262 Temperature of Mercury in a Siphon Barometer. 
sad ose 
In like manner, if we compare the elements (a,, 6,, t,,) 
(A, By) ty) With (a, 6, t,) successively, we shall have 
2 R?2 . 
7a (%—4) = (On b)=«(—pty’] (t” —t,) (28) 
R?2 R2 
: Ban 4)—(0,,—8) =" =P +P’) (sh) (29) 
R2 
Eliminating «(=p +P’) from the last three equations, we shall 
have 
= (a, -a) —(b,—B) ; 
ee LOR CE OP ra . 
r 2 4#—* (30:) 
R?2 t’—t 
(4-4) (by—B) 
et, 
—(a,,—a)—(b,,-6),, 
r? / t’—zt 
R? = (31.) 
pe (Gun me a) == (Dy p-— b) 
Solving (31) for (=) we have 
R*  (#—1) (b, = b)—(#"—2) b,,— 
(#”=t) (by ~B)—(#"—t) by =8)_ fgg 
7 (01) a, ~a)= (0 =) (yy =@) 
The readings in (30) and (32) are those of the supposed inex- 
pansible scale. 'T'o change these equations into terms of the 
readings taken from the brass scale, after being corrected for the 
height of the meniscuses, we have 
from (3/) a, =a’ +¢(t/ —t)(a’ +f) 
: ts =a" fe’ (te! —t) (a” + 
Q,,,= al! Le (tl —t mt 
and from (3°) baw Lele <i Cy “p (33) 
b,, =f" +e’ t” —t) (b” —f 
ge 6,,,= 0" ’’—t) (b” —f) 
Substituting in (32,) and reducing, we have 
t +t # tes t 
R: x BT b+s (¢ ‘—t) (6 —b/”) — he" ois b) (34) 
a” — ae (t/ —t) (a a") (a""—2) 
Neglecting the third terms in the numerator and denominator, 
as they are nearly equal and very small, we have 
7? 
