Combined and ideal gas laws

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Combined and ideal gas laws
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Combined gas law

• If we combine all of the relationships from the 3
  laws covered thus far (Boyles, Charless, and
  Gay-Lussacs) we can develop a mathematical
  equation that can solve for a situation where 3
  variables change

PVk1
V/Tk2
P/Tk3
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Combined gas law

• Amount is held constant
• Is used when you have a change in volume,
  pressure, or temperature

P1V1T2 P2V2T1
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Example problem
A gas with a volume of 4.0L at STP. What is its
volume at 2.0atm and at 30C?
1atm
2.0 atm
?
4.0 L
273K
30C 273
303K
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Example problem
(1 atm)
(4.0L)
(2 atm)
( V )
2

(273K)
(303K)
2.22L V2
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Classroom Practice
P1 V1 T1 P2 V2 T2
1.5 atm 3.0 L 20?C 2.5 atm 30?C
720mmHg 256 ml 240 ml
2.5 L 22?C 1.8 L
95 kPa 4.0L 101 kPa 6.0 L 198?C
850mmHg 15?C 30?C
1.00 atm 150 ml 0.762 atm
125 kPa 100?C 100kPa 100 ml 75?C
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Avogadros Law

• So far weve compared all the variables except
  the amount of a gas (n).
• There is a lesser known law called Avogadros Law
  which relates V n.
• It turns out that they are directly related to
  each other.
• As of moles increases then V increases.

V/n k
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Ideal Gas Law

• Which leads us to the ideal gas law
• The fourth and final variable is amount
• We have been holding it constant.
• We can set up a much more powerful eqn, which can
  be derived by combining the proportions expressed
  by the previous laws.

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Ideal Gas Law

• If we combine all of the laws together including
  Avogadros Law mentioned earlier we get

Where R is the universal gas constant
Normally written as
PV nRT
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Ideal Gas Constant (R)

• R is a constant that connects the 4 variables
• R is dependent on the units of the variables for
  P, V, T
• Temp is always in Kelvin
• Volume is in liters
• Pressure is in either atm or mmHg or kPa

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Ideal Gas Constant

• Because of the different pressure units there are
  3 possibilities for our ideal gas constant

• If pressure is given in atm

• If pressure is given in mmHg

• If pressure is given in kPa

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Using the Ideal Gas Law
What volume does 9.45g of C2H2 occupy at STP?
1atm
P ?
R ?
?
V ?
T ?
273K
n ?
.3635 mol
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(1.0atm)
(V)

(.3635mol)
(273K)
V 8.15L
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A camping stove propane tank holds 3000g of C3H8.
How large a container would be needed to hold
the same amount of propane as a gas at 25C and a
pressure of 303 kpa?
303kPa
P ?
R ?
?
V ?
T ?
298K
n ?
68.2 mol
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(303kPa)
(V)

(298K)
(68.2 mol)
V 557.7L
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Classroom Practice

1. Use the Ideal Gas Law to complete the following
   table for ammonia gas (NH3).

Pressure Volume Temp Moles Grams
2.50 atm 0?C 32.0
75.0 ml 30?C 0.385
768 mmHg 6.0 L 100?C
195 kPa 58.7 L 19.8
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Ideal Gas Law Stoichiometry
What volume of hydrogen gas must be burned to
form 1.00 L of water vapor at 1.00 atm
pressure and 300C?
(1.00 atm)
(1.00 L)
nH2O
(573K)
(.0821L atm/mol K)
nH2O .021257 mols
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Ideal Gas Law Stoichiometry
2H2 O2 ? 2H2O
.021257 mol
.476 L H2
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Classroom Practice
To find the formula of a transition metal
carbonyl, one of a family of compounds having the
general formula Mx(CO)y, you can heat the solid
compound in a vacuum to produce solid metal and
CO gas. You heat 0.112 g of Crx(CO)y Crx(CO)y(s)
? x Cr(s) y CO(g) and find that the CO
evolved has a pressure of 369 mmHg in a 155 ml
flask at 27?C. What is the empirical formula of
Crx(CO)y?
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Variations of the Ideal Gas Law

• We can use the ideal gas law to derive a version
  to solve for MM.
• We need to know that the unit mole is equal to m
  MM, where m is the mass of the gas sample

PV nRT
n m/MM
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Variations of the Ideal Gas Law

• We can then use the MM equation to derive a
  version that solves for the density of a gas.
• Remember that D m/V

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Classroom Practice 1

1. A gas consisting of only carbon and hydrogen has
   an empirical formula of CH2. The gas has a
   density of 1.65 g/L at 27?C and 734 mmHg.
   Determine the molar mass and the molecular
   formula of the gas.
2. Silicon tetrachloride (SiCl4) and trichlorosilane
   (SiHCl3) are both starting materials for the
   production of electro-nics-grade silicon.
   Calculate the densities of pure SiCl4 and pure
   SiHCl3 vapor at 85?C and 758 mmHg.

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Real Vs. Ideal

• All of our calculations with gases have been
  assuming ideal conditions and behaviors.
• We assumed that there was no attraction
  established between particles.
• We assumed that each particle has no volume of
  its own
• Under normal atmospheric conditions gases tend to
  behave as we expect and as predicted by the KMT.

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Real Vs. Ideal

• However, under high pressures and low
  temperatures, gases tend to deviate from ideal
  behaviors.
• Under extreme conditions we tend to see a
  tendency of gases to not behave as independently
  as the ideal gas law predicts.
• Attractive forces between gas particles under
  high pressures or low temperature cause the gas
  not to behave predictably.

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Loose Ends of Gases

• There are a couple more laws that we need to
  address dealing with gases.
• Daltons Law of Partial Pressures
• Grahams Law of Diffusion and Effusion.

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Daltons Law of Partial Pressure

• States that the total pressure of a mixture of
  gases is equal to the sum of the partial
  pressures of the component gases.

PTP1P2P3

• What that means is that each gas involved in a
  mixture exerts an independent pressure on its
  containers walls

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Simple Daltons Law Calculation

• Three of the primary components of air are CO2,
  N2, and O2. In a sample containing a mixture of
  these gases at exactly 760 mmHg, the partial
  pressures of CO2 and N2 are given as PCO2 0.285
  mmHg and PN2 593.525 mmHg. What is the partial
  pressure of O2?

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Simple Daltons Law Calculation
PT PCO2 PN2 PO2
760mmHg .285mmHg 593.525mmHg PO2
PO2 167mmHg
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Daltons Law of Partial Pressure

• Partial pressures are also important when a gas
  is collected through water.
• Any time a gas is collected through water the gas
  is contaminated with water vapor.
• You can determine the pressure of the dry gas by
  subtracting out the water vapor

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Atmospheric Pressure
Ptot Patmospheric pressure Pgas PH2O

• The waters vapor pressure can be determined from
  a list and subtract-ed from the atmospheric
  pressure

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WATER VAPOR PRESSURES WATER VAPOR PRESSURES WATER VAPOR PRESSURES
Temp (C) (mmHg) (kPa)
0.0 4.6 .61
5.0 6.5 .87
10.0 9.2 1.23
15.0 12.8 1.71
15.5 13.2 1.76
16.0 13.6 1.82
16.5 14.1 1.88
17.0 14.5 1.94
17.5 15.0 2.00
18.0 15.5 2.06
18.5 16.0 2.13
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WATER VAPOR PRESSURES WATER VAPOR PRESSURES WATER VAPOR PRESSURES
Temp (C) (mmHg) (kPa)
19.0 16.5 2.19
19.5 17.0 2.27
20.0 17.5 2.34
20.5 18.1 2.41
21.0 18.6 2.49
21.5 19.2 2.57
22.0 19.8 2.64
22.5 20.4 2.72
23.0 21.1 2.81
23.5 21.7 2.90
24.0 22.4 2.98
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WATER VAPOR PRESSURES WATER VAPOR PRESSURES WATER VAPOR PRESSURES
Temp (C) (mmHg) (kPa)
24.5 23.1 3.10
25.0 23.8 3.17
26.0 25.2 3.36
27.0 26.7 3.57
28.0 28.3 3.78
29.0 30.0 4.01
30.0 31.8 4.25
35.0 42.2 5.63
40.0 55.3 7.38
50.0 92.5 12.34
60.0 149.4 19.93
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WATER VAPOR PRESSURES WATER VAPOR PRESSURES WATER VAPOR PRESSURES
Temp (C) (mmHg) (kPa)
70.0 233.7 31.18
80.0 355.1 47.37
90.0 525.8 70.12
95.0 633.9 84.53
100.0 760.0 101.32
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Simple Daltons Law Calculation

• Determine the partial pressure of oxygen
  collected by water displace-ment if the water
  temperature is 20.0C and the total pressure of
  the gases in the collection bottle is 730 mmHg.

PH2O at 20.0C 17.5 mmHg
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Simple Daltons Law Calculation
PT PH2O PO2
PH2O 17.5 mmHg
PT 730 mmHg
730mmHg 17.5468 PO2
PO2 712.5 mmHg
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Your Turn

1. A mixture of 1.00 g H2 and 1.00 g He is placed in
   a 1.00 L container at 27?C. Calculate the
   partial pressure of each gas and the total
   pressure in atm.
2. Helium is collected over water _at_ 25?C and 1.00
   atm total pressure. What total volume of He must
   be collected to obtain 0.586 g of He?

1) PH2 12.3 atm PHe 6.16 atm 2) 3.7 L
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Grahams Law

• Thomas Graham studied the effusion and diffusion
  of gases.
• Diffusion is the mixing of gases through each
  other.
• Effusion is the process whereby the molecules of
  a gas escape from its container through a tiny
  hole

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Diffusion
Effusion
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Grahams Law

• Grahams Law states that the rates of effusion
  and diffusion of gases at the same temperature
  and pressure is dependent on the size of the
  molecule.
• The bigger the molecule the slower it moves the
  slower it mixes and escapes.

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Grahams Law

• Kinetic energy can be calculated with the
  equation ½ mv2
• m is the mass of the object
• v is the velocity.
• If we work with two different at the same
  temperature their energies would be equal and the
  equation can be rewritten as

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½ MAvA2 ½ MBvB2

• M represents molar mass
• v represents molecular velocity
• A is one gas
• B is another gas

• If we want to compare both gases velocities, to
  determine which gas moves faster, we could write
  a ratio of their velocities.
• Rearranging things and taking the square root
  would give the eqn

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vA
MB

vB
MA

• This shows that the velocities of two different
  gases are inversely propor-tional to the square
  roots of their molar masses.
• This can be expanded to deal with rates of
  diffusion or effusion

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Grahams Law

• The way you can interpret the equation is that
  the number of times faster A moves than B, is the
  square root of the ratio of the molar mass of B
  divided by the Molar mass of A
• So if A is half the size of B than it effuses or
  diffuses 1.4 times faster.

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Grahams Law Example Calc.
If equal amounts of helium and argon are placed
in a porous container and allowed to escape,
which gas will escape faster and how much faster?
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Grahams Law Example Calc.
Rate of effusion of He

Rate of effusion of Ar
Helium is 3.16 times faster than Argon.
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Your Turn

1. Calculate the average rate of effusion of a H2
   molecule at 0 C if the average rate of effusion
   of an O2 molecule at this temperature is 500 m/s.
2. The rate of effusion of a gas was meas-ured to be
   24.0 ml/min. Under the same conditions, the rate
   of effusion of pure CH4 gas is 47.8 ml/min. What
   is the molar mass of the unknown gas?

1) 2000 m/s 2) 63.5 g/mol