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Question
Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):-
the number of air molecules in your classroom.
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Solution 1
Let the volume of the room be V.
One mole of air at NTP occupies 22.4 l i.e., 22.4 × 10–3 m3 volume.
Number of molecules in one mole = 6.023 × 1023
∴Number of molecules in room of volume V
=`(6.023xx10^(23))/(22.4xx10^(-3))xxV`
= 134.915 × 1026 V
= 1.35 × 1028 V
Solution 2
We can determine the volume of the class-room by measuring its length, breadth and height. Consider a class room of size 10 m x 8 m x 4 m.
Volume of this room is 320 m3.
We know that 22.4l or 22.4 x 10-3 m3 of air has 6.02 x 1023 molecules (equal to Avogadro’s number).
Number of molecules of air in the class room =(6.02 x 1023 /22.4 x 10-3 ) x 320 =8.6 x 1027
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Larger unit for measuring time is ______.
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