Transcript Avogadro’s Hypothesis

EMPIRICAL GAS LAWS
Boyle’s Law
P1V1 = P2V2
Charles’ Law
V1 / T1 = V2 / T2
Guy-Lussac’s Law
P1 / T1 = P2 / T2
Avogadro’s Law
V1 / n1 = V2 / n2
Combined Gas Law
P1V1 / T1 = P2 V2 / T2
Ideal Gas Law
PV = nRT
P = pressure (atm)
V = volume (L)
n = chemical amount (mol)
T = Temperature (K)
R = ideal gas constant = 0.08206 L-atm / mol-K
AvogAdro’s HypotHesis
• Avogadro pictured the moving molecule as occupying a
small portion of the larger space apparently occupied by
the gas. Thus the “volume” of the gas is related to the
spacing between particles and not to the particle size itself.
• Imagine 3 balloons each filled with a different gas (He, Ar,
& Xe). These gases are listed in increasing particle size,
with Xe being the largest atom. According to Avogadro’s
Hypothesis, the balloon filled with one mole of He will
occupy that same volume as a balloon filled with one mole
of Xe.
• So for a gas, the “volume” and the moles are directly
related.
V a n
AvogAdro’s HypotHesis
• A sample of N2 gas at 3.0 atm and 20.0oC is known to occupy a
volume of 1.43 L. What volume would a 0.179 mole sample of
NH3 gas occupy at the same pressure and temperature?
First calculate the number of moles of nitrogen gas:
PV = nRT
where P = 3.0 atm, V = 1.43 L, R = 0.082 L-atm/mol-K, and T =
20.0 oC + 273 = 293K
n = PV / RT
= (3.0 atm x 1.43L) / (0.082 L-atm/mol-K x 293K)
= 0.179 moles of N2
So since the moles of N2 is 0.179 mol and the moles of ammonia is
0.179 mol according to Avogadro’s hypothesis the volume of NH3
at that pressure and that temperature is 1.43 L, the same!!!
At STP, gas molecules are so far
apart that for 1 mole of gas, the
overall volume does not change.
STP :
P = 1 atm & T = 273 K
GAS
WEIGHT
H2
2.0 g/mol
MOLAR
VOLUME
22.4 L/mol
N2
28.0 g/mol
22.4 L/mol
Xe
131.3 g/mol
22.4 L/mol
EMPIRICAL GAS LAWS & STP
1. The gases in a rigid Helium filled ball at 25oC exerts a
pressure of 1.2 atm. If the ball is placed in a freezer and
the pressure decreases to 1/8 of its original value, what is
the temperature inside the ball?
2. A balloon containing 2.50 moles of He has a volume of
850 mL at a certain temperature and pressure. How
many grams of He would have to be removed from the
balloon in order for the volume to decrease to 250 mL
under the same conditions?
3. A sample of Ne gas at 21oC & 760 mmHg had a volume
of 250.0 mL. What would be the volume of the gas under
STP conditions?
4. How many atoms are present in 1.0 L of SO3 gas at
STP?
IDEAL GAS LAW
Q. What is the pressure inside a gas balloon
if it filled with 852 g of Xe gas at 25.0oC
and occupies a volume of 7.00 L?
P=?
V = 7.00 L
852 g Xe ( 1 mol / 131 g) = 6.50 mol
T = 25oC + 273 = 298 K
P = nRT
V
P = (6.50 mol) (0.082 L-atm / mol-K) (298 K)
7.00 L
P = 22.7 atm
IDEAL GAS LAW
What would be the temperature of 100 g
of Ar gas contained in a 500 L sealed
container at 0.8976 atm.
T2 = 1914 oC
How much would a balloon weigh if it
contained 40.0 L of O2 gas at 987 mmHg
and 45.3 oC?
mass O2 = 63.7 g
DENSITY OF A GAS
The density of a gas at STP can be calculated by
dSTP = molar mass/molar volume
Calculate the density of hydrogen
sulfite gas at STP.
d (STP) = (82 g/mol) / 22.4 L/mol) = 3.66 g/L
Identify an unknown homonuclear
diatomic gas that was found to have a
density of 3.165 g/L at STP.
Cl2
DENSITY OF A GAS
The density of a gas not at STP can be calculated by
d = (MM) P / R T
Calculate the density of hydrogen sulfite
gas at 587 torr and 56.9 oC.
d = (82 g/mol*0.772 atm) / (0.082 L-atm/mol-K*330K)
Identify
= 2.34 g/L an unknown homonuclear
diatomic gas that was found to have a
density of 1.950 g/L at 2.57 atm and 177oC.
N2
Properties of Gases
DIFFUSION
Diffusion is the ability of two or more gases to mix spontaneously until
a uniform mixture is formed.
Example: A person wearing a lot of perfume walks into an enclosed
room, eventually in time, the entire room will smell like the perfume.
 EFFUSION
Effusion is the ability of gas particles to pass through a small opening
or membrane from a container of higher pressure to a container of
lower pressure.
The General Rule is: The lighter the gas, the faster it moves.
Graham’s Law of Effusion:
Rate of effusion of gas A
Rate of effusion of gas B
=
√(molar mass B / molar mass A)
The rate of effusion of a gas is inversely proportional to the square root
of the molar mass of that gas.
DALTON’S LAW OF PARTIAL
PRESSURES
If there is more than one gas present in a
container, each gas contributes to the
total pressure of the mixture.
Ptotal = Pgas A + P gas B + Pgas C …
If the total pressure of a system was 2.5
atm, what is the partial pressure of
carbon monoxide if the gas mixture also
contained 0.4 atm O2 and 1.48 atm of N2?
PT - PO2 - PN2 = PCO 2.5 atm - 0.4 atm - 1.48 atm = 0.62 atm
STOICHIOMETRY & THE GAS
LAWS
1. Write a balanced chemical equation
2. Convert to moles (if gas, use PV=nRT or Molar Volume)
3. Use the mole ratio to convert from moles of “A” to moles of “B”.
4. Convert moles of “B” to desired measurement, if a gas use
PV=nRT.
1. What volume of O2 is needed to combust 348.0 L of C3H8?
C3H8 + 5 O2  3 CO2 + 4 H2O
Due to Avogadro’s Hypothesis, the moles of a gas are directly
related to the volume of a gas therefore it is possible to use the
mole ratio on volumes of gas.
348.0 L C3H8 (5 mol O2 / 1 mol C3H8) = 1740 L O2
STOICHIOMETRY & THE GAS
LAWS
2. How many grams of CO2 is produced from 348.0 L of C3H8 if
the temperature is 40.0oC and the pressure is 654 torr?
C3H8 + 5 O2  3 CO2 + 4 H2O
P = 654 torr (1 atm / 760 torr) = 0.861 atm
T = 40oC + 273 = 313 K
PV / RT = n
= (0.861 atm) (348.0 L) /(0.082 L-atm/mol-K) (313 K)
= 11.67 mol of C3H8
11.67 mol C3H8 (3 mol CO2 / 1 mol C3H8) = 35.02 mol CO2
35.02 mol CO2 (44 g / 1 mol) = 1541 g of CO2
STOICHIOMETRY & THE GAS
LAWS
3. In lab, you decomposed potassium chlorate into oxygen and
potassium chloride. What volume of O2 at STP can be formed
from 3.65 g of potassium chlorate?
2 KClO3  3 O2 + 2 KCl
3.65 g (1 mol / 122.6g) = 0.02977 mol KClO3
0.02977 mol KClO3 (3 mol O2 / 2 mol KClO3) = 0.04466 mol O2
0.04466 mol O2 ( 22.4 L / 1 mol) = 1.00 L
PRACTICE PROBLEM # 20b
1. Both hydrogen and helium have been used as buoyant gas in blimps. If a
small leak were to occur, which gas would effuse more rapidly and by what
factor?
Hydrogen effuses first by a factor of 1.41
2. At STP, 560 mL of a gas has a mass of 1.08 g. What is the molecular weight
of the gas?
43.2 g/mol
3. A gas is known to be either sulfur dioxide or sulfur trioxide. Its density at
98oC and 1.08 atm is 2.84 g/L. Which gas is it?
SO3
4. A 50.0 L cylinder of nitrogen has a pressure of 17.1 atm at 23oC. What is the
mass of nitrogen in the cylinder?
986 g
5. When a 2.0 L bottle of concentrated HCl was spilled, 3.0 kg of CaCO3 was
required to neutralize the spill.
CaCO3 (s) + 2HCl (aq)  CaCl2 (aq) + H2O (l) + CO2 (g)
What volume of CO2 gas was released by the neutralization at 735 mmHg and
20 oC?
745 L
Group Study Problem # 20b
1. Measured at 65 oC and 500.0 torr, the mass of 3.21 L of a gas is 3.5 g.
What is the molar mass of this gas
2. A 100.0 mL sample of air is analyzed and found to contain 0.835 g N2, 0.0640
g CO2 and 0.197 g O2 at 35 oC. What is the total pressure of the sample and
the partial pressure of each component?
3. What volume would 5.30 L of H2 gas at STP occupy if the temperature was
increased to 70oF and the pressure to 830 torr?
4. Carbon monoxide is produced by the reaction of coke with oxygen from
preheated air. 2 C + O2  2 CO . How many liters of atmospheric oxygen at
an effective pressure of 182 torr and a temperature of 29.0oC are required to
produce 895 L of carbon monoxide at 846 torr and 1700oC?
5. Hydrogen gas is produced by the complete reaction of 8.34 g of aluminum
metal with an excess of gaseous hydrogen sulfate. How many liters of
hydrogen will be produced if the temperature is 50.0 oC and the pressure is
0.950 atm?
Group Study Problem
Answers:
1. 45.9 g/mol
2.
PN2=7.53 atm, PO2=1.55 atm,
PCO2=0.380 atm; PT = 9.46 atm
3.
5.23 L
4.
320 L
5.
12.9 L