Find the number of molecules in a room of volume $27 m^{3}$ at temperature $27^{o} C$ and $1$ atm pressure $(k_{g} = 1.38 \times 10^{- 23} J K^{- 1})$

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Solution

Given : $T = 27 + 273 = 300 K$ $V = 27 m^{3}$ $P = 10^{5} P a$
Number of moles $n = \frac{P V}{R T} = \frac{10^{5} \times 27}{8.314 \times 300} = 1083$
Number of molecules $N = n N_{A} = 1083 \times 6 \times 10^{23} = 6.5 \times 10^{26}$

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Find the number of molecules in a room of volume 27m3 at temperature 27oC and 1atm pressure (kg=1.38×10−23JK−1)

Given : T=27+273=300 K V=27m3 P=105 PaNumber of moles n=PVRT=105×278.314×300=1083Number of molecules N=nNA=1083×6×1023=6.5×1026

Find the number of molecules in a room of volume $27 m^{3}$ at temperature $27^{o} C$ and $1$ atm pressure $(k_{g} = 1.38 \times 10^{- 23} J K^{- 1})$

Given : $T = 27 + 273 = 300 K$ $V = 27 m^{3}$ $P = 10^{5} P a$
Number of moles $n = \frac{P V}{R T} = \frac{10^{5} \times 27}{8.314 \times 300} = 1083$
Number of molecules $N = n N_{A} = 1083 \times 6 \times 10^{23} = 6.5 \times 10^{26}$