(a) Explain why it is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure. (b) State the effect of an increase...
Question 1 Report
(a) Explain why it is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure.
(b) State the effect of an increase in pressure on the (i) boiling point; and (ii) melting point of water.
(c) The graph below shows the saturated vapour pressure (s.v.p.) of water plotted against temperature.
Pure water is known to boil at 100°C and at an atmospheric pressure of 760 mmHg. What general conclusion can be drawn from the information given above?
(d) A thread of mercury of length 20 cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end. With the tube vertical and the open end uppermost, the length of the trapped air column is 15cm. Calculate the length of the air column when the tube is held: (i) horizontally; (ii) vertically with the open end underneath. [Atmospheric pressure = 76 cmHg]
Answer Details
(a) A clinical thermometer has a small temperature range, approximately \(35\,^{\circ}\mathrm{C}\) to \(43\,^{\circ}\mathrm{C}\). In boiling water, the mercury expands excessively and may produce enough pressure to crack or burst the glass thermometer.
(b)
- An increase in pressure increases the boiling point of water.
- An increase in pressure decreases the melting point of ice.
(c) A liquid boils when its saturated vapour pressure is equal to the external atmospheric pressure. Thus, at \(100\,^{\circ}\mathrm{C}\), the saturated vapour pressure of pure water is \(760\ \mathrm{mmHg}\).
(d) The arrangements of the trapped air and mercury thread are represented below.
Since the capillary tube has uniform cross-sectional area, volume is proportional to length. Hence, Boyle's law may be written as:
\[P_1L_1=P_2L_2\]
Initially, with the open end uppermost, the pressure of the trapped air is
\[P_1=76+20=96\ \mathrm{cmHg}\]
and \(L_1=15\ \mathrm{cm}\). Therefore,
\[P_1L_1=96\times15=1440\ \mathrm{cmHg\,cm}.\]
(i) Tube held horizontally
The pressure of the trapped air is atmospheric pressure:
\[P_2=76\ \mathrm{cmHg}\]
\[96\times15=76L_2\]
\[L_2=\frac{1440}{76}=18.95\ \mathrm{cm}\approx\boxed{18.9\ \mathrm{cm}}\]
(ii) Tube held vertically with the open end underneath
The pressure of the trapped air is
\[P_3=76-20=56\ \mathrm{cmHg}\]
\[96\times15=56L_3\]
\[L_3=\frac{1440}{56}=25.7\ \mathrm{cm}\]
\[\boxed{L_3=25.7\ \mathrm{cm}}\]
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