A container of just 10cc has 2.69×1020 gas molecules at O0C what is the pressure Exerted _____

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A container of just 10cc has $2.69 \times 10^{20}$ gas molecules at $O^{0} C$ what is the pressure Exerted ______

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Detailed Solution

$\begin{array}{l} P V = n R T \\ P = \frac{n R T}{V} = \frac{(\frac{2.69 \times 10^{20}}{6.02 \times 10^{23}}) \times 0.081 \times 273}{\frac{10}{1000}} \\ P = \frac{2.69 \times 10^{- 3}}{6.02 \times 10} \times 0.821 \times 273 \times 1000 \\ = \frac{2.69 \times 0.0821 \times 273}{602} \\ = 1 a t m \end{array}$