Gas Laws - PowerPoint PPT Presentation

Loading...

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



Loading


The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
About This Presentation
Title:

Gas Laws

Description:

Gas Laws What to do when conditions are ideal – PowerPoint PPT presentation

Number of Views:61
Avg rating:3.0/5.0
Slides: 49
Provided by: sfi72
Category:
Tags: gas | laws | schoolwires

less

Write a Comment
User Comments (0)
Transcript and Presenter's Notes

Title: Gas Laws


1
Gas Laws
  • What to do when conditions are ideal

2
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

3
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

4
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

5
Gas Laws Charles Law
  • Constant Pressure
  • Relationship Temperature is directly
    proportional to volume
  • Temp a Volume
  • Written As V1/T1 V2/T2

6
Charles Law
  • What unit of measure is needed for these
    calculations? C or K?
  • Temperature is in K (K 273 C)

7
Gas Laws Gay-Lussacs Law
  • Constant Volume
  • Relationship Pressure is directly proportional
    to temperature
  • Pressure a Temperature
  • Written As P1/T1 P2/T2

8
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

9
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.

10
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

11
Combined Gas Law
  • P1V1 P2V2
  • T1 T2
  • Isolate V2
  • P1V1T2 P2V2T1
  • V2 P1V1T2
  • P2T1

12
Learning Check G15
  • What is the value of R when the STP value for P
    is 760 mmHg?

13
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

14
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?

15
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 ? ?

16
  • 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

17
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?

18
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

19
Learning Check C1
  • Solve the combined gas laws for T2.

20
Solution C1
  • Solve the combined gas law for T2.
  • (Hint cross-multiply first.)
  • P1V1 P2V2
  • T1 T2
  • P1V1T2 P2V2T1
  • T2 P2V2T1
  • P1V1

21
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?

22
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
    ??

??
23
Solution
  • Solve for T2
  • Enter data
  • T2 302 K x atm x mL
    K
  • atm mL
  • T2 K - 273 C

24
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

25
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?

26
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

27
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?

28
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.

29
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.

30
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

31
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)

32
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

33
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

34
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

35
Molar Volume Factor
  • 1 mole of a gas at STP 22.4 L
  • 22.4 L and 1 mole
  • 1 mole 22.4 L

36
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

37
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

38
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 .....

39
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

40
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

41
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

42
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
43
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.

44
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?

45
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

46
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

47
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

48
  • 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
About PowerShow.com