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Q2.22(e):
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.

Solution :

Step 1: Establishing Classroom Dimensions

To estimate the number of air molecules in a classroom, we must first approximate the volume of the space. Assuming a standard classroom of dimensions:

  • Length ($l$) = $10\text{ m}$
  • Width ($w$) = $8\text{ m}$
  • Height ($h$) = $3\text{ m}$

The volume of the room ($V_{room}$) is calculated as: $V_{room} = l \times w \times h = 10 \times 8 \times 3 = 240\text{ m}^3$ [Per the definition of rectangular volume].

Step 2: Determining Molecular Density via Avogadro’s Hypothesis

Air behaves as an ideal gas under Standard Temperature and Pressure (STP) conditions. According to [Avogadro’s Law], one mole of any ideal gas at STP occupies a volume of $22.4\text{ liters}$ ($22.4 \times 10^{-3}\text{ m}^3$).

The number of moles ($n$) in the classroom is given by the ratio of the total volume to the molar volume:

$n = \frac{V_{room}}{V_{molar}} = \frac{240\text{ m}^3}{22.4 \times 10^{-3}\text{ m}^3/\text{mol}} \approx 1.07 \times 10^4\text{ mol}$

Step 3: Calculating Total Molecules

To find the absolute number of molecules ($N$), we multiply the number of moles by [Avogadro’s Constant] ($N_A = 6.022 \times 10^{23}\text{ molecules/mol}$):

$N = n \times N_A$ $N = (1.07 \times 10^4\text{ mol}) \times (6.022 \times 10^{23}\text{ molecules/mol})$ $N \approx 6.44 \times 10^{27}\text{ molecules}$ V = 240 m³ Estimation: ~ 10²⁷ - 10²⁸ molecules

Step 4: Analytical Justification and Order of Magnitude

While the temperature and pressure in a classroom vary slightly from STP, the order of magnitude remains robust. Considering atmospheric fluctuations, the volume occupied per molecule ($v_m$) at standard conditions is approximately $3.7 \times 10^{-26}\text{ m}^3$. Dividing the volume of the room by the volume per molecule confirms the magnitude:

$N = \frac{240}{3.7 \times 10^{-26}} \approx 6.5 \times 10^{27}$

Final Solution: The number of air molecules in a typical classroom of $240\text{ m}^3$ is estimated to be on the order of $10^{27}$ to $10^{28}$ molecules.


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