396 
Proceedings of the Royal Society of Edinburgh. [Sess. 
experimental condition has influenced Olszewski’s observations. This 
question is discussed in the present paper, in all cases on the presumption 
that there is no appreciable difference of kinetic energy in the initial and 
the final states of the gas. 
3. Under the latter condition it is not possible for Van der Waal’s 
equation to give a positive value of dt/dp for any suitable values of v and 
v'. For we have 
R— 
dn 
which is always positive ; and 
dp 
dv 
1 
( 1 , 
1 M 
— - 
— 
b 
V v 
V / J 
which is always negative so long as v and v' exceed 2b. 
4. The modified form of Van der Waal’s equation given by Clausius is 
/m _ Ri a 
P ^v^b~t{v + a) 2 ' 
From this, along with (1) and the condition Q = 0, we obtain the expression 
t 2 = — ( v ~ ~ fy{3 vv ' + 2a(y + v) + a 2 } 
R6 ( v + a) 2 (v + a) 2 ’ 
where t is the inversion temperature. Hence 
9 j_dt _ a , . _ jdv + a)[(v - b)a + (a+ b)v] + (2v + a)(a + b)(v + a) 
dv R& (v + a f(v + a) 2 
so that dt/dv is essentially positive. 
We have also 
dp _ R / t _ a / 2 + 1 d/\ 
dv v-b\v-b dv) t(v + a) 2 \v + a t dv) 
The first two terms on the right-hand side of the equation determine the 
sign of dp/dv. Evaluation of them gives 
b v — b 
a v—b 
dp_ 1 
dv 2(v + a ) 3 (F + a) 2 , 
(v + a)(3vv' + 2aV + 2av + a 2 ) + (v - b)( 3vv + av + 2a^)J , 
so that dp/dv is essentially negative, provided that the two terms involving 
a do not affect this result, and it is easily shown that the ratio of their sum 
to the sum of the first two terms is less than unity under conditions of 
volume and temperature such as those used by Olszewski. 
Thus Clausius’ modification of Van der Waal’s formula also makes dt/dp 
negative under these conditions when we can neglect the difference of the 
initial and the final kinetic energies of the gas. 
