Chapter 5 Gases -2 Gas Laws - Manu's Adventuresdrsapnag.manusadventures.com/chemistry/general... ·...
Transcript of Chapter 5 Gases -2 Gas Laws - Manu's Adventuresdrsapnag.manusadventures.com/chemistry/general... ·...
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Chapter 5Gases -2Gas Laws
Dr. Sapna Gupta
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Gas Laws
• Gas laws – empirical relationships among gas parameters. • Volume (V)
• Pressure (P)
• Temperature (T)
• Amount of a gas (n)
• By holding two of these physical properties constant, it becomes possible to show a simple relationship between the other two properties.
Dr. Sapna Gupta/Gas Laws 2
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Boyle’s Law• The volume of a sample of gas is inversely proportional to pressure at
constant temperature. (one increases if the other decreases and vice versa)
PV = constant
P1V1=P2V2
PV
1
Dr. Sapna Gupta/Gas Laws 3
When a 1.00-g sample of O2 gas at 0oC is placed in a container at a pressure of 0.50 atm, it occupies a volume of 1.40 L.
When the pressure on the O2 is doubled to 1.0 atm, the volume is reduced to 0.70 L, half the original volume.
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Boyle’s Law Contd….
Graphical representation
Dr. Sapna Gupta/Gas Laws 4
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Solved Problem: A volume of oxygen gas occupies 38.7 mL at 751 mmHg and 21°C. What is the volume if the pressure changes to 359 mmHg while the temperature remains constant?
Vi = 38.7 mL
Pi = 751 mmHg
Ti = 21°C
Vf = ?
Pf = 359 mmHg
Tf = 21°C
f
iif
P
VPV
mmHg)(359
mmHg)mL)(751(38.7f V
= 81.0 mL
(3 significant figures)
Dr. Sapna Gupta/Gas Laws 5
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Charles’s Law
• The volume of a sample of gas at constant pressure is directly proportional to the absolute temperature (K).
TV
At room Temp At – 72oC
Dr. Sapna Gupta/Gas Laws 6
f
f
i
i
constant
T
V
T
V
T
V
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Charles’s Law contd….
Graphical Representation
The temperature -273.15°C is called absolute zero. It is the temperature at which the volume of a gas is hypothetically zero.
This is the basis of the absolute temperature scale, the Kelvin scale (K).
Dr. Sapna Gupta/Gas Laws 7
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Solved Problem:You prepared carbon dioxide by adding HCl(aq) to marble chips, CaCO3. According to your calculations, you should obtain 79.4 mL of CO2 at 0°C and 760 mmHg. How many milliliters of gas would you obtain at 27oC?
Vi = 79.4 mL
Pi = 760 mmHg
Ti = 0°C = 273 K
Vf = ?
Pf = 760 mmHg
Tf = 27°C = 300. K
i
iff
T
VTV
K)(273
mL)K)(79.4(300.f V
= 87.3 mL
(3 significant figures)
Dr. Sapna Gupta/Gas Laws 8
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Avogadro’s Law
The volume of a gas sample is directly proportional to the number of moles in the sample at constant pressure and temperature.
Dr. Sapna Gupta/Gas Laws 9
𝑉
𝑛= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
V a n Vi
ni=Vf
nf
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Dalton’s Law of Partial Pressure
• Dalton found that in a mixture of unreactive gases, each gas acts as if it were the only gas in the mixture as far as pressure is concerned.
• The sum of the partial pressures of all the different gases in a mixture is equal to the total pressure of the mixture:
P = PA + PB + PC + . . .
Partial Pressure
The pressure exerted by
a particular gas in a mixture
Ptotal = PN2+ O
2PN2
Dr. Sapna Gupta/Gases-Stoichiometry 10
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Solved Problem:A 100.0-mL sample of air exhaled from the lungs is analyzed and found to contain 0.0830 g N2, 0.0194 g O2, 0.00640 g CO2, and 0.00441 g water vapor at 35°C. What is the partial pressure of each component and the total pressure of the sample?
•
•
mL10
L1mL100.0
K308Kmol
atmL0.08206
Ng28.01
Nmol1Ng0.0830
3
2
22
N2P
•
•
mL10
L1mL100.0
K308Kmol
atmL0.08206
Og32.00
Omol1Og0.0194
3
2
22
O2P
•
•
mL10
L1mL100.0
K308Kmol
atmL0.08206
COg44.01
COmol1COg0.00640
3
2
22
CO2P
•
•
mL10
L1mL100.0
K308Kmol
atmL0.08206
OHg18.01
OHmol1OHg0.00441
3
2
22
OH2P
atm0.749
atm0.153
atm0.0368
atm0.0619
OHCOON 2222PPPPP = 1.00 atm
Dr. Sapna Gupta/Gases-Stoichiometry 11
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Key Points
• The gas laws
• Boyle’s law
• Charles’ law
• Avogadro’s law
• Dalton’s law
Dr. Sapna Gupta/Gas Laws 12