The Laws of Thermodynamics

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The Laws of Thermodynamics

• Chapter 15

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What is temperature? Heat? Thermal/Internal
Energy?

• Temperature (T) A measure of the average kinetic
  energy of individual molecules
• Heat Amount of energy transferred from one body
  to another at different temperature
• Thermal/Internal Energy (U) Total energy of all
  molecules in an object
• The sum of the translational kinetic energies of
  all the atoms

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Average Kinetic Energy and RMS velocity
The higher the temperature of a gas, the faster
the molecules move!
R Gas constant 8.315 J/(mol K) M Molar mass µ
mass of molecule
kB Boltzmanns constant 1.38 x 10-23 J/K T
temperature
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Internal Energy of an Ideal Gas
N number of molecules n number of
moles kBoltzmanns Constant RGas Constant T
Temperature
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Ideal Gas Law and Combined Gas Law
P Pressure V Volume n number of moles R Gas
Constant T temperature
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Types of systems

• A closed system No mass enters/leaves but energy
  may be exchanged with the environment
• An open system Mass and energy may enter/leave
• Isolated System No energy in any form
  enters/leaves the boundaries

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Thermodynamics

• The study of processes in which energy is
  transferred as heat and as work
• Heat is a transfer of energy due to a difference
  in temperature
• Work is a transfer of energy that is not due to a
  temperature difference

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First Law of Thermodynamics

• The change in internal energy of a closed system,
  ?U, is equal to the heat added to the system
  minus the work done by the system
• Q net heat added to the system (Q)
• If heat leaves the system, Q is negative
• W net work done by the system (W)
• The work done on a system (-W) is the opposite of
  the work done by the system (W)!

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How does a gas do work?

• Lets say there is a gas in a container with a
  movable piston
• As you heat the temperature, the gas expands and
  it causes the piston to move upward
• Since W Fd, the gas does work on the piston
• Work done by a gas is equal to the product of
  pressure and volume!

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What if the gas is compressed?

• If the gas is compressed by the piston, that
  means work is done on the gas, so W is negative!

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Types of Processes Isothermal

• Isothermal process An idealized process that is
  carried out at constant temperature. Since U
  depends on T, there is no change in internal
  energy in this process
• Ideal Gas Law PVnRTconstant
• 1st Law QW because ?U0

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Types of Processes Adiabatic

• Adiabatic Process No heat is allowed to flow
  into or out of a system so Q0, but work is done
  on the system. This happens if the system is
  extremely well insulated or if it happens so
  quickly that heat has no time to flow in or out
• 1st Law ?U-W
• Internal Energy decreases if the gas expands
  (because the gas does work when it expands) so
  the temperature must also decrease because U is
  proportional to T
• Internal Energy increases if the gas is
  compressed (because work is done on the gas when
  it is compressed), so the temperature must also
  increase

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Isobaric and Isochoric/Isovolumetric Processes

• In an isobaric process, the pressure is kept
  constant.
• In an isochoric/isovolumetric process, the volume
  is kept constant

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PV Diagram for multiple processes (p. 447)
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Using PV Diagram to Find Work

• The work done by a gas is equal to the area under
  the PV curve

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Sample Problem p. 472 9

• Consider the following two step process. Heat is
  allowed to flow out of an ideal gas at constant
  volume so that its pressure drops from 2.2 atm to
  1.4 atm. Then the gas expands at constant
  pressure from a volume of 6.8 L to 9.3 L where
  the temperature reaches its original value.
  Calculate (a) the total work done by the gas in
  the process, (b) the change in internal energy of
  the gas in the process and (c) the total heat
  flow into or out of the gas.

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Solve the problem. Part (a)

• How much work is done by the gas?
• From point A to point B, what kind of process is
  this?
• Volume remains constant. This is isovolumetric.
  WP ? V0
• From point B to point C, what kind of process is
  this?
• Pressure remains constant. This is isobaric
• W P ? V Area under the curve

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Solve the problem

• Area under the curve is P?V
• P 1.4 atm 1.42 x 105 Pa
• ?V 9.3 L- 6.8 L 2.5 L 2.5 x 10-3 m3
• W P ?V 355 J
• Part Bwhat is the change in internal energy?
• Initial and final temperature are the same
• Since ?T0, then ?U 0

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Finish the Problem

• Part C. What is the total heat flow into or out
  of the gas?

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Practice Problem p. 472 10

• The PV diagram in Fig 15-28 shows two possible
  states of a system containing two moles of a
  monatomic ideal gas. (P1P2450 Pa, V1 2m3 ,
  V2 8m3)
• A. Draw the process which depicts an isobaric
  expansion from state 1 to state 2 and label this
  process (A)
• B. Find the work done by the gas and the change
  in internal energy of the gas in process A.
• C. Draw the process which depicts an isothermal
  expansion from state 1 to the volume V2 followed
  by an isochoric increase in temperature to state
  2 and label this process (B).
• D. Find the change in internal energy of the gas
  for the two step process (B)

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Practice Problem p. 472 10

• A. A. Draw the process which depicts an isobaric
  expansion from state 1 to state 2 and label this
  process (A)

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Practice Problem p. 472 10

• B. Find the work done by the gas and the change
  in internal energy of the gas in process A.
• Work P ?V (450 Pa)(8m3 6m3) 2700 J
• Change in internal energy?
• How can we find the change in T?
• Use the Ideal Gas Law! PVnRT

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Practice Problem p. 472 10

• C. Draw the process which depicts an isothermal
  expansion from state 1 to the volume V2 followed
  by an isochoric increase in temperature to state
  2 and label this process (B).

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Practice Problem p. 472 10

• D. Find the change in internal energy of the gas
  for the two step process (B)
• Since both paths have the same initial and final
  temperatures, ?U is the same 4050 J