Lecture 5

Thermal  Physics   Lecture  5  –  Adiaba6c  Processes   Textbook  reference:  21.3,  20.4,  22.5  (not  examinable)  

Project  Orion  –  nuclear-­‐powered  spacecraK,  because  AMERICA!!!   Popularised  in  FooSall  (Larry  Niven  &  Jerry  Pournelle),  and  Ark  (Stephen  Baxter)  

Last  6me  in  Physics…   W =−

�

Vf

P dV Vi

• Work  done  on  a  gas:  nega6ve  for  expansions  and     posi6ve  for  contrac6ons  (think  pistons).   • Molar  specific  heats  for  ideal  gases.   • The  heat  added  to  a  system  to  get  from  state   with  Vi,  Pi,  Ti  to  a  state  with  Vf,  Pf,  Tf  changes,     CV depending  on  the  path  (intermediate  states).  

f = R 2

CP − CV = R

Ques6ons,  comments,  complaints?   Dr  Rob  Wibenmyer   Office:  Old  Main  130,  in  Astrophysics  area   [email protected]   Visitors  welcome!      It’s  good  to  feel  needed  

Warm-­‐up  Ques6on   The  temperature  of  2.00  mol  of  an  ideal   monatomic  gas  is  raised  15.0  K  at  constant   volume.  What  are:   (a)  the  work,  W,  done  by  the  gas?   (b)  the  energy  transferred  as  heat,  Q   (c)  the  change,  ∆E                int          ,  in  the  internal  energy   of  the  gas   (d)  the  change,  ∆K                ,  in  the  average  kine6c   energy  per  atom?    

Adiaba6c  Process   An  adiaba6c  process  is  one  in  which  no  energy   enters  or  leaves  a  system  as  heat,  Q.  

Q=0

∆Eint = W

Adiaba6c  Process   Two  ways  to  mimimise  heat  transfer  are:   •  Insulate  the  walls  of  the  container   •  Perform  the  process  quickly  

Examples:  Adiaba6c  Free  Expansion  

In  this  case  W=0  as  well.  No  force  is  being   applied.   What  does  this  tell  us  about  the  temperature?  

Adiaba6c  Free  Expansion   •  Break  the  membrane  so  the  gas   expands  to  fill  the  vacuum.    An   example  of  adiaba6c  free  expansion.   •  The  process  is  adiaba6c  because  it   takes  place  in  an  insulated  container.   •  No  work  is  done,  so  W  =  0   •  Since  Q  =  0  and  W  =  0,  ΔEint  =  0  and   the  ini6al  and  final  energies  are  the   same   –  Hence  no  change  in  temperature   occurs.  

Adiaba6c  Processes  and  Temperature   •  Since  Q  =  0,  ΔEint  =  Q+W  =  W   •  If  the  gas  expands  adiaba6cally,  the  temperature  of  the   gas  decreases.  

Adiaba6c  Processes  and  Temperature   •  If  the  gas  is  compressed  adiaba6cally,  W  is  posi6ve  so   ΔEint  is  posi6ve  and  the  temperature  of  the  gas   increases.   •  This  is  why  spacecraK  have  heat  shields.  NOT  because   of  “fric6on  with  the  air.”    The  air  cannot  get  out  of  the   way  fast  enough    rapid  compression  and  hea6ng.  

D  

Demo  Unit  Hb13:  adiaba6c   expansion   •  As  the  gas  expands  rapidly  it  cools,  leading  to   condensa6on  and  cloud  forma6on.  

Homework:  Have  a  beer   •  Pressure  released  when   you  open  the  boble.   •  This  is  an  adiaba6c   expansion!   •  Rapid  cooling    vapor   forms   •  Disclaimer:  If  you  are   under  18,  do  not  have  a   beer.  Beer  is  bad,   mmkay  

 Gasoline  Engine   See active figure 22.11

A  gasoline  engine  converts  chemical  poten6al  energy  to   work.   Note:  this  bit  not  examinable.    See  “Real  Engines”  sec6on  22.5  in  book.  

Rela6onship  between  P  and  V   During  an  adiaba6c  process  P,  V  and  T  all   change.  

∆Eint = −P ∆V ∆Eint = nCV ∆T −P dV = nCV dT

Rela6onship  between  P  and  V  

Rela6onship  between  P  and  V   P V = nRT P and V are functions of T Take the derivative, use the product rule dV dP P +V = nR dT dT P dV + V dP = nRdT

Rela6onship  between  P  and  V   P dV + V dP = nRdT −P dV = nCV dT −P dV P dV + V dP = nR × nCV R P dV + V dP = − P dV CV

Rela6onship  between  P  and  V   R P dV + V dP = − P dV CV Substitute R = CP − CV and divide by P V dV dP CP − CV dV + = −( ) V P CV V dV dP dV + = (1 − γ) V P V

Rela6onship  between  P  and  V   dV dP dV + = (1 − γ) V P V dP dV = −γ P V � � dP dV = −γ P V ln P + γ ln V = constant

Rela6onship  between  P  and  V   ln P + γ ln V = constant ln P V γ = constant ⇒ PV

γ

= constant

Let’s  take  a  short  break  

Bombardment  of  defini6ons! Isothermal   An  isothermal  process  is  one  carried  out  at   constant  temperature.  For  an  isothermal   process:  

∆Eint = 0 And  

P V = constant Any  energy  that  enters  the  system  by  heat  must   leave  the  system  by  work.  

Isobaric   A  process  carried  out  at  constant  pressure.   Heat  and  work  are  generally  not  zero.  

T = a constant V and  

Q = nCP ∆T

Isovolumetric   A  process  that  takes  place  at  a  constant  volume   is  called  an  isovolumetric    process.  When  the   volume  does  not  change  no  work  is  done.  

W =0 ∆Eint = Q and  

Q = nCV ∆T

ΔEint=Q+W  

In  the  empty  columns  of  the  table,   fill  in  the  correct  signs  (–,  +,  0)  for  Q ,   W  and  ΔEint.  

Situation

System

(a) Rapidly pumping up a bicycle tyre

Air in the pump

(b) Pan of room temperature water sitting on a hot stove

Water in the pan

(c) Air leaking quickly out of a balloon

Air originally in the balloon

Q

W

ΔEint

ΔEint=Q+W  

In  the  empty  columns  of  the  table,   fill  in  the  correct  signs  (–,  +,  0)  for  Q ,   W  and  ΔEint.  

Situation

System

Q

W

ΔEint

(a) Rapidly pumping up a bicycle tyre

Air in the pump

0

+

+

(b) Pan of room temperature water sitting on a hot stove

Water in the pan

+

0

+

(c) Air leaking quickly out of a balloon

Air originally in the balloon

0

–

–

Special  Processes,  Summary    ΔEint  =  Q  +  W   •  Adiaba6c  

–  No  heat  exchanged   –  Q  =  0  and  ΔEint  =  W   –  PVγ  =  constant  

•  Isovolumetric  

–  Constant  volume   –  W=0  so  ΔEint  =  Q  

•  Isobaric  

–  Constant  pressure   –  W  =  P  (Vf  –  Vi)  and  ΔEint  =  Q  +  W  

•  Isothermal  

–  Constant  temperature   –  ΔEint  =  0  and  Q  =  -­‐W   –  PV  =  constant  

Name  each  of  the  paths:  A,  B,  C  and  D   on  the  PV  diagram  below.  

A.  B.  C.  D. 

Isovolumetric   Adiaba6c   Isothermal   Isobaric  

Ques6on   An  ideal  gas  ini6ally  at  300  K  undergoes  an   isobaric  expansion  at  2.50  kPa.  If  the  volume   increases  from  1.00  m3  to  3.00  m3  and  12.5  kJ   is  transferred  to  the  gas  by  heat,  what  are   a)  the  change  in  internal  energy   b)  its  final  temperature  

Ques6on   The  figure  on  the  slide  aKer  this  ques6on  shows   a  cycle  undergone  by  1.00  mol  of  an  ideal   monatomic  gas.  The  temperatures  are  T1  =   300  K,  T2  =  600  K,  and  T3  =  455  K.     For  12,  what  are   a)  Heat,  Q     b)  Change  in  internal  energy   c)  the  work  done,  W?  

Ques6on  cont.   For  23  what  are   d)  Q   e)  change  in  internal  energy   f)  W?   For  31  what  are   g)  Q   h)  change  in  internal  energy   i)  W?  

Ques6on  cont.  

adiaba6c  

Pressure  

Volume  

Another  Ques6on   An  ideal  diatomic  gas,  with  rota6on  but  no   vibra6ons  undergoes  an  adiaba6c   compression.  Its  ini6al  pressure  and  volume   are  1.20  atm  and  0.200  m3.  Its  final  pressure  is   2.40  atm.  How  much  work  is  done  by  the  gas?