246 Mr Searle, The Expansion of a Gas, etc. 



From such statements a student would not unnaturally draw 

 the conclusion that, if the pressure at entry had exceeded that on 

 emergence by the pressure due to a few centimetres of water, the 

 heat absorbed from or given out to the calorimeter while one 

 gramme passed would have differed appreciably from the product 

 of the change of temperature and the mean specific heat. But 

 we have seen that, for a " perfect " gas the difference of pressure 

 is absolutely without effect upon the result and thus we are led 

 to expect that, for gases which are almost perfect, the effects of 

 differences of pressure will be practically negligible. 



The correction which is necessary in the case of an imperfect 

 gas can be at once written down, when the results of the Thomson- 

 Joule porous plug experiment for that gas are known. 



If H ergs of heat be absorbed from the calorimeter while one 

 gramme of gas passes through the spiral, we have 



pv-p'v' + H= TJ'- TJ (11), 



where p, v, TJ refer to the gas as it enters and p ', v', TJ' refer to the 

 gas as it leaves the spiral. Here we are only concerned with the 

 initial and final states of the gas and with the total amount of 

 heat absorbed from the calorimeter during the passage of one 

 gramme of gas through the spiral. We may therefore suppose 

 that the gas passes through a porous plug before it reaches the 

 calorimeter, and that there is no fall of pressure in the spiral 

 itself. If we further suppose that no heat enters or leaves the 

 gas as it passes through the porous plug, the heat absorbed from 

 the calorimeter is the same as in the actual experiment. The gas, 

 which enters the plug at temperature t, leaves it at a tempera- 

 ture t", the difference t" — t being calculable from the results of 

 the Thomson-Joule experiment in terms of the fall of pressure 

 p-p. 



If p, v" , TJ" refer to the gas as it leaves the porous plug, the 

 energy equation, corresponding to the passage of the gas through 

 the spiral at the constant pressure p', becomes 



p'v"-p'v' + H=U'-U". 



In finding TJ' — TJ" we may take the gas from one state to the 

 other by any path on the pv diagram, since TJ' — TJ" depends only 

 on the two states. The most convenient path is obviously the 

 one in which the pressure has the constant value p '. Then 

 TJ' — TJ" is equal to the heat absorbed by one gramme of gas 

 along this path minus the work done by it along the same path. 

 Hence, we have 



p'v" -p'v' + H= T C p ,dt-p' (v' -v"). 



it" 



