an introduction to gases chapter 13. kinetic molecular theory postulate #1 –gases consist of tiny...

An Introduction to Gases

Chapter 13

Kinetic Molecular Theory

• Postulate #1– Gases consist of tiny particles (atoms or

molecules)

• Postulate #2– These particles are so small, compared

with the distances between them, that the volume (size) of the individual particles can be assumed to be negligible (zero).

– Gases are COMPRESSIBLE

Kinetic Molecular Theory

• Postulate #3– The particles are in constant random

motion, colliding with the walls of the container. These collisions with the walls cause the pressure exerted by the gas.

Kinetic Molecular Theory

• Postulate #4– The particles are assumed not to

attract or to repel each other.

• Postulate #5– The average kinetic energy of the gas

particles is directly proportional to the Kelvin temperature of the gas.

Avogadro’s Hypothesis

• At the same temp & pressure, equal volumes of gas hold same number of molecules.

• V and n are directly related.V and n are directly related.

twice as many molecules

Pressure

Pressure of air is Pressure of air is measured with a measured with a BAROMETERBAROMETER(developed by Torricelli in (developed by Torricelli in 1643)1643)

Pressure

• What is Pressure?What is Pressure?

• What tool do we use to measure it?What tool do we use to measure it?

Pressure

1.1. mmHg (or Torr)mmHg (or Torr)2.2. Atmospheres (atm)Atmospheres (atm)3.3. Pascals (used in physics: Pascals (used in physics:

1 pascal = 1 newton per square meter)1 pascal = 1 newton per square meter)4. 4. psi psi

Equivalences:Equivalences:1 atm = 760 mmHg1 atm = 760 mmHg1 atm = 101,325 Pa = 101.325 kPa1 atm = 101,325 Pa = 101.325 kPa1 atm = 14.7 psi

Pressure Calculation

What is 475 mm Hg expressed in atm?

475 mm Hg=

1 atm

760 mm Hg0.625 atm

The pressure of a tire is measured as 294,000 pascals. What is this pressure in atm?

Pressure Calculation

294, 000 Pa

101, 325 Pa

1 atm= 2.90 atm

Dalton’s Law

“The Law of Partial Pressure”

• The total pressure of a mixture of gases is the sum of the partial pressures of the gases in the mixture.

Ptotal = PA + PB + PC

Gas Laws Calculations

Get out a calculator!!!

The Gas Law

PV=nRT

P = pressure ( atm or kPa )V= volume ( L )

n= number of moles (mol)T= temperature (K)

R – The Proportionality Constant

Value depends on units

8.314L (kPa)mol (K)

0.0821L (atm)mol (K)

Or

The Gas Law – Problem If 7.0 moles of an ideal gas has a volume

of 12.0 L with a temperature of 300. K, what is the pressure in kPa?

P (12.0 L) =(7.0 mol)(300

K)8.31

4

L (kPa)

mol (K)

PV = nRT

P = 1454.95 kPa P = 1500 kPa

Combined Gas Law

fff

iii

ff

ii

TRn

TRn

VP

VP

Let’s say we have some O2 gas AND we change some conditions. Would there be anything similar between the two gases?

Combined Gas Law – Problem

You have 3 moles of a solution at 300. K and 15 atm in a 2 L container. If the container is heated to 350. K and the volume decreased to 1 L, what will the new pressure be?

P1 15 atm P2 want

V1 2 L V2 1 L

n1 3 moles n2 3 moles

R1 constant R2 constant

T1 300. K T2 350. K

Combined Gas Law – Problemc

P1V1=

n1R1T1

P2V2 n2R2T2

P1V1=

T1

P2V2 T2

If we know that R1 = R2 and the mass is constant then

(15 atm)(2 L) = (300. K)

P2(1L) (350. K)

Replace with numbers

Combined Gas Law – Problem

(15 atm)(2 L) = (300. K)

P2(1L) (350. K)

(15 atm)(2 L)(350. K) =

P2(1L)(300. K)

P2 = 35 atm

Pressure & Volume

• At constant Temperature• Pressure and Volume vary

inversely.– Why? – More collisions More pressure

P1V1 = P2V2

P & V – Example Problem If you start with 0.500 L of a gas at 7.0 atm and you move the gas to a container with 3.5 L available, how much pressure will the gas exert?

7 atm (0.500 L)= P2

3.5 L

P1 (V1) = P2 (V2)

7.0 atm (0.500 L) = P2 (3.5 L)

1.0 atm = P2

Put a few drops of water in a can. Heat the can until the water boils. What is happening to the gas inside? Now flip the can over

into cold water. Predict what do you predict will happen?

Demo

On a Larger Scale

On a Larger Scale

Temperature & Volume

At constant PressureVolume & Temperature vary directly.– Why?– More collisions More Volume

V1=

V2

T1 T2

T & V – Example Problem

If a gas is in a balloon with a volume of 12.0 L and at a temperature of 300. K, what will the volume be if you place the balloon in a freezer at 250. K?

V1=

V2

T1 T2

12.0 L=

V2

300. K250.

K

12.0 L (250. K) = V2

300. K

10.0 L = V2

S.T.P.

• Standard Temperature and PressureStandard Temperature and Pressure

These are conditions that are universalThese are conditions that are universal

Standard Temperature: Standard Temperature:

0ºC or 273K0ºC or 273K

Standard Pressure: Standard Pressure:

1atm or 101.325kPa1atm or 101.325kPa

S.T.P. – Example Problem

What is the volume What is the volume of 1 mole of a gas of 1 mole of a gas at STP?at STP?

P 1 atm

V want

n 1 mole

R 0.0821 (L)(atm)/(K)(mole)

T 273 KPV = nRT

(1atm)V = (1 mole)(0.0821 [Latm/Kmole])(273K)

V= 22.4 L