EXPANSION OF GASES. 45 



pressure-increase when the volume is constant. The manometer was, 

 however, somewhat modified in its details, and the water-bath was 

 removed. As before, the bulb was filled with dry air through the 

 tube p, and it was first exposed to the boiling temperature. The 

 mercury being brought to the same level a in both tubes, p was 

 sealed. The temperature was now allowed to fall, and ultimately A was 

 surrounded by melting ice. The air tended to contract and draw the 

 mercury above A, but the pressure was diminished by allowing mercury 

 to run out at P', so that the level in M was maintained at a, while it fell 

 to y in M'. The barometric height was read for each of the two 

 temperatures. We will suppose it the same throughout. The tempera- 

 ture of the manometer is that of the surrounding air t, not being dis- 

 turbed by the introduction of hot air from the bulb. Let the volume of 

 the bulb at be V, and at the boiling-point T let it become V (1 + *T). 

 K may be measured by using the bulb as a mercury thermometer in a 

 preliminary experiment. Let the volume of the connecting-tube and 

 manometer down to the level a be v. Let the barometric height be H 

 and the difference of the levels ay be h. Let j3 be the co-efficient of 

 pressure-increase at constant- volume assumed to be constant, that is 

 to say we use the gas scale of temperature. Then, in the first part 

 of the experiment, we have a volume V(l +*T) of air at T and v at t, 

 all at a pressure H, which would become 



1 -4- aT 

 V(l + K!) +v-= - , if all were at the one temperature T.* 



1 T" & 



In the second part of the experiment, we have a volume V of air at 

 0, and v at t, all at pressure (H A), which would become 



V + , v , if all were at 0. 

 1 +at 



Now, these volumes are nearly equal, since K and v, are both small. 

 Then, without sensible error, we may use Boyle's Law to find what H 

 would become if the former volume were reduced to the latter. It would 

 obviously be 



and now we have the two pressures (H h) and H' with the same 

 volume. 



Hence H' = (1 + /3T)(H - h) 



, - orp H 



and 1 + pT = x 



H-fc 



l+at 



* We have omitted the correction for increase of volume through increase of 

 internal pressure as this is in general too small to come into account (Callendar, 

 Phil. Trans., A., 1887, p. 170). 



