Gases 12 with notes by 7LiH18j

VIEWS: 4 PAGES: 7

									                                   Gases
Recall that an ideal gas is a gas that behaves according to the kinetic
molecular theory. Also recall that any gas behaves relatively ideally at low
pressure and high temperature.

We can look at the following four variables to describe the state of an ideal
gas:

1. Pressure (atm)

2.Temperature (K)

3. Volume (L)

4. Amount (moles)


Pressure: Force exerted over a unit of area. (F/A) In a sample of gas, the
pressure is caused by collisions of the particles with the walls of the
container.

Units of Pressure: atm, torr, mmHg, Pa, kPa, psi

Pressure Conversions: 1 atm = 760 torr= 760 mmHg = 101.325kPa= 1.01325
x105Pa= 14.7 psi.

Pressure Measurement:

1. Barometer – measures air pressure (barometric pressure)
http://www.barometerplanet.com/the-mercury-barometer/




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2. Manometer – measures pressure of an enclosed gas.

                            P0 = atmospheric pressure
                    Pgas   = Pressure of the enclosed gas.




                                 The Gas Laws




1. Boyle’s Law – The pressure and volume of a sample of gas are inversely
proportional to each other at constant temperature. P= k (1/V) where k is a
constant. Mathematical form: P1V1=P2V2
Sample problem C on page 370.

2. Charles’ Law - The volume of a fixed amount of gas is directly
proportional to its Kelvin temperature, at constant pressure. V= kT where k
is a constant. Mathematical form: V1/T1=V2/T2
Sample problem D on page 372.

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3. Gay – Lussac’s Law The pressure of a fixed mass of gas is directly
proportional to its Kelvin temperature, at constant volume. P= kT where k is
a constant. Mathematical form: P1/T1=P2/T2
Sample problem E on page 373.

4. The Combined Gas Law: Expresses the relationship between pressure,
volume, and Kelvin Temperature of a fixed amount of gas.

PV/T =K so (P1V1/T1) = (P2V2/T2)

Sample problem F on page 375.



5. Ideal Gas Law
Describes the physical behavior of an ideal gas in terms of pressure,
temperature, volume, and # of moles of the gas.

                                    PV= nRT
Units are specific!
P in atm, V in L, n in moles, T in Kelvin, R= (0.0821 L- atm)/ (K-mole)

Problems in HW packet (page 11) 15b, 14c

6. Applications of the Ideal Gas Law: Using PV=nRT to calculate Molar Mass
(M) or Density (D)
1. Molar Mass (M)
M= m/n (grams/mol) so n= m/ M Substitute n into PV=nRT 
                               M= mRT/ PV
                either memorize or know how to derive this!

2. Density(D)
D= m/V so V= m/D Substitute n (m/ M ) and V (m/D) into PV=nRT 
P (m/D) = (m/ M )RT:

                              Get D= P M/ RT




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7. Avogadro’s Law: Equal volumes of gases contain the same # of particles
at the same Temperature and Pressure. STP means Standard Temperature
and Pressure. Standard Temperature = 0 C. Standard Pressure = 1 atm.
At STP 1 mole of any gas occupies a volume of 22.4 L

                        So at STP: 1 mole gas = 22.4 L



8. Applications of Avogadro’s Law: Only at STP!!!!!!

Since D= m/V, for a gas

D= Molar Mass/ Molar Volume [M(g/mol)/ 22.4 L/mol]

And

M = Dx 22.4L/mol




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9. Dalton’s Law – In the absence of a chemical reaction, the total pressure
of a mixture of gases is equal to the sum of the partial pressures of the
individual gases.

                              PT = P1 + P2 + P3 …etc

PT = Total Pressure, P1 = Partial Pressure of Gas 1 P2 = Partial Pressure of Gas 2, etc.

To determine the partial pressure, calculate the mole fraction (X gas) of the
individual gas:
                                  X gas =    n gas
                                            n total
n gas = moles of individual gas, n total = total moles of gas. See example

The Partial Pressure of Gas 1 = Total Pressure x X        gas 1


            See problem 16 on worksheet 13-3 Practice Problems:
                         Total pressure = 762 mmHg
                                 X N2= 0.78
          This means that the N2 exerts 78% of the total pressure:
                    PN2 = 0.78 x 762 mmHg = 594 mmHg

10. Applications of Dalton’s Law – Collecting a gas over water.




Gases produced in lab are usually collected over water. The gas is bubbled
into a bottle that is filled with water, and inverted in a dish or trough of


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water. As the gas is produced, it bubbles through the water and displaces
it. (pushes it out of the bottle and into the dish) The gas occupies the
volume over the water level. When the reaction is finished, the water level
inside and outside the bottle are equalized. (lined up) This means that the
pressure on the inside of the bottle is the same as the pressure on the
outside of the bottle (atmospheric pressure.) The pressure on the inside of
the bottle is due to 2 gases: The one that was produced in the experiment,
and WATER vapor, since there is always water vapor in the space over liquid
water. The total pressure in the bottle is then equal to the atmospheric
pressure, (which we can read from a barometer) and it is equal to the sum of
the pressure of the water vapor and the other gas. We can then use a water
vapor pressure table to subtract the pressure due to the water from the
atmospheric pressure to calculate the pressure of the gas we produced.

Read pages 366- 367 and do example problem B, and the practice problem
that follows it.



                        Collecting a Gas Over Water

A certain mass of Oxygen was collected over water when potassium Chlorate
was decomposed by heating. The volume of the gas collected was 720. mL
at 25.0 C and a barometric pressure of 755 mmHg. What would the volume
of dry oxygen be at STP?

To solve you need the Vapor Pressure of water, which is listed in Appendix
A-8 in your textbook. Answer = 635 mL.




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                         Stoichiometry and Gases



Use gas laws in stoichiometry problems to solve quantitative problems about
reactions.

Ex. Tungsten, a metal that is used to make light bulb filaments, is produced
industrially by the following reaction of Tungsten Oxide with hydrogen:

WO3   (s)   + 3H2 (g)  W   (s)   + 3 H20 (l)

How many liters of hydrogen at 350  C and 745 mm Hg are needed to react
completely with 875 g of Tungsten Oxide?




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