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Chapter 16: Macroscopic Description of Matter. Stop to think 16.2 page 485 Stop to think 16.3 page 487 Stop to think 16.5 page 494 Stop to think 16.6 page 498 Stop to think 16.7 page 498 Example 16.2 page 485 Example 16.3 page 492 Example 16.6 page 495 Example 16.10 page 498.

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chapter 16 macroscopic description of matter
Chapter 16: Macroscopic Description of Matter
  • Stop to think 16.2 page 485
  • Stop to think 16.3 page 487
  • Stop to think 16.5 page 494
  • Stop to think 16.6 page 498
  • Stop to think 16.7 page 498
  • Example 16.2 page 485
  • Example 16.3 page 492
  • Example 16.6 page 495
  • Example 16.10 page 498
chapter 16 macroscopic description of matter2

Chapter 16: Macroscopic Description of Matter

Temperature: T is related to a system’s thermal energy.

Temperature scale:

1) CelsiusC: The world most common temperature scale As the old centigradescale, has 100 degree between melting point and boiling point of water, taken to occur 0 and 100 degrees, respectively.

2) Kelvin K : Temperature is measured as thermal energy of system. Takes its zero to be absolute zero.

3) Fahrenheit F: used primarily in US

solids liquids and gases
Solids, Liquids and Gases
  • Solid: a rigid macroscopic system with a definite shape and volume. An atom is not free to move around inside the solid: Solids are nearly incompressible.
  • Liquid: Liquid flows and deform to fit the shape of its container. It is nearly incompressible
  • Gas: a system in which each molecule moves through space as a free noninteracting particle until it collides with another molecule or with the wall of the container. A gas is also highly compressible.
density
Density
  • Mass density
pressure
Pressure
  • Pressure P, as defined
  • Units of P, 1pascal = 1 Pa = 1 N/m2
  • Atmospheric Pressure:
  • 1 standard atmosphere = 1 atm = 101,300 Pa =101.3kpa
atoms and moles
Atoms and Moles
  • Atomic massunit: m(12C) = 12u
  • Molecular mass m(O2) = 32u
  • Mole, the amount of any substance containing 6.02×1023 particle is defined as one mole.
  • Avogadro’s number NA= 6.02×1023
  • Number of mole
ideal gases
Ideal gases
  • The ideal-gas law
  • R = 8.31 J/mol•K Universal gas constant
  • P– pressure, units is Pa
  • V—volume, units is m3
  • PV– units is Pam3= Nm= joules.
  • n– the number of moles
  • T—temperature in Kelvin scale.
application of ideal gas law 16 20
Application of ideal gas law (16.20)
  • A 20 cm-diameter cylinder that is 40 cm long contains 50 g of oxygen gas at 20 C?
  • How many moles of oxygen are in the cylinder?
  • Notice the molar mass of oxygen is 32 g, therefore the number of mole for 50 g oxygen is 50g/32 g = 1.56 mole
  • How many oxygen molecules are in cylinder

N = 1.56/mol x NA= 9.4×1023

  • What is the number density of oxygen:
  • What is the reading of pressure gauge attached to the tank
  • Using the ideal gas law to get pressure inside the tank, P = nRT/V
  • Notice the reading of pressure gauge is the difference of pressure inside of tank from the pressure of outside.
ideal gas processes
Ideal-gas processes
  • The PV Diagram: Pressure –vs – volume
  • Each point on the graph represents a single unique state of the gas
  • Assuming that n is known, from ideal-gas law PV=nRT, we can get T for each point.
constant volume process
Constant Volume Process
  • Constant-volume process---isochoric process, volume does not change during the process.
constant pressure process
Constant-Pressure Process
  • Constant-pressure Process—isobaric process
  • The pressure does not change during the process
constant temperature process
Constant-temperature process
  • Constant-temperature process – isothermal process.
problem 16 31
Problem 16.31
  • 0.0040mol of gas undergoes the process as Fig.
  • What type of process is this?
  • The Fig. shows that the pressure is inversely proportional to the volume. So the process is isothermal. PV = nRT
  • The initial and final temperatures should be same T = PV/nR

Notice 1atm = 1.013×105 Pa,

1cm3=10-6 m3

masteringphysics up up and away
Masteringphysics Up Up and away
  • For the balloon float, at least the buoyant force should equal the total weight (basket, fabric, hot air inside etc), that is:
  • With
  • We get