﻿366 
  Mr. 
  R. 
  A. 
  Lehfeldt 
  on 
  a 
  Numerical 
  Evaluation 
  of 
  

  

  But 
  v 
  + 
  He 
  = 
  v(l 
  +ju»e/T); 
  so 
  that, 
  as 
  e 
  is 
  very 
  small, 
  unless 
  

   the 
  pressure 
  be 
  much 
  greater 
  than 
  is 
  actually 
  used 
  in 
  gas 
  

   thermometers 
  pe 
  will 
  be 
  less 
  than 
  1 
  per 
  cent, 
  of 
  T 
  (for 
  

   T=273°, 
  say) 
  ; 
  and 
  as 
  e 
  is 
  not 
  known 
  to 
  1 
  per 
  cent., 
  we 
  may 
  

   safely 
  neglect 
  the 
  second 
  term 
  of 
  the 
  denominator 
  and 
  write 
  

   the 
  gain 
  of 
  internal 
  energy 
  = 
  — 
  K 
  F 
  epdv/v. 
  But 
  as 
  the 
  tem- 
  

   perature 
  of 
  the 
  gas 
  is 
  the 
  same 
  at 
  the 
  end 
  as 
  at 
  the 
  beginning 
  

   of 
  the 
  process, 
  its 
  heat 
  energy 
  is 
  unchanged, 
  and 
  the 
  quantity 
  

   found 
  is 
  the 
  amount 
  of 
  internal 
  work 
  done 
  by 
  the 
  gas 
  in 
  

   expansion. 
  

  

  Reverting 
  to 
  the 
  main 
  argument, 
  we 
  find 
  that 
  when 
  a 
  gram 
  

   of 
  hydrogen 
  is 
  heated 
  at 
  constant 
  pressure, 
  the 
  heat 
  required, 
  

   being 
  the 
  sum 
  of 
  the 
  increase 
  of 
  heat 
  energy, 
  the 
  external 
  

   work, 
  and 
  the 
  internal 
  work, 
  amounts 
  to 
  

  

  whence 
  

  

  K 
  

  

  It 
  is 
  in 
  terms 
  of 
  that 
  quantity 
  that 
  we 
  wish 
  to 
  find 
  a 
  relation 
  

   between 
  the 
  absolute 
  temperature 
  T 
  and 
  the 
  temperature 
  t 
  

   according 
  to 
  the 
  gas 
  scale. 
  The 
  relation 
  is 
  clearly 
  expressed 
  

   by 
  Boltzmann 
  in 
  the 
  paper 
  cited 
  ; 
  and 
  his 
  argument 
  is 
  sub- 
  

   stantially 
  the 
  same 
  as 
  the 
  following. 
  

  

  We 
  start 
  with 
  the 
  well-known 
  thermodynamic 
  relation 
  

  

  BT 
  BT 
  " 
  ^BTyJ 
  dp' 
  * 
  ' 
  " 
  {lU) 
  

  

  But 
  as 
  the 
  specific 
  heats 
  used 
  in 
  the 
  preceding 
  paragraphs 
  

   are 
  in 
  terms 
  of 
  the 
  normal 
  scale, 
  in 
  introducing 
  them 
  we 
  must 
  

   multiply 
  the 
  left-hand 
  side 
  of 
  the 
  above 
  equation 
  by 
  St, 
  and 
  

   the 
  right-hand 
  side 
  by 
  ST, 
  or 
  

  

  (K 
  P 
  -K 
  V 
  )^=-T^) 
  2 
  |^8T. 
  

  

  Now 
  write 
  djt)/c)T 
  as 
  (dpfdt) 
  (cWdT) 
  in 
  this 
  equation, 
  and 
  

  

  dT 
  

   remembering 
  that 
  ST 
  = 
  -=- 
  8t, 
  we 
  get 
  

  

  (K 
  P 
  -K 
  v)& 
  =-T(|f) 
  2 
  ^J 
  &; 
  

  

  or 
  

  

  ST 
  V 
  d* 
  ) 
  ^p 
  

  

  T 
  ~ 
  Kp-Kv 
  

  

  Sl 
  

  

  