Constant temp

1. Constant
Temperature of
Gas
PREPARED BY: TYPE YOUR NAME HERE

2. S9MT -IIj - 20
 Investigate the relationship
between:
 1 volume and pressure at constant
temperature of a gas
 2 volume and temperature at
constant pressure of a gas
 3 explains these relationships using
the kinetic molecular theory

3. A. Kinetic Molecular Theory
 Particles in an ideal gas…
have no volume.
have elastic collisions.
are in constant, random,
straight-line motion.
don’t attract or repel each other.
have an avg. KE directly related
to Kelvin temperature.

4. B. Real Gases
 Particles in a REAL gas…
 have their own volume
 attract each other
 Gas behavior is most ideal…
 at low pressures
 at high temperatures
 in nonpolar atoms/molecules

5. C. Characteristics of Gases
 Gases expand to fill any container.
 random motion, no attraction
 Gases are fluids (like liquids).
 no attraction
 Gases have very low densities.
 no volume = lots of empty space

6. C. Characteristics of Gases
Gases can be compressed.
no volume = lots of empty space
Gases undergo diffusion &
effusion.
random motion

7. D. Temperature
 Always use absolute temperature (Kelvin) when working with gases.
ºF
ºC
K
-459 32 212
-273 0 100
0 273 373
 
32
F
C 9
5



 K = ºC + 273

8. E. Pressure
area
force
pressure 
Which shoes create the most pressure?

9. E. Pressure
 Barometer
 measures atmospheric pressure
Mercury Barometer
Aneroid Barometer

10. E. Pressure
 Manometer
 measures contained gas pressure
U-tube Manometer Bourdon-tube gauge

11. E. Pressure
 KEY UNITS AT SEA LEVEL
101.325 kPa (kilopascal)
1 atm
760 mm Hg
760 torr
14.7 psi
2
m
N
kPa 

12. F. STP
Standard Temperature & Pressure
0°C 273 K
1 atm 101.325 kPa
-OR-
STP

13. Gas Laws
BOYLES
CHARLES
GAY-
LUSSAC
Ch. 12 - Gases

14. A. Boyle’s Law
P
V
PV = k
Volume
(mL)
Pressure
(torr)
P·V
(mL·torr)
10.0 760.0 7.60 x 103
20.0 379.6 7.59 x 103
30.0 253.2 7.60 x 103
40.0 191.0 7.64 x 103

15. A. Boyle’s Law
 The pressure and volume of a gas are inversely
related
 at constant mass & temp
P
V
PV = k

16. A. Boyle’s Law

17. k
T
V

V
T
B. Charles’ Law
Volume
(mL)
Temperature
(K)
V/T
(mL/K)
40.0 273.2 0.146
44.0 298.2 0.148
47.7 323.2 0.148
51.3 348.2 0.147

18. k
T
V

V
T
B. Charles’ Law
 The volume and absolute temperature (K) of a gas are
directly related
 at constant mass & pressure

19. B. Charles’ Law

20. GIVEN:
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
P1V1T2 = P2V2T1
E. Gas Law Problems
 A gas occupies 473 cm3 at 36°C. Find its volume at 94°C.
CHARLES’ LAW
T V
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3

21. GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1T2 = P2V2T1
E. Gas Law Problems
 A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.
BOYLE’S LAW
P V
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL

22. III. Ideal Gas Law

23. k
n
V

V
n
A. Avogadro’s Principle
Gas
Volume
(mL)
Mass
(g)
Moles, n
V/n
(L/mol)
O2 100.0 0.122 3.81  10-3
26.2
N2 100.0 0.110 3.93  10-3
25.5
CO2 100.0 0.176 4.00  10-3
25.0

24. k
n
V

V
n
A. Avogadro’s Principle
 Equal volumes of gases contain
equal numbers of moles
at constant temp & pressure
true for any gas

25. PV
T
B. Ideal Gas Law
V
n
PV
nT
= k
UNIVERSAL GAS
CONSTANT
R=0.0821 Latm/molK
R=8.315 dm3kPa/molK
= R

26. B. Ideal Gas Law
UNIVERSAL GAS
CONSTANT
R=0.0821 Latm/molK
R=8.315 dm3kPa/molK
PV=nRT

27. GIVEN:
P = ? atm
n = 0.412 mol
T = 16°C = 289 K
V = 3.25 L
R = 0.0821Latm/molK
WORK:
PV = nRT
P(3.25)=(0.412)(0.0821)(289)
L mol Latm/molK K
P = 3.01 atm
B. Ideal Gas Law
 Calculate the pressure in atmospheres of
0.412 mol of He at 16°C & occupying 3.25
L.
IDEAL GAS LAW

28. GIVEN:
V = ?
n = 85 g
T = 25°C = 298 K
P = 104.5 kPa
R = 8.315 dm3kPa/molK
B. Ideal Gas Law
Find the volume of 85 g of O2 at
25°C and 104.5 kPa.
= 2.7 mol
WORK:
85 g 1 mol = 2.7 mol
32.00 g
PV = nRT
(104.5)V=(2.7) (8.315) (298)
kPa mol dm3kPa/molK K
V = 64 dm3
IDEAL GAS LAW

29. WORK:
PV = nRT
(97.3 kPa) (15.0 L)
= n (8.315dm3kPa/molK) (294K)
n = 0.597 mol O2
B. Gas Stoichiometry Problem
 How many grams of Al2O3 are formed from
15.0 L of O2 at 97.3 kPa & 21°C?
GIVEN:
P = 97.3 kPa
V = 15.0 L
n = ?
T = 21°C = 294 K
R = 8.315 dm3kPa/molK
4 Al + 3 O2  2 Al2O3
15.0 L
non-STP ? g
Given liters: Start with
Ideal Gas Law and
calculate moles of O2.
NEXT 

30. 2 mol
Al2O3
3 mol O2
B. Gas Stoichiometry Problem
How many grams of Al2O3 are formed
from 15.0 L of O2 at 97.3 kPa & 21°C?
0.597
mol O2
= 40.6 g Al2O3
4 Al + 3 O2  2 Al2O3
101.96 g
Al2O3
1 mol
Al2O3
15.0L
non-STP
? g
Use stoich to convert moles
of O2 to grams Al2O3.

31. A. Dalton’s Law
 The total pressure of a
mixture of gases equals the
sum of the partial pressures
of the individual gases.
Ptotal = P1 + P2 + ...
Patm = PH2 +
PH2O

32. GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 2.72 kPa
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH2 + 2.72 kPa
PH2 = 91.7 kPa
A. Dalton’s Law
 Hydrogen gas is collected over water at
22.5°C. Find the pressure of the dry gas if
the atmospheric pressure is 94.4 kPa.
Look up water-vapor pressure
on p.899 for 22.5°C.
Sig Figs: Round to least number
of decimal places.
The total pressure in the collection bottle is equal to atmospheric
pressure and is a mixture of H2 and water vapor.

33. GIVEN:
Pgas = ?
Ptotal = 742.0 torr
PH2O = 42.2 torr
WORK:
Ptotal = Pgas + PH2O
742.0 torr = PH2 + 42.2 torr
Pgas = 699.8 torr
A. Dalton’s Law
 A gas is collected over water at a temp of 35.0°C
when the barometric pressure is 742.0 torr. What
is the partial pressure of the dry gas?
DALTON’S LAW
Look up water-vapor pressure
on p.899 for 35.0°C.
Sig Figs: Round to least number
of decimal places.
The total pressure in the collection bottle is equal to barometric
pressure and is a mixture of the “gas” and water vapor.

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infringement intended.