chapter 5 chem notes

chapter 5 chem notes - The Chemistry of Gases...

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The Chemistry of Gases Classification of matter Chemical constitution Physical state solid liquid gas Water at <0 o C at 20 o C at >100 o C Chemical reactions can lead to the formation of gaseous substances. Example: CaCO 3 ( s ) CaO( s ) +CO 2 ( g ) CaCO 3 ( s )+2HCl( aq ) CaCl 2 ( aq )+H 2 O ( l ) +CO 2 ( g ) Acid rain effect on marble S( s ) + O 2 ( g ) SO 2 ( g ) Elemental sulfur produces SO 2 (g) when burned in oxygen.
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The State of a gas is defined by Pressure(P), Volume(V) and Temperature(T) Volume is expressed in L, mL or cc Pressure : Molecules in a container are in constant motion and collide with the walls of the container and among themselves. Force exerted during these collisions is related to pressure. vacuum 760 mm Hg Liquid Hg Torricelli’s experiment (1608-1647): fill a capillary tube with Hg, close the open end with thumb, invert it into a beaker of Hg and release the thumb. Pressure exerted by outside air Pressure exerted by Hg column Pressure exerted by Hg column in the tube = pressure exerted by outside air vacuum
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Force(F) = mass (m) x accelaration due to gravity(g); F = mg So, P= F/A= mg/A But, Area (A) x Height(h) = Volume (V); Ah = V or A = V/h Then, P=mg/(V/h) = mgh/V But m/V=d, density; Therefore, P = dgh Pressure exerted by Hg column of height h If some other liquid is used, in place of liquid Hg, then density of Hg will be replaced by density of that liquid and height of Hg column will be replaced by the height of that liquid column. Pressure exerted by Hg column of height h= density of Hg x accln due to gravity x height of mercury column For a given pressure, d 1 gh 1 =d 2 gh 2 , subscripts 1 and 2 represent two different liquids Pressure(P) = Force (F) Area (A)
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At 0 o C, d Hg = 13.5951 g/cc At sea level,under normal conditions, h = 760 mm Hg = 0.760 m Hg P = (13.5951 g/cc)(9.80665 m/s 2 )(0.760 m) = 101.325 g.m 2 /cc.s 2 1 g= 1g x (1kg/1000g)=10 -3 kg P=(101.325)(10 -3 kg) m 2 / (10 -6 m 3 .s 2 )= 1.01325 x10 5 kg. m -1 .s -2 Therefore, 760 mm of Hg column corresponds to a pressure of 1.01325 x10 5 Pa = 1.01325 bar (1 bar = 10 5 Pa) A pressure of 1 atm supports 760 mm Hg at 0 o C. 1 atm = 760 mm Hg = 1.01325 x10 5 Pa = 1.01325 bar 1 torr = (1/760) atm 1 torr = 1 mm Hg at 0 o C Units for Pressure: atm, pascal, torr or bar 1 atm =760 torr
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Boyle’s law Liquid Hg Open to atmosphere trapped air Volume V 1 Pressure P 1 =1 atm Volume V 1 h mm, additional height of Hg P =1 atm P =1 atm Pressure P 2 = 1 atm + h mm x (atm/760 mm) Volume, V 2 Volume V 2 Boyle discovered that P 1 V 1 = P 2 V 2 at a constant temperature (a). plot P vs 1/V P 1/V at T 1 at T 2 Boyle’s law: PV = constant C (at a fixed temp and amount of gas) Slope of the line =constant C (b). Plot PV vs P PV P at T 1 at T 2 Product PV is independent of P
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Temperature Temperature is often associated with “hot” and “cold”. Temperature where water freezes is colder than that where water boils. Two different temperature scales are in use. Scale freezing pt boiling pt of H 2 O of H 2 O Celsius 0 o C 100 o C Fahrenheit 32 o F 212 o F ice 0 o C 32 o F steam (a). But what happens to measuring temperature when Hg changes its physical state to gas or solid ? (b).What happens if we use a different liquid in place of Hg, which may not expand linearly with temperature?
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This note was uploaded on 04/13/2008 for the course CHEM 102a taught by Professor Hanusa during the Fall '06 term at Vanderbilt.

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chapter 5 chem notes - The Chemistry of Gases...

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