Gases have some interesting characteristics that have fascinated scientists for 300 years....
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Transcript of Gases have some interesting characteristics that have fascinated scientists for 300 years....
Gases have some interesting characteristics that have fascinated scientists for 300 years.
Regardless of their chemical identity, gases tend to exhibit similar physical behaviors
The Nature of GasesThe Nature of Gases
Gas particles can be monatomic Gas particles can be monatomic (Ne), diatomic (N(Ne), diatomic (N22), or ), or polyatomic (CHpolyatomic (CH44) – but they all ) – but they all have these characteristics in have these characteristics in common:common:
Gas particles can be monatomic Gas particles can be monatomic (Ne), diatomic (N(Ne), diatomic (N22), or ), or polyatomic (CHpolyatomic (CH44) – but they all ) – but they all have these characteristics in have these characteristics in common:common: 1) Gases have mass.1) Gases have mass. 1) Gases have mass.1) Gases have mass.
2) Gases are compressible. 2) Gases are compressible. 3) Gases fill their containers. 3) Gases fill their containers. 4) Gases diffuse 4) Gases diffuse 5) Gases exert pressure. 5) Gases exert pressure. 6) Pressure is dependent on Temp. 6) Pressure is dependent on Temp.
KINETIC MOLECULAR THEORY (KMT)KINETIC MOLECULAR THEORY (KMT)Deriving the Ideal Gas LawDeriving the Ideal Gas Law
Theory used to explain gas laws. KMT Theory used to explain gas laws. KMT assumptions areassumptions areGases consist of molecules in constant, Gases consist of molecules in constant, random motion(Brownian motion).random motion(Brownian motion).P arises from collisions with container P arises from collisions with container walls.walls.No attractive or repulsive forces No attractive or repulsive forces between molecules. Collisions elastic.between molecules. Collisions elastic.Volume of molecules is negligible.Volume of molecules is negligible.
Properties of GasesProperties of GasesGas properties can be modeled Gas properties can be modeled
using math. Model depends onusing math. Model depends on——V = volume of the gas (L)V = volume of the gas (L)T = temperature (K)T = temperature (K)n = amount (moles)n = amount (moles)P = pressureP = pressure (atmospheres) (atmospheres)
Common Units of Pressure
Unit Atmospheric Pressure Scientific Field
pascal (Pa); 1.01325 x 105 Pa SI unit; physics, kilopascal(kPa) 101.325 kPa chemistry
atmosphere (atm) 1 atm* Chemistry millimeters of mercury 760 mmHg Chemistry, medicine, ( mm Hg ) biology
torr 760 torr* Chemistry
bar 1.01325 bar Meteorology, chemistry, physics
Converting Units of Pressure
Problem: A chemist collects a sample of carbon dioxide from the decomposition of limestone (CaCO3) in a closed end manometer, the height of the mercury is 341.6 mm Hg. Calculate the CO2 pressure intorr, atmospheres, and kilopascals.Plan: The pressure is in mmHg, so we use the conversion factors to find the pressure in the other units.Solution:
PCO2 (torr) = 341.6 mm Hg x = 341.6 torr 1 torr1 mm Hg
converting from mmHg to torr:
converting from torr to atm:
PCO2( atm) = 341.6 torr x = 0.4495 atm 1 atm760 torr
converting from atm to kPa:
PCO2(kPa) = 0.4495 atm x = 45.54 kPa101.325 kPa 1 atm
Gas Pressure Gas Pressure It can be defined as the force exerted by a
gas per unit surface area of an object
Due to: a) force of collisions, and b) number of collisions
No particles present? Then there cannot be any collisions, and thus no pressure – called a vacuum
Since the Kinetic Molecular Theory states that the volume of the gas particles is zero, then the equation simplifies. As a result, the amount of available space for the gas particles to move around in is approximately equal to the size of the container. Thus, as stated before, the variable V is the volume of the container.
1 L = 1dm3 = 1000 mL = 1000 cm3 = 10-3 m3
The temperature variable is always symbolized as T (absolute temperature that is in K). Consequently, T values must be converted to the Kelvin Scale. To do so;Kelvin = Celsius + 273
The Gas Laws are a mathematical interpretation of the behavior of gases.
1. Boyle- 1. Boyle- Mariotte’s Mariotte’s LawLawIf n and T are
constant, thenPV = k This means, for
example, that P goes up as V goes down and k is dependent on n and T.
Robert Boyle (1627-1691). Son of Early of Cork, Ireland.
Effect of changing pressure on volumeEffect of changing pressure on volume
The pressure can be changed by adding or removing green weights from the top of the red piston. This is an example of Boyle's law.
Boyle’s LawBoyle’s LawHyperbolic Relation Between Pressure and Volume
(courtesy F. Remer)
n constant
Boyle’s LawBoyle’s Law
Hyperbolic Relation Between Pressure and Volume
p
V
p – V Diagramp – V Diagram
n1 n2 n3 n3 >n2>n1
(courtesy F. Remer)
T constant
PV
P or V
Variation of gas volume with temperatureat constant pressure.
V T
T (K) = t (0C) + 273
n constant
Charles’ Law
Temperature must bein Kelvin!!!V/T = k
Charles’s Mathematical Law:Charles’s Mathematical Law:
since V/T = ksince V/T = k
Eg: A gas has a volume of 3.0 L at 127°C. What is its volume at
227 °C?
Eg: A gas has a volume of 3.0 L at 127°C. What is its volume at
227 °C?
V1 V2
T1 T2
=
What if we had a change in conditions?What if we had a change in conditions?
1)determine which variables you have:
1)determine which variables you have:
T and V = Charles’s LawT and V = Charles’s Law
2)determine which law is being represented:
2)determine which law is being represented:
T1 = 127°C + 273 = 400K
V1 = 3.0 L T2 = 227°C + 273 =
5ooK V2 = ?
T1 = 127°C + 273 = 400K
V1 = 3.0 L T2 = 227°C + 273 =
5ooK V2 = ?
4) Plug in the variables:4) Plug in the variables:
(500K)(3.0L) = V2 (400K)(500K)(3.0L) = V2 (400K)
V2 = 3.8LV2 = 3.8L
3.0L V23.0L V2
400K 500K400K 500K = =
5) Cross multiply and chug5) Cross multiply and chug
absolute zero = no molecular motion
no molecular motion = zero force in the container
Gay Lussac’s LawGay Lussac’s Law
Old man Lussac determined the relationship between temperature and pressure of a gas.He measured the temperature of air at different pressures, and observed a pattern of behavior which led to his mathematical law.During his experiments volume of the system and amount of gas were held constant.
Old man Lussac determined the relationship between temperature and pressure of a gas.He measured the temperature of air at different pressures, and observed a pattern of behavior which led to his mathematical law.During his experiments volume of the system and amount of gas were held constant.
Temp
Pre
ssu
re
How does Pressure and Temperature of gases relate
graphically?
How does Pressure and Temperature of gases relate
graphically?
P/T = k
Volume, # of particlesremain constant
Volume, # of particlesremain constant
Lussac’s Mathematical Law:Lussac’s Mathematical Law:
What if we had a change in conditions?What if we had a change in conditions?
since P/T = ksince P/T = k
P1 P2
T1 T2
=
Eg: A gas has a pressure of 3.0 atm at 127º C. What is its pressure at 227º C?
Eg: A gas has a pressure of 3.0 atm at 127º C. What is its pressure at 227º C?
T and P = Gay-Lussac’s LawT and P = Gay-Lussac’s Law
T1 = 127°C + 273 =
400K P1 = 3.0 atm T2 = 227°C + 273 =
500K P2 = ?
T1 = 127°C + 273 =
400K P1 = 3.0 atm T2 = 227°C + 273 =
500K P2 = ?
1)determine which variables you have:
1)determine which variables you have:
2)determine which law is being represented:
2)determine which law is being represented:
4) Plug in the variables:4) Plug in the variables:
(500K)(3.0atm) = P2 (400K)(500K)(3.0atm) = P2 (400K)
P2 = 3.8atmP2 = 3.8atm
3.0atm P23.0atm P2
400K 500K400K 500K = =
5) Cross multiply5) Cross multiply
Gay-Lussac’s LawGay-Lussac’s Law
Gay-Lussac’s LawGay-Lussac’s Law
Linear Relation Between Temperature and Pressure
P
T (K)0 100 200 300
P – T DiagramP – T Diagram
V1 V2
V3
V1 <V2 <V3
(courtesy F. Remer)
The mass is changed by injecting molecules at the left. The density (mass/volume) remains a constant for constant pressure and temperature.
Effect of changing mass on volumeEffect of changing mass on volume
LAWLAW RELAT-RELAT-IONSHIPIONSHIP LAWLAW CON-CON-
STANTSTANT
Boyle’sBoyle’s PP V V PP11VV1 1 = P= P22VV22 T, nT, n
CharlesCharles’’
VV T TVV11/T/T11 = = VV22/T/T22
P, nP, n
Gay-Gay-Lussac’Lussac’
ssPP T T
PP11/T/T11 = = PP22/T/T22
V, nV, n
Avogadro’Avogadro’ss nnVV
VV11/n/n11 = = VV22/n/n22
P,TP,T
The equality for the four variables involved in Boyle’s Law, Charles’ Law, Gay-Lussac’s Law and Avogadro’s law can be written
R = ideal gas constant
PV = nRT
R is known as the universal gas constant
Using STP conditions P V
R = PV = (1.00 atm)(22.4 L) nT (1mol) (273K)
n T
= 0.0821 L-atm /mol-K
Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mmHg) in the tank in the dentist office?
What is the molar mass of a gas if 0.250 g of the gas occupy 215 mL at 0.813 atm and 30.0°C?
Calculate the density in g/L of O2 gas at STP. From STP, we know the P and T.
How many L of O2 are need to react 28.0 g NH3 at 24°C and 0.950 atm?
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
GAS DENSITYGAS DENSITY
PV = nRTPV = nRTnV
= P
RT
nV
= P
RT
mM• V
= P
RT
where M = molar mass
mM• V
= P
RT
where M = molar mass
d = mV
= PMRT
d = mV
= PMRT
d and M proportional
Density (d) Calculations
d = m
V=
PM
RT
m is the mass of the gas in g
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRTPM = d is the density of the gas in g/L