Thermodynamics -------------------------------------  Joseph F. Alward, PhD      Department of  Physics    University of the Pacific

 

Thermodynamics:  "thermo":     Greek  therme heat      "dynamics":  Greek  dynamikos powerful    Physics that deals with the mechanical  action or relations between heat and work  Example 1:  Heat to work  Heat Q from flame provides energy  to do work ----------------------------------------------------------  Example 2:  Work to heat.  Work done by person is converted   to heat energy via friction.

Internal Energy U:      (measured in joules)    Sum of random translational,  rotational, and vibrational  kinetic energies DU:  change in U DU > 0 is a gain of internal              energy DU < 0 is a loss of internal              energy ---------------------------------------- Thermal Energy:    same as internal energy

Vibrational kinetic energy  in solids The hotter the object, the larger the vibrational kinetic energy

Motions of a diatomic  molecule in a fluiid

    Ideal Gases

Molecules are point-size.  Gas is low-density.     Elastic collisions are the only interactions. -----------------------------------------------------------------  Average kinetic energy  per molecule = (3/2)kT   k = Boltzmann's constant      = 1.38 x 10-23 J/K   U = (3/2) NkT   N = number of molecules In an ideal gas, internal energy is proportional to the absolute (Kelvin) temperature.

 

Heat     ... is the amount of internal energy entering or leaving a system ... occurs by conduction, convection, or radiation. ... causes a substance's temperature to change ... is not the same as the internal energy of a substance   ... is positive if thermal energy flows into the substance ... is negative if thermal energy flows out of the substance ... is measured in joules -------------------------------------------------------------------------------------    improper:  "heat flow" is redundant

 

Thermal Equilibrium:    Systems (or objects) are said to be in thermal equilibrium  if there is no net flow of thermal energy from one to the other.  A thermometer is in thermal equilibrium with the medium  whose temperature it measures, for example.  If two objects are in thermal equilibrium, they are at the  same temperature.  Are you now in thermal equilibrium with your environment?

  Positive and Negative Work   (Recall:  Work = Fcos q)

W is positive if work is done by system. Air does work on the environment:   W > 0.

W is negative if work is done on the system.  Environment (man) does work on system:  W < 0  (Alternative:  system does negative work because   force by air pressure on thumb is opposite to the   direction of motion of the thumb.)

  The First Law of Thermodynamics
  (Conservation of Energy)

"Energy can neither be created nor destroyed, but only transferred from one system to another and transformed from one form to another." or "The internal energy of an isolated system is constant (even though that energy may be transformed from one type to another)." For thermodynamic systems, the 1st Law is: DU = Q - W

   Types of Thermodynamic
   Processes

Greek    isos:              equal               baros:           weight               adiabatos:    not passable -----------------------------------------------------      Isothermal:   Same temperature      Isobaric:       Same pressure      Isochoric:     Same volume      Adiabatic:     Zero heat flow (Q = 0)

   First Law Example

Example:  1000 J of thermal energy flows into a system (Q = 1000 J).  At the same time, 400 J of work is done by the system (W = 400 J). What is the change in the system's internal energy U? ---------------------------------------------------------- Solution: DU = Q - W       = 1000 J - 400 J       = 600 J

   First Law Example

Example:  800 J of work is done on a system (W = -800 J) as 500 J of thermal energy is removed from the system (Q = -500 J). What is the change in the system's internal energy U? -------------------------------------------------------------- Solution: DU = Q - W        = -500 J - (-800 J)        = -500 J + 800 J        =   300 J

  Work Done by an Expanding Gas

W   =  Fs = (PA) s = PDV  DV =  Vf - Vi W   =  P (Vf - Vi)

Area under pressure-volume curve is     the work done   ----------------------------------------------------   Isobaric Process: "same pressure"         Greek:  barys, heavy

    Work and the Pressure-Volume Curve            (Related to Problems 10 and 14)

Work Done = Area Under PV curve ------------------------------------- How much work is done by the system when the system is taken from: (a)  A to B  (900 J) (b)  B to C  (0 J) (c)  C to A  (-1500 J) ------------------------------------- Each "rectangle" has an area of 100 Pa-m3 = 100 (N/m2)-m3                   = 100 N-m                   = 100 Joules

   Expanding Gas     (First Law)

Example:   If a gas expands at a constant   pressure, the work done by the gas is:     W = Fs = (PA)s = P(As) = PDV

10 grams of steam at 100 C at constant  pressure rises to 110 C:  P   = 4 x 105 Pa                  DT = 10 C   DV = 30.0 x 10-6 m3        c = 0.48 cal/g What is the change in internal energy?   DU = Q - W ------------------------------------------------------ W = (4 x 105)(30.0 x 10-6) = 12 J Q = mcDT = (10)(.48)(10) = 48 cal     = (48 cal)(4.186 J/cal) = 201 J  DU = Q - W = 201 J - 12 J = 189 J ------------------------------------------------------ What if the steam were compressed while it was absorbing heat?

    First Law Example

Example: Aluminum cube of side L is heated in a chamber at atmospheric pressure.  What is the change in the cube's internal energy?  DU = Q - W Go to Solution

Key Ideas: 1.  DU = Q - W 2.  Q = mcDT 3.  m = rV0 4.  V0 = L3 5.  W = PDV 6.  DV = b   V0DT

   First Law Example

DU = Q - W       Q is zero.    W = -0.03 J DU = 0 - (-0.03 J) =  0.03 J (positive)                              Assuming ideal gas:  U = (3 / 2) N k T DT is positive since DU is positive

DU = Q - W       Q is zero W is positive    DU is negative Temperature drops from 41F to - 31 F causing vapor to condense into a cloud of tiny droplets.

  First Law Examples

DU = Q - W      Q is zero W is positive    DU is negative Steam temperature drops and is cool to the touch.

DU = Q - W       Q is zero.   W is negative Why does paper catch on fire?

   Work, Rubber Bands, and Internal Energy

DU = Q - W Expand rubber band:  W < 0, Q = 0 DU > 0     temperature increases --------------------------------------------------------- Press thick rubber band to lip and expand it rapidly.  The warming should be obvious. Now allow the band to contract quickly; cooling will also be evident.

  Heating a Gas at
  Constant Volume
    Isochoric Process   "same volume"

DU = Q - W = Q

    Conceptual Questions

1.  100 Joules of heat is added to a gas, and the gas expands at constant pressure.          Is it possible that the internal energy increases by 200 J?     No.  2.  A gas is compressed isothermally, and its internal energy increases.  Is the gas       an ideal gas?   No.   For an ideal gas, U = (3 / 2) N k T.  3.  State how U changes in each of the following processes (DU = Q - W)       (a)  W = -500 J    and    Q =  0         ................. DU = 0 - (- 500 J) = 500 J               (b)  W = 0            and    Q = -200 J  ................. DU = - 200 J - 0 = - 200 J       (c)  W =   100 J    and    Q =  100 J       (d)  W = -100 J    and    Q = -100 J       (e)  W =  300 J     and    Q =  500 J

   Adiabatic Expansion of a Ideal Gas    
     Key Idea #1:  For an ideal gas, U depends only on the absolute temperature T.
     Key Idea #2:  If there's nothing to push against, an expanding gas does zero work.

Gas in Chamber A suddenly rushes into Chamber B.  How does the final temperature of gas compare to its initial temperature? ---------------------------------------- DU = Q - W Q    = 0 W   = 0 DU = (3 / 2) N k DT = 0 DT = 0

Second Law of Thermodynamics Heat flows naturally from a region at high temperature to a region at low temperature.  By itself, heat will not flow from a cold to a hot body. When an isolated system undergoes a change, passing from one state to another, it will do so in such a way that its entropy (disorder) will increase, or at best remain the same. The 2d Law applied to creationism:   "The earth must have been created supernaturally, since it is a highly-ordered system."  Rebuttal:  The earth is not isolated;  the sun provides the energy to create the order.

 

Example:  Aluminum cube of side L is heated in a chamber at atmospheric pressure.  What is the change in the cube's internal energy? ---------------------------------------------------------------------------  DU = Q - W =  mcDT        - PDV                       =  ( rV0 ) cDT - P ( b   V0DT )                               = ( rc - Pb )   V0DT --------------------------------------------------------------------------- L = 20 cm       P  = 1.01 x 105 Pa       DT = 100 C r = 2.7 x 103 kg/m3         c = 0.90 x 103 J/kg-C b = 72 x 10-6 C-1 (coefficient of volume expansion)  DU = ( rc - Pb )   V0DT        = [( 2.7 x 103 )( .48 ) - (1.01 x 105) (72 x 10-6) ] (0.20)3 (100)