Find the number of molecules in $1 c m^{3}$ of an ideal gas at $0^{0} C$ and at a pressure of $10^{- 5} m m$ of mercury.

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Solution

$V = 1 c m^{3} = 10^{- 3} m^{3}$

$T = 0^{0} C = 273 K$

$P = 10^{- 5} m m H g$

$= 133.32 \times 10^{- 5} N / m^{2}$

$R = 8.314 J / m o l . k$

$P V = n R T$ (ideal gas eq.)

$\Rightarrow 133.32 \times 10^{- 5} \times 10^{- 6} = n \times 8.314 \times 273$

$\Rightarrow n = 0.0056 \times 10^{- 11} m o l$

$1 m o l = 6.023 \times 10^{23}$ molecules

$∴ 6.023 \times 10^{23} \times 0.056 \times 10^{- 11}$

$= 3.37 \times 10^{11}$ molecules.

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Find the number of molecules in $1 c m^{3}$ of an ideal gas at $0^{0} C$ and at a pressure of $10^{- 5} m m$ of mercury.