Question #1cd39

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Question #1cd39

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Answer 1 · 2015-03-15T13:23:06

You have approximately $1.02 \cdot 10^{27}$ $1.02 * 10^(27)$ molecules of air in that volume under thos specific conditions.

In order to determine how many molecule of air you have, you must determine how many moles you have in the room. Before calculating the number of moles you have, convert the volume from cubic feet to liters

$V_{room} = 12.0 \cdot 12.0 \cdot 10.0 = {1440 ft}^{3}$ $V_("room") = 12.0 * 12.0 * 10.0 = "1440 ft"^(3)$

${1440 ft}^{3} \cdot \frac{28.2 L}{{1 ft}^{3}} = 40,608 L$ $"1440 ft"^(3) * "28.2 L"/"1 ft"^(3) = "40,608 L"$

Now just use the ideal gas law equation to solve for the number of moles

$P V = n R T \Rightarrow n = \frac{P V}{R T}$ $PV = nRT => n = (PV)/(RT)$

$n = \frac{1.00 atm \cdot 40608 L}{0.082 \frac{atm \cdot L}{mol \cdot K} \cdot (273.15 + 20) K} = 1689.2 moles of air$ $n = ("1.00 atm" * "40608 L")/(0.082("atm" * "L")/("mol" * "K") * (273.15 + 20)"K") = "1689.2 moles of air"$

Since 1 mole of a substance is defined as containing $6.022 \cdot 10^{23}$ $6.022 * 10^(23)$ molecules of that substance - this is known as Avogadro's number - the total number of molecules of air in the room will be

$1689.2 moles \cdot \frac{6.022 \cdot 10^{23} molecules}{1 mole} = 1.02 \cdot 10^{27} molecules$ $"1689.2 moles" * (6.022 * 10^(23)"molecules")/"1 mole" = 1.02 * 10^(27)"molecules"$

The answer is rounded to three sig figs.

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Question #1cd39

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