﻿i?. 
  deSaussure 
  — 
  Graphical 
  Thermodynamics. 
  37 
  

  

  dTT 
  s 
  c?T_ 
  g> 
  

  

  <fo 
  ~~ 
  KE 
  an 
  ~^s 
  "" 
  KE 
  

  

  Hence 
  

  

  f 
  .. 
  1 
  / 
  dV 
  dV\ 
  

  

  << 
  „ 
  1 
  / 
  dP 
  cZP\ 
  

  

  These 
  formulae 
  give 
  the 
  value 
  of 
  A, 
  JS, 
  G 
  corresponding 
  to 
  

   any 
  physical 
  state 
  of 
  the 
  substance 
  defined 
  by 
  the 
  coordinates 
  

   c 
  and 
  s. 
  

  

  If 
  we 
  now 
  express 
  each 
  one 
  of 
  the 
  physical 
  coefficients 
  in 
  

   terms 
  of 
  A, 
  J5, 
  G 
  we 
  shall 
  be 
  able 
  to 
  compute 
  the 
  value 
  of 
  

   these 
  coefficients 
  for 
  any 
  state 
  of 
  the 
  substance, 
  by 
  means 
  of 
  

   equations 
  (8), 
  since 
  A, 
  jB, 
  G 
  are 
  depending 
  upon 
  equations 
  (8) 
  

   by 
  means 
  of 
  equations 
  (11). 
  

  

  When 
  the 
  thermodynamic 
  surface 
  is 
  given 
  under 
  the 
  ordi- 
  

   nary 
  form 
  T 
  — 
  f 
  (P, 
  V), 
  it 
  is 
  always 
  possible 
  to 
  introduce 
  two 
  

   auxiliary 
  variables, 
  by 
  putting 
  : 
  

  

  (P 
  = 
  u 
  

  

  I 
  T 
  =f(u, 
  v] 
  

  

  whence 
  : 
  ( 
  T 
  =/*(", 
  v) 
  

  

  Under 
  this 
  form, 
  the 
  area 
  de 
  of 
  the 
  elementary 
  parallelograms 
  

   cut 
  off 
  by 
  the 
  curves 
  u 
  = 
  constant 
  and 
  v 
  = 
  constant, 
  is 
  : 
  

  

  de 
  — 
  (A/ 
  + 
  B; 
  + 
  C^dudv 
  

   in 
  which 
  : 
  

  

  dT 
  _ 
  dT 
  _, 
  

  

  The 
  element 
  de 
  is 
  not 
  the 
  same 
  as 
  the 
  element 
  dco, 
  but 
  it 
  has 
  

   the 
  same 
  tangent 
  plane. 
  If 
  we 
  measure 
  off 
  on 
  the 
  normal 
  to 
  

   the 
  surface 
  a 
  length 
  : 
  

  

  X=(A 
  2 
  + 
  B 
  2 
  -fC 
  2 
  )^ 
  

  

  the 
  components 
  of 
  JV 
  parallel 
  to 
  the 
  axes 
  of 
  coordinates 
  shall 
  

   be 
  respectively 
  : 
  A. 
  B 
  and 
  C. 
  Now, 
  since 
  the 
  normal 
  and 
  the 
  

   axes 
  of 
  coordinates 
  are 
  the 
  same 
  for 
  both 
  elements 
  and 
  since 
  

   the 
  components 
  G 
  and 
  C\ 
  are 
  equal 
  (C 
  ==fc 
  1 
  = 
  CJ, 
  the 
  other 
  

   two 
  components 
  must 
  be 
  equal 
  to 
  each 
  other, 
  thus 
  : 
  

  

  dT 
  

  

  B 
  = 
  B 
  1=±rfy 
  

  

  In 
  other 
  words 
  

  

  