Elements that exist as gases at 25 0 C and 1 atmosphere.
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Transcript of Elements that exist as gases at 25 0 C and 1 atmosphere.
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Elements that exist as gases at 250C and 1 atmosphere
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Pressure
Remember:• Pressure = Force / area or
mass x gravity / area• Force = mass x gravity
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Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa
Barometer
Pressure = ForceArea
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Measuring PressureDevices used:• Barometer-Invented by Torricelli• Pressure Gauge• Tire Gauge• Sphygmomanometer
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Units of Pressure
• Standard atmospheric pressure (1 atm) = 760 mm Hg or 760 torr (1 mm Hg = 1 torr)
• In the United States we use inches in atmospheric pressure.
• Pascal- used in science as a unit of pressure (Pa).
• Kilopascals are used as well (KPa).
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PressureColumn height measures
Pressure of atmosphere• 1 standard atmosphere
(atm) *
= 760 mm Hg (or torr) *
= 29.92 inches Hg *
= 14.7 pounds/in2 (psi) *HD only
= 101.3 kPa (SI unit is PASCAL) * HD only
= about 34 feet of water!
* Memorize these!
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Practice
• How many torrs are in 4.5 atm?
• The barometer reads 1,760 torr, how many atm?
• I have a reading of 3700 torr, convert to atm.
• You have 3 atm, how many torr?
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Boyle’s Law
• Shows the relationship between Pressure and Volume.
• The Law assumes that the temperature does not change.
• Proposed by Robert Boyle.
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Boyle’s LawP1V1 = P2V2
• P1= Pressure 1• V1= Volume 1• P2= Pressure 2• V2= Volume 2• This formula works with any unit of Pressure (torr, Pa)• Any unit of volume works (mL, L).• Make sure you have the same units on both sides! If not,
change one to match the other.• If one side has mL and the other has L, convert one to
match the other.
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Boyle’s Law• A gas measured a volume of 100 mL under
pressure of 740 torr. What would be the volume under a pressure of 780 torr with constant temperature?
Step 1: Figure out what you have.• P1 = 740 torr• V1= 100 mL• P2= 780 torr• V2= X
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Boyle’s Law
• Step 2: Plug the numbers into the formula
P1V1 = P2V2
(740 torr)(100 mL) = (780 torr)(X)
• Step 3: Solve for x
74000 = 780X
X= 94.87 mL
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Boyle’s Law• So the new volume in this case is 94.87 mL• Since we increased the pressure, the volume is decreased. • The formula is a proportion. If something increases, something
else will decrease!
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Practice
1. A sample of gas is confined to a 100 mL flask under pressure of 740 torr. If the same gas were transferred to a 50 mL flask, what’s the new pressure?
P1 =
V1 =
P2 =
V2 =
P1V1 = P2V2
740 torr
100 ml
X
50 ml
P2 = P1V1/V2
P2 = (740)(100)/50
P2 = 74000/50
P2 = 1480 torr
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Practice
2. You are given a gas that you measure under a pressure of 720 Pa. When the pressure is changed to 760 Pa, the volume became 580 mL. What is the first volume?
P1V1 = P2V2
P1 =
V1 =
P2 =
V2 =
720 Pa
X
760 Pa
580 ml
V1 = P2V2/P1
V1 = (760)(580)/720
V1 = 440800/720
V1 = 612.2 ml
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Practice3. A pressure on 134 mL of air is changed to
1200 torr at a constant temperature, if the new volume is 45 mL what is the original pressure?
P1V1 = P2V2
P1 =
V1 =
P2 =
V2 =
X
134 ml
1200 torr
45 ml
P1 = P2V2/V1
P1 = (1200)(45)/134
P1 = 54000/134
P1 = 402.9 torr
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Practice4. An amount of Oxygen occupies 2 L when
under pressure of 680 torr. If the volume is increased to 3 L what is the new pressure?
P1V1 = P2V2
P1 =
V1 =
P2 =
V2 =
680 torr
2 L
X
3L
P2 = P1V1/V2
P2 = (740)(100)/50
P2 = 74000/50
P2 = 1480 torr
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Reminders• Check your units• Temperature must remain constant in
Boyle’s Law.• P= Pressure (torr, Pa, KPa).• V= Volume (mL, L).
“If you don’t exhale on your way up Cookie, your lungs will explode!
Robert De Niro as Chief Saturday
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When the temperature increases, the volume increases.
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As T increases V increases
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Charles Law
• Shows a relationship between Volume and Temperature. One thing increase, so does the other. Both behave the same way.
• Pressure remains constant.
• V= Volume
• T= Temperature
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Few things first…
• ALL TEMPERATURES MUST BE CONVERTED TO KELVIN!
• The Law only works with Kelvin
• If you have Celsius, make sure you convert it to Kelvin.
• Remember Kelvin = Celsius + 273
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Temperature Conversion Practice
1. Convert 85°C to K. ____________________
2. Convert 376K to °C. ____________________
3. Convert 154K to °C. ____________________
4. Convert -65°C to K. ____________________
5. Convert 0°C to K. _____________________
6. Convert 0K to °C. _____________________
85 + 273 =
376 - 273 =
154 - 273 =
-65 + 273 =
0 + 273 =
0 - 273 =
358K
103 °C
-119 °C
208K
273K
-273 °C
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Few Practice Problems
You heat 100 mL of a gas at 25 C to 80 C. What is the new volume of the gas?
V1 =
T1 =
V2 =
T2 =
V1/T1 = V2/T2
100 ml
25ºC
X
80ºC
+ 273 =
+ 273 =
298K
353K
V2 = V1T2/T1
V2 = (100)(353)/298
V2 = 35300/298
V2 = 118.5 ml
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Few Practice Problems
You have 200 mL of a gas at 55 C and you freeze it to 0 C. What is the new volume?
V1/T1 = V2/T2
V1 =
T1 =
V2 =
T2 =
200 ml
55ºC
X
0ºC
+ 273 =
+ 273 =
328K
273K
V2 = V1T2/T1
V2 = (200)(273)/328
V2 = 54600/328
V2 = 166.5 ml
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Few Practice Problems
A 150 mL sample of gas is at 125 K. After cooling the new volume is 80 mL. What is the new temperature?
V1/T1 = V2/T2
V1 =
T1 =
V2 =
T2 =
150 ml
125K
80 ml
X
T2 = V2T1/V2
T2 = (80)(125)/150
T2 = 10000/150
T2 = 66.67K
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Gay-Lussac’s LawP1/T1 = P2/T2
• Shows relationship between pressure (P) and temperature (T). If Temperature increases, so does the pressure.
• Based on Charles’ Law.• Volume remains constant.• Like Charles’ Law, all temperature units must be in
KELVIN.• Used in pressure cookers and autoclaves.
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Practice
A soda bottle with a temperature of 25ºC and 3 atm of pressure was put into a freezer with a temperature of –1°C. What is the pressure on the bottle inside the freezer?
P1 =
T1 =
P2 =
T2 =
3 atm
25ºC
X
-1ºC
+ 273 =
+ 273 =
298K
272K
P1/T1 = P2/T2
P2 = P1T2/T1
P2 = (3)(272)/298
P2 = 816/298
P2 = 2.74 atm
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Combined Gas Law
• Combines Boyle’s Law, Gay-Lussac’s Law, and Charles’ Law.
• Shows the relationship of Pressure, Volume and Temperature.
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Combined Gas Law
• Derived from Boyle’s, Gay-Lussac, and Charles Law.• Used when both temperature and pressure changes.• Can be used to find a constant.• If two things increase, one thing will decrease.• Math was used to derive the law.
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Combined Gas Law (P, V, T)
Combined Gas Law: nR = PV
T
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1. A gas has a volume of 800.0 mL at -23.00 °C and 300.0 torr. What would the volume of the gas be at 227.0 °C and 600.0 torr of pressure?
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =
300 torr
800 ml
-23ºC
600 torr
X
227.0ºC
+ 273 =
+ 273 =
250K
500K
P1V1 = P2V2
T1 T2
V2 = P1V1T2
T1 P2
V2 = (300)(800)(500)
(250)(600)
V2 = 800 ml
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Remember this…
• When you are working with Charles’ Law and Combined Gas Law, make sure you have the proper units!
• The laws help us predict the behavior of gases under different circumstances.
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How is this possible?
KC-135 Jet
Rail Car
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Avogadro’s LawV1/n1 = V2/n2
• Allows us to calculate the number of moles (n) of a gas with in a volume (V).
• Pressure and Temperature remains the same.• If volume increases, so does the number of moles.
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Practice
In 150 mL sample, there are 2.5 grams of Cl2 gas, if I double the volume to 300 mL how much Cl2 gas in moles, is in the new sample?
V1 =
n1 =
V2 =
n2 =
150 ml
2.5g
300 ml
x
÷ 17g/m = 0.147 m
V1/n1 = V2/n2
n2 = V2 n1/ V1
n2 = (300)(0.147)/150
n2 = 44.1/150
n2 = 0.294 m
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Ideal gas law
• Formula: PV= nRT
• P= Pressure in atmospheres• V= Volume in Liters• n= number of moles of a gas• R= constant, 0.0821 L • atm / mol • K• T= Temperature in Kelvin
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Ideal Gas Law
• An ideal gas is a gas that obeys Boyle’s and Charles’ Law.
• Based on the Kinetic Theory.• Shows the relationship between Pressure,
Volume, number of moles and Temperature.• Make sure everything has the proper unit.• Can be used to figure out density of gas or
determine the behavior of a tire.
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Practice
How many moles of gas are contained in 890.0 mL at 21.0 °C and 750.0 mm Hg pressure?
PV = nRT
P =
V =
n =
R = 0.0821 Latm/molK
T =
750.0 mm Hg
890.0 ml
X
21.0ºC + 273 = 294 K
n = PV/RT
n = (0.987)(0.89)/(0.0821)(294)
n = 0.878/24.1
n = 0.036 m
÷ 760 mmHg/atm = 0.987 atm
÷ 1000 ml/L = 0.89 L
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And now, we pause for this commercial message from STP
OK, so it’s really not THIS kind of STP…
STP in chemistry stands for Standard Temperature and Pressure
Standard Pressure = 1 atm (or an equivalent)
Standard Temperature = 0 deg C (273 K)
STP allows us to compare amounts of gases between different pressures and temperatures
STP allows us to compare amounts of gases between different pressures and temperatures
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The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).
PV = nRT
R = PV
nT=
(1 atm)(22.414L)
(1 mol)(273.15 K)
R = 0.082057 L • atm / (mol • K)
Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.
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What is the volume (in liters) occupied by 49.8 g of HCl at STP?
PV = nRT
V = nRT
P
T = 0 0C = 273.15 K
P = 1 atm
n = 49.8 g x 1 mol HCl36.45 g HCl
= 1.37 mol
V =1 atm
1.37 mol x 0.0821 x 273.15 KL•atmmol•K
V = 30.6 L
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GAS DENSITYGAS DENSITY
High density
Low density
22.4 L of ANY gas AT STP = 1 mole
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Density (d) Calculations
d = mV
=PMRT
m is the mass of the gas in g
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRT
PM = d is the density of the gas in g/L
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Gas Stoichiometry
What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
g C6H12O6 mol C6H12O6 mol CO2 V CO2
5.60 g C6H12O6
1 mol C6H12O6
180 g C6H12O6
x6 mol CO2
1 mol C6H12O6
x = 0.187 mol CO2
V = nRT
P
0.187 mol x 0.0821 x 310.15 KL•atmmol•K
1.00 atm= = 4.76 L
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Gases and Stoichiometry
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP?
Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.
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Deviations from Ideal Gas Law
• Real molecules have volume.The ideal gas consumes the entire
amount of available volume. It does not account for the volume of the molecules themselves.
• There are intermolecular forces.
An ideal gas assumes there are no attractions between molecules. Attractions slow down the molecules and reduce the amount of collisions.– Otherwise a gas could
not condense to become a liquid.
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Gases in the Air
The % of gases in air Partial pressure (STP)
78.08% N2 593.4 mm Hg
20.95% O2 159.2 mm Hg
0.94% Ar 7.1 mm Hg
0.03% CO2 0.2 mm Hg
PAIR = PN + PO + PAr + PCO = 760 mm Hg
Total Pressure 760 mm Hg
2 22
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Dalton’s Law of Partial Pressures
V and T are constant
P1 P2 Ptotal = P1 + P2
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Dalton’s Law
John Dalton1766-1844
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Consider a case in which two gases, A and B, are in a container of volume V.
PA = nART
V
PB = nBRT
V
nA is the number of moles of A
nB is the number of moles of B
PT = PA + PB XA = nA
nA + nB
XB = nB
nA + nB
PA = XA PT PB = XB PT
Pi = Xi PT
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A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?
Pi = Xi PT
Xpropane = 0.116
8.24 + 0.421 + 0.116
PT = 1.37 atm
= 0.0132
Ppropane = 0.0132 x 1.37 atm = 0.0181 atm
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2KClO3 (s) 2KCl (s) + 3O2 (g)
Bottle full of oxygen gas and water vapor
PT = PO + PH O2 2
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Collecting a gas “over
water”
• Gases, since they mix with other gases readily, must be collected in an environment where mixing can not occur. The easiest way to do this is under water because water displaces the air. So when a gas is collected “over water”, that means the container is filled with water and the gas is bubbled through the water into the container. Thus, the pressure inside the container is from the gas AND the water vapor. This is where Dalton’s Law of Partial Pressures becomes useful.
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Table of Vapor Pressures for Water
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Solve This!
A student collects some hydrogen gas over water at 20 degrees C and 768 torr. What is the pressure of the gas?
768 torr – 17.5 torr = 750.5 torr
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GAS DIFFUSION AND EFFUSION
• diffusion is the gradual mixing of molecules of different gases.
• effusion is the movement of molecules through a small hole into an empty container.
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GAS DIFFUSION AND EFFUSION
Graham’s law governs effusion and diffusion of gas molecules.
Thomas Graham, 1805-1869. Professor in Glasgow and London.
Rate of effusion is inversely proportional to its molar mass.
Rate of effusion is inversely proportional to its molar mass.
M of AM of B
Rate for B
Rate for A
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GAS DIFFUSION AND EFFUSION
Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is
• proportional to T• inversely proportional to M.
Therefore, He effuses more rapidly than O2 at same T.
He
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Gas Diffusionrelation of mass to rate of diffusion
Gas Diffusionrelation of mass to rate of diffusion
• HCl and NH3 diffuse from opposite ends of tube.
• Gases meet to form NH4Cl
• HCl heavier than NH3
• Therefore, NH4Cl forms closer to HCl end of tube.
• HCl and NH3 diffuse from opposite ends of tube.
• Gases meet to form NH4Cl
• HCl heavier than NH3
• Therefore, NH4Cl forms closer to HCl end of tube.