) mol-K 0.0446 mol

O2/L - (0.0821 L-atm) (273 K)
- Change moles/L to g/L
- 0.0446 mol O2 x 32.0 g O2 1.43

g/L - 1 L 1 mol O2
- Therefore the density of O2 gas at STP is
- 1.43 grams per liter