Ideal Gas LawIdeal Gas Law
PV=nRTPV=nRT
Remember…Remember…
Boyle’s Law Boyle’s Law
Charles’ Law: Charles’ Law:
Combined Gas Law:Combined Gas Law:(Units MUST Match(Units MUST Match Temp in Kelvin!!!) Temp in Kelvin!!!)
2
2
1
1
T
V
T
V
2
22
1
11
T
VP
T
VP
2211 VPVP
A gas with a volume of 350 ml is collected A gas with a volume of 350 ml is collected at 15at 15o o C and 120 kPa. If the temperature C and 120 kPa. If the temperature changes to 30changes to 30o o C, what pressure would be C, what pressure would be required to put this gas in a 300 ml required to put this gas in a 300 ml container?container?
K
mlP
K
mlkPa
303
300
288
350120 2
kPaP 3.1472
A balloon has a volume of 500 ml at a A balloon has a volume of 500 ml at a temperature of 22temperature of 22ooC and a pressure of 755 C and a pressure of 755 mmHg. If the balloon is cooled to 0mmHg. If the balloon is cooled to 0o o C and C and a pressure of 145 mmHg, what is its new a pressure of 145 mmHg, what is its new volume?volume?
mlV 3.24092
K
VmmHg
K
mlmmHg
273
145
295
500755 2
Avogadro’s PrincipleAvogadro’s Principle
Under similar conditions (same Temp and Under similar conditions (same Temp and Pressure) equal volumes of gases contain Pressure) equal volumes of gases contain equal numbers of particles.equal numbers of particles.
10 L of H10 L of H22 (g) and 10 L of O (g) and 10 L of O22 (g) (g)
Both at Standard Temperature and Both at Standard Temperature and Pressure (STP) containPressure (STP) contain……
The same number of particles!The same number of particles!
Molar VolumeMolar Volume
The volume of 1 mole of gas particles The volume of 1 mole of gas particles
at STP at STP
is is 22.4 L22.4 L
Ideal Gas Law Animation:Ideal Gas Law Animation:
Try this:Try this:
1 mole of gas occupies 22.4 L at STP1 mole of gas occupies 22.4 L at STP
= __________ ml= __________ ml
(22400)(22400)
= ___________ moles of gas= ___________ moles of gas
(1 mole)(1 mole)
= ___________ particles= ___________ particles
(6.02 x 10(6.02 x 102323))
How many particles in 11.2 dmHow many particles in 11.2 dm33 of gas at STP? of gas at STP? 0.5 moles = 3.01 x 100.5 moles = 3.01 x 102323 particles particles22,400 cm22,400 cm3 3 of NHof NH33 gas at STP weighs? gas at STP weighs? = 22.4 L = 1 mole = 17 grams (add up MW)= 22.4 L = 1 mole = 17 grams (add up MW)44.8 L of NH44.8 L of NH33 at STP weighs? at STP weighs? = 2 moles = 34 grams= 2 moles = 34 grams
_____ grams = 1 mole of nitrogen gas _____ grams = 1 mole of nitrogen gas
= _____ L at STP? = _____ L at STP?
mole
LL
4.22
8.44
28.00
22.4
How many NHow many N22 molecules are molecules are
in 22.4 dmin 22.4 dm3 3 at STP?at STP?
= 1 mole = 6.02 x 10= 1 mole = 6.02 x 102323
What volume will 1.2 x 10What volume will 1.2 x 102424 H H2 2 molecules occupy molecules occupy
at STP?at STP?
= 2 moles = 44.8 L at STP= 2 moles = 44.8 L at STP
Ideal Gas EquationIdeal Gas EquationUse when Use when NOTNOT at STP!!! at STP!!!
PV= nRTPV= nRTP = Pressure (in kPa)P = Pressure (in kPa)
V = Volume (V = Volume (in Liters or dmin Liters or dm33))
n = number of molesn = number of moles
T = Temperature (T = Temperature (in Kelvinin Kelvin))
R =R = 8.31 8.31 L• kPaL• kPa
mole • Kmole • K
Development of R inDevelopment of R in
Kmole
LkPa
Kmole
LkPa
nT
PVR
31.82731
4.223.101
Kmole
LkPa
1. What volume will 2 moles of NO1. What volume will 2 moles of NO22 occupy occupy at 300 Kelvin and 90 kPa?at 300 Kelvin and 90 kPa?
LV
KmolesVkPa
nRTPV
4.5590
30031.82
30031.8290
What will be the temp of What will be the temp of 2 grams2 grams of of HH22 if 5000 cm if 5000 cm33 is at 5 atm? is at 5 atm?
KT
TmoleLkPa
nRTPV
8.30431.81
55.506
31.8155.506
Finding Molecular Weight of a GasFinding Molecular Weight of a Gas
Remember: MW = grams / molesRemember: MW = grams / moles
Converting grams to molesConverting grams to moles– Divide grams by the molecular weightDivide grams by the molecular weight
1) 5.0 L of a gas weighs 30.00 g at 201) 5.0 L of a gas weighs 30.00 g at 20o o C & C & 92 kPa. What is the mole weight of the gas?92 kPa. What is the mole weight of the gas?
g/mol 581mol 0.19
g 30.00MW
PV=nRT92kPa • 5.0 L= n • 8.31 • 293 K
n = 0.19 mol
mol ?
g 30.00
moles
gramsMW
2)If the mole weight of a gas is 26 g/mol and 2)If the mole weight of a gas is 26 g/mol and 18.00 g of the gas is 30 L at 2118.00 g of the gas is 30 L at 21oo C, what is the C, what is the pressure of the gas?pressure of the gas?
PV= nRT
P x 30 L=0.69 mol x 8.31 x 294 K
P = 56.2kPa
mol 0.69 g/mol 26
g 18.00
MW
g n
StoichiometryStoichiometry
Solving Steps.Solving Steps.– Balance the EquationBalance the Equation– Change grams to molesChange grams to moles– Use mole ratio to solveUse mole ratio to solve– Change moles to volumesChange moles to volumes
Use mole ratio (coefficients)
Use 22.4 L/mol @ STPOr PV = nRT
Use MWon P.T.
Use MWon P.T.
Use 22.4 L/mol@ STPOr PV = nRT
Mg (c) + HCl (ag) Mg (c) + HCl (ag) MgCl MgCl22 + H + H22 (g) (g)
If 2.43 g Mg react what volume of HIf 2.43 g Mg react what volume of H22 is is
produced? (at STP)produced? (at STP)
Mg (c) + HCl (ag) Mg (c) + HCl (ag) MgCl MgCl22 + H + H22 (g) (g)
2.43 g Mg2.43 g Mg
0.1 mole0.1 mole
molgg
3.24
43.2
1:1
molesmol
L1.04.22
0.1 moles0.1 moles
2
2.24 L H2.24 L H22? L
)313)(31.8)(1.0())(85( KmolVkPa
nRTPV
Mg (c) +Mg (c) +22 HCl (ag) HCl (ag) MgCl MgCl22 + H + H22 (g) (g)
If 2.43 g Mg react what volume of HIf 2.43 g Mg react what volume of H22 is produced at is produced at
4040ooC and 85 kPa?C and 85 kPa?
Mg (c) + HCl (ag) Mg (c) + HCl (ag) MgCl MgCl22 + H + H22 (g) (g)
2.43 g Mg2.43 g Mg
0.1 mole0.1 mole
molgg
3.24
43.2
1:10.1 moles0.1 moles
2
3.06 L H3.06 L H22? L
If 250 ml of HIf 250 ml of H22 is produced at 20 is produced at 20oo C & C &
100 kPa, what mass of Mg reacted?100 kPa, what mass of Mg reacted?
KnLkPa 29331.825.0100
2H 01.0 molesn
Mg (c) + HCl (ag) Mg (c) + HCl (ag) MgCl MgCl22 + H + H22 (g) (g)2
? g 250 mlnRTPV
0.01 moles0.01 moles0.01 mole0.01 mole 1:1
0.243 g Mg0.243 g Mg
molesmol
g01.03.24
Volume to VolumeVolume to Volume
THE MOLE RATIO IS THE SAME AS THE THE MOLE RATIO IS THE SAME AS THE VOLUME RATIO.VOLUME RATIO.
Liters BUse mole ratio (coefficients)
Liters A
Burning of methane:Burning of methane:
What vol. of oxygen is needed to completely What vol. of oxygen is needed to completely burn 1 L of methane?burn 1 L of methane?
CHCH44 + O + O22 CO CO22 (g) + H (g) + H22O (l)O (l)
2.0 L2.0 L 1:2
1.0 L1.0 L
2 2
Burning of methane:Burning of methane:
To produce 11.2 L of COTo produce 11.2 L of CO22 requires how requires how
many moles of Omany moles of O22 at STP? at STP?
CHCH44 + O + O22 CO CO22 (g) + H (g) + H22O (l)O (l)
22.4 L22.4 L 2:111.2 L11.2 L1 mole = 1 mole =
2 2
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