The bulb of a Callender's compensated constant pressure air...

In Callender's compensated constant pressure air thermometer, the main system and the compensating system have equal pressures and equal number of moles of air.

The equation governing the thermometer is given by:

$\dfrac{V}{T} + \dfrac{v_1}{T_0} = \dfrac{V}{T_0} + \dfrac{v_2}{T_0}$

where $V$ is the volume of the bulb, $T$ is the temperature of the vessel, $T_0$ is the temperature of the ice bath ($273 \text{ K}$), and $v_1 - v_2 = v$ is the volume of mercury poured out to maintain constant pressure.

Rearranging the equation, we get:

$\dfrac{V}{T} = \dfrac{V}{T_0} - \dfrac{v_1 - v_2}{T_0}$

$\dfrac{V}{T} = \dfrac{V - v}{T_0}$

\Rightarrow T = T_0 \left(\dfrac{V}{V - v} \right)

Given:
$V = 1800 \text{ cc}$
$v = 200 \text{ cc}$
$T_0 = 273 \text{ K}$

Substituting the values into the formula:

$T = 273 \times$\dfrac{1800}{1800 - 200}

$T = 273 \times$\dfrac{1800}{1600}

$T = 273 \times$\dfrac{9}{8}

$T = 307.125 \text{ K}$

Rounding to the nearest integer, we get $T \approx 307 \text{ K}$.