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Kinetic Theory

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Question : 6 of 14
Marks: +1, -0
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m325.0\ \mathrm{m}^3 at a temperature of 27°C and 1 atm pressure.
Solution:  
Here, volume of air molecules, V=25.0 m3V = 25.0\ \mathrm{m}^3
Temperature of air molecules, T=27∘C=27+273=300 KT = 27^{\circ}\mathrm{C} = 27 + 273 = 300\ \mathrm{K}
Pressure of air molecules, P=1 atm=1.013×105 PaP = 1\ \mathrm{atm} = 1.013 \times 10^{5}\ \mathrm{Pa}
Now, PV=nRTPV = nRT
P=nRTVP = \frac{nRT}{V}
P=n(Nk)TVP = \frac{n(Nk)T}{V} [∵RN=k=1.38×10−23 J K−1=Boltzmann constant]\left[\because \frac{R}{N} = k = 1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}} = \text{Boltzmann constant}\right]
P=(nN)kTVP = \frac{(nN)kT}{V}
P=N′kTVP = \frac{N' k T}{V}[∵N′=nN=Total number of air molecules][\because N' = nN = \text{Total number of air molecules}]
N′=PVkTN' = \frac{PV}{kT} =(1.01×105)×25(1.38×10−23)×300= \frac{(1.01 \times 10^{5}) \times 25}{(1.38 \times 10^{-23}) \times 300} =6.10×1026= 6.10 \times 10^{26}
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