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