680  Mr.  E.  Buckingham  on 
unit  mass  of  the  gas.  Let  D  (coordinates:  p  +  dp,  v  +  Av, 
T  +  AT)  represent  the  state  after  the  irreversible  adiabatic 
flow  through  the  porous  plug.  Let  B  (coordinates  :  p  +  dp, 
v-i-dv,  T)  represent  the  state  the  gas  would  have  been  in,  if 
the  same  fall  of  pressure  had  occurred  without  change  of 
temperature.  The  temperature  T.  is  to  be  understood  as 
measured  on  the  scale  of  a  constant-pressure  thermometer 
filled  with  the  gas  under  investigation. 
§  3.  Consider  first  the  Joule-Thomson,  or  adiabatic,  plug 
experiment.  If  the  state  of  flow  is  steady  and  no  exchange 
of  heat  with  other  bodies  than  the  plug  is  taking  place,  the 
change  of  internal  energy  of  unit  mass  of  gas  in  passing 
through  the  plug  is 
or,  if  we  let  the  fall  of  pressure  be  infinitesimal; 
Ae=—  A(pv)  =  —  p>Av  —  vdp,  .  .  .  .  (1) 
Ap  and  dp  being  identical,  as  is  seen  from  the  figure.  Let 
the  gas  which  has  thus  passed  from  the  state  A  to  the  state  D 
be  now  brought,  at  constant  pressure,  to  the  state  B.  If  de 
be  the  excess  of  the  specific  internal  energy  at  B  over  that 
at  A,  we  now  have 
de=Ae+Q  +  W,        (2) 
in  which  Q  and  W  are,  respectively,  the  heat  added  to  and 
the  work  done  on  the  unit  mass  of  gas  during  the  isopiestic 
change  DB,  which  brings  it  back  from  (T  +  AT)  to  its 
original  temperature  T.  We  suppose,  for  simplicity,  that  all 
quantities  of  heat  are  expressed  in  ergs. 
If  we  let  fju  be  the  ratio  of  the  fall  of  temperature  to  the 
fall  of  pressure,  i.  e.  AT/dp,  we  have  AH=ndp,  and  the  value 
of  Q  is  evidently  * 
Q,-=—fjbCpdp, (3) 
•Cp  being  the  specific  heat  at  constant  pressure.  The  value  of 
W  is  given  by  the  equation 
W=-p(dv-Av) (4) 
We  also  have 
^-Al'=(-AT)(lX=-^(rr)/  •  •  ^ 
so  that 
Ar=A'+  (Jr)/'^ (6) 
*  The  figure  as  drawn  corresponds  to  the  ordinary  case  in  which  there 
is  a  fall  of  temperature  during  the  flow  through  the  plug.  The  sign  of  it,, 
as  here  defined,  is  then  positive,  while  dp  is  negative,  so  that  the  value 
of  Q,  as  given  in  (3)  is  in  fact  positive,  as  it  is  evident  from  the  figure 
that  it  must  be. 
