(b) The electrical resistances of the element in a platinum restistance thermometer at 100\(^o\)C, 0\(^o\) and room temperature are 75.000, 63.000 and 64.992 \(\Omega\) respectively. Use these data to determine the room temperature.
(ii) A uniform capillary tube, closed at one end contained dry air trapped by a thread of mecury 8.5 x 10\(^{-2}\)m long. When the tube was held horinzontally, the length of the air column was 5.0 x 10\(^{-2}\)m, when it was held vertically with the closed end downwards, the length was 4.5 x 10\(^{-2}\)m, Determine the value of the atmospheric pressure. [g = 10ms\(^{-2}\), density of mecury = 1.36 x 10\(^4\) kg m\(^{-3}\)]
(a) Upper fixed point: the temperature of pure boiling water (steam) at standard atmospheric pressure, taken as \(100^{\circ}\text{C}\). Lower fixed point: the temperature of pure melting ice at standard atmospheric pressure, taken as \(0^{\circ}\text{C}\).
(b) For a platinum resistance thermometer, temperature on the resistance scale is:
\[ \theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100 \]
With \(R_0 = 63.000\ \Omega\), \(R_{100} = 75.000\ \Omega\), \(R_\theta = 64.992\ \Omega\):
\[ \theta = \frac{64.992 - 63.000}{75.000 - 63.000} \times 100 = \frac{1.992}{12.000} \times 100 \]
\[ \theta = 16.6^{\circ}\text{C} \]
The room temperature is \(16.6^{\circ}\text{C}\).
(c)(i) Boyle's law: the volume of a fixed mass of gas is inversely proportional to its pressure, provided the temperature is kept constant; i.e. \(PV = \text{constant}\).
(c)(ii) Let the atmospheric pressure be \(H\) (expressed as a length of mercury). The mercury thread is \(8.5\ \text{cm}\) long.
Horizontal: the mercury adds nothing, so the air pressure \(= H\) and the length is \(5.0\ \text{cm}\).
Vertical, closed end downwards: the mercury column presses down on the trapped air, so its pressure \(= H + 8.5\), and the length is \(4.5\ \text{cm}\).
Applying Boyle's law (constant cross-section, so length replaces volume):
\[ H \times 5.0 = (H + 8.5) \times 4.5 \]
\[ 5.0H = 4.5H + 38.25 \Rightarrow 0.5H = 38.25 \Rightarrow H = 76.5\ \text{cm of mercury} \]
Converting to pascals with \(\rho = 1.36 \times 10^{4}\ \text{kg m}^{-3}\), \(g = 10\ \text{ms}^{-2}\), \(H = 0.765\ \text{m}\):
\[ P = \rho g H = 1.36 \times 10^{4} \times 10 \times 0.765 = 1.04 \times 10^{5}\ \text{Pa} \]
The atmospheric pressure is \(76.5\ \text{cmHg} \approx 1.04 \times 10^{5}\ \text{Pa}\).
