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Physics Paper 2, MAy/June. 2009  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main

General Comments

Question 12

(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)        Diagram:

The graph shown above is that of the saturated vapour pressure (s.v.p.) of water against temperature.
Pure water is known to boil at 100oC 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 ]


Generally, candidates’ responses fell short of expectation in part (a) but performance was very high in parts (b) & (d).  Majority of candidates did not answer part (c) of the question.  The few that gave responses to it did more of guess work than state the required general conclusion.

The expected answers were as follows:

(a)    A clinical thermometer has small temperature range. The glass will crack/burst due to excessive pressure created by expansion of mercury.  
(b)     Increase in pressure
          (i)   increases the boiling point;                                                            
          (ii)  decreases the melting point.                                                                      

  1. At the boiling point of pure water, the saturated vapour pressure (s.v.p)

of the water is equal to the external atmospheric pressure.         

(d)    Diagram

Volume V is proportional to length Ɩ stated or implied                                             

(i)         P1V1 = P2 V2    implies   P1  Ɩ1 = P2Ɩ2                                                  
(76 + 20) 15  = 76 x Ɩ2                                                                                               

            Ɩ2  =  96 x 15
            Ɩ2   =  18.95 cm                                                                                               

(ii)        P1  Ɩ1 = P3 Ɩ3                                                                                                    
(76 + 20) 15 = (76 – 20) x Ɩ3                                                                          

            Ɩ3  =  96 x 15            
Ɩ3  =  25.7l cm                                                                                   

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