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