powerpoint to accompany chapter 9a gases. brown, lemay, bursten, murphy, langford, sagatys:...
TRANSCRIPT
PowerPoint to accompany
Chapter 9a
Gases
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Characteristics of Gases Like liquids and solids, they:
Still have mass and momentum Have heat capacities & thermal conductivities Can be chemically reactive
Unlike liquids and solids, they: Expand to fill their containers Are highly compressible Have extremely low densities
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Pressure
Pressure is the amount of force applied to an area.
Atmospheric pressure is the weight of air per unit of area.
P =FA
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Units of Pressure
Pascals 1 Pa = 1 N/m2 = 1 Kg/ms2
Bar 1 bar = 105 Pa = 100 kPa
mm Hg or torr The difference in the heights in mm (h) of
two connected columns of mercury Atmosphere
1.00 atm = 760 torr
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Standard Pressure
Normal atmospheric pressure at sea level.
This is ambient pressure, equal all around us, so not particularly felt
It is equal to: 1.00 atm 760 torr (760 mm Hg) 101.325 kPa 14.696 psi
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Manometer
Figure 9.2
Used to measure the difference between atmospheric pressure & that of a gas in a vessel.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Manometer calculation:
Patm= 102 kPa
Pgas> Patm
Pgas= Patm + h
Pgas= 102 kPa + (134.6mm – 103.8mm)Hg
Pgas= 102 kPa + (32.6mm Hg) Note 1mmHg = 133.322 Pa Pgas= 102 kPa + (32.6x133.322x10-3 kPa)
Pgas= 106 kPa = 1.06 bar =1.06/1.013 atm
Pgas= 1.046 atm
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Boyle’s Law
The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure: V ~ 1/P
Figure 9.4
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
As P and V are inversely proportional
A plot of V versus P results in a curve.
Since
V = k (1/P)
This means a plot of V versus 1/P will be a straight line.
PV = k
Figure 9.5
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Charles’s Law: initialwarm balloon, T big, V big
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Charles’s Law: finalcold balloon T & V little
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Charles’s Law
The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.
VT = k
A plot of V versus T will be a straight line.
Figure 9.7V = kT
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Avogadro’s Law The volume of a gas at constant temperature
and pressure is directly proportional to the number of moles of the gas.
Mathematically, this means: V = kn
Figure 9.8
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Ideal-Gas Equation
So far we’ve seen that:V 1/P (Boyle’s law)
V T (Charles’s law)
V n (Avogadro’s law)
Combining these, we get, if we call the proportionality constant, R:
nTP V = R
i.e. PV = nRT (where R = 8.314 kPa L/mol K)
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
V = k (1/P)
nTP V = R V
P
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Ideal Gas Law Calculations
Heat Gas in Cylinder from 298 K to 360 K at fixed piston position. (T incr. so P increases)
Same volume, same moles but greater P P1 = (nR/V1)T1 or P1/T1 = P2 / T2
P2 = P1T2 / T1
P2 = P1(T2 /T1 ) = P1(360/298) = 1.2081P1
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Ideal Gas Law Calculations
Move the piston to reduce the Volume from 1 dm3 to 0.5 dm3. (V decreases so P increases)
Same moles, same T, but greater P P1V1 = (nRT1) = P2V2
P2 = P1V1 / V2 = P1 (V1 / V2)
P2 = P1(1.0/0.5) = P1(2.0)
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Ideal Gas Law Calculations
Inject more gas at fixed piston position & T. the n increases so P increases)
Same volume, more moles but greater P P1 = n1(RT1/V1) or P1/n1 = P2 / n2
P2 = P1 n2 / n1 for n2 = 2 n1
P2 = P1(2)
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Ideal Gas Law Calculations
R = (PV / nT) When kPa and dm3 are used as is common R = 8.314 kPa dm3 / mol K or J/mol K It is useful to write the ideal gas law this way
so that R is the constant and other variables can change from initial to final conditions
This again reduces the gas law to a simple arithmetic ratio calculation
The only tricks are to ensure all units are consistent or converted as needed.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
V = (nRT/P) = 1mol (8.314kPa dm3 /molK)(273.15K)/100 kPaV/mol = 22.71 dm3 = 22.71L
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Gas Densities and Molar Mass
If we divide both sides of the ideal-gas equation by V and by RT, we get:
nV
PRT=
Most of our gas law calculations involve proportionalities of these basic commodities.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Gas Densities and Molar Mass
We know that moles molar mass = mass
So multiplying both sides by the molar mass () gives:
n = m
PRT
mV = i.e. =
PRT
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Molecular Mass
We can manipulate the density equation to enable us to find the molar mass of a gas:
Becomes
PRT =
RTP =
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
End of Part 9a Ideal Gas Law