﻿the 
  Absolute 
  Scale 
  of 
  Temperature. 
  365 
  

  

  be 
  heated 
  at 
  constant 
  volume 
  through 
  the 
  range 
  St 
  {t 
  being 
  

   the 
  temperature 
  on 
  the 
  constant 
  volume 
  or 
  " 
  normal 
  " 
  scale). 
  

   The 
  heat 
  required 
  is 
  K 
  v 
  &. 
  But 
  let 
  it 
  be 
  heated 
  through 
  the 
  

   same 
  range 
  at 
  constant 
  pressure, 
  and 
  it 
  will 
  expand 
  by 
  an 
  

   amount 
  

  

  8v=(d 
  p 
  vfdt)Bt. 
  

  

  It 
  will 
  consequently 
  do 
  an 
  amount 
  of 
  external 
  work 
  in 
  the. 
  

   expansion 
  represented 
  by 
  

  

  p(d 
  P 
  vj'dt)St. 
  

  

  Further, 
  internal 
  work 
  is 
  done 
  by 
  the 
  gas 
  (t. 
  e, 
  its 
  internal 
  

   energy 
  is 
  increased) 
  by 
  an 
  amount 
  which 
  is 
  a 
  function 
  of 
  the 
  

   initial 
  and 
  final 
  states 
  of 
  the 
  gas, 
  and 
  which 
  therefore 
  we 
  may 
  

   find 
  by 
  any 
  method 
  in 
  which 
  the 
  same 
  expansion 
  is 
  consi- 
  

   dered, 
  whatever 
  the 
  external 
  circumstances 
  may 
  be. 
  We 
  

   proceed 
  to 
  find 
  it 
  in 
  the 
  following 
  way. 
  

  

  Lemma. 
  — 
  Let 
  one 
  gram 
  of 
  hydrogen 
  in 
  the 
  state 
  specified 
  

   by 
  p, 
  v, 
  T 
  expand 
  by 
  an 
  amount 
  Sv 
  without 
  doing 
  external 
  

   work 
  and 
  without 
  acquiring 
  any 
  appreciable 
  kinetic 
  energy 
  (as 
  

   in 
  Joule 
  and 
  Thomson's 
  experiment). 
  In 
  this 
  case 
  it 
  is 
  well 
  

   known 
  that 
  the 
  relation 
  SIT 
  + 
  8(pv) 
  = 
  holds 
  ; 
  but 
  as 
  pu=RT 
  

   nearly, 
  we 
  may 
  put 
  8U 
  + 
  RST 
  = 
  0, 
  where 
  6T 
  is 
  the 
  rise 
  in 
  

   temperature 
  that 
  occurs 
  during 
  the 
  expansion. 
  Next 
  let 
  the 
  

   gas 
  be 
  cooled 
  at 
  constant 
  volume 
  to 
  its 
  original 
  temperature 
  ; 
  

   then 
  it 
  gives 
  out 
  K 
  V 
  8T 
  of 
  heat. 
  Hence 
  its 
  internal 
  energy 
  is 
  

   less 
  than 
  at 
  starting 
  by 
  (K 
  V 
  + 
  R)3T 
  = 
  K 
  P 
  6T 
  to 
  the 
  same 
  degree 
  

   of 
  approximation 
  as 
  before. 
  

  

  Now 
  according 
  to 
  Joule 
  and 
  Thomson, 
  ST= 
  — 
  e8p, 
  where 
  

   8p 
  is 
  the 
  change 
  (increase) 
  of 
  pressure 
  involved 
  in 
  the 
  free 
  

   expansion, 
  and 
  e 
  is 
  a 
  constant, 
  which 
  for 
  hydrogen 
  is 
  positive, 
  

   Hence 
  the 
  gain 
  of 
  internal 
  energy 
  is 
  4- 
  'K-peSp. 
  We 
  wish 
  to 
  

   express 
  this 
  in 
  terms 
  of 
  the 
  change 
  of 
  volume. 
  

  

  But 
  

  

  = 
  _ 
  L 
  b 
  p 
  + 
  5&T 
  

  

  P 
  P 
  

  

  v 
  ^ 
  R 
  « 
  

  

  = 
  — 
  — 
  Op 
  — 
  —€Op. 
  

  

  p 
  1 
  p 
  r 
  

  

  Hence 
  the 
  gain 
  of 
  internal 
  energy 
  may 
  be 
  written 
  

  

  = 
  — 
  Kpex 
  , 
  -p 
  St'. 
  

  

  v 
  + 
  Re 
  

  

  fldl. 
  Nay. 
  S. 
  5. 
  Vol. 
  45. 
  No. 
  275. 
  April 
  1898. 
  2 
  C 
  

  

  