Question #73fe8

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Question #73fe8

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Answer 1 · 2015-11-26T06:00:02

The volume of the gas at STP is 7.58 L.

Explanation:

Use the ideal gas law, $P V = n R T$ $PV=nRT$ , where $n$ $n$ is moles and $R$ $R$ is the gas constant.

Convert molecules to moles by dividing by $\frac{6.022 \times 10^{23} molecules}{1 mol}$ $(6.022xx10^23"molecules")/(1"mol")$ .

$2.01 \times 10^{23} molecules \times \frac{1 mol}{6.022 \times 10^{23} molecules} = 0.33378 mol$ $2.01xx10^23"molecules"xx(1"mol")/(6.022xx10^23"molecules")="0.33378 mol"$

(I am keeping a couple of guard digits to reduce rounding errors.)

$STP$ $"STP"$ is $273.15 K$ $"273.15 K"$ and $100 kPa$ $"100 kPa"$ .

Given/Known
$P = 100 kPa$ $P="100 kPa"$
$n = 0.33378 mol$ $n="0.33378 mol"$
$R = {8.3144598 L kPa K}^{- 1} {mol}^{- 1}$ $R="8.3144598 L kPa K"^(-1) "mol"^(-1)"$
https://en.m.wikipedia.org/wiki/Gas_constant
$T = 273.15 K$ $T="273.15 K"$

Unknown
$V$ $V$

Equation
$P V = n R T$ $PV=nRT$

Solution
Rearrange the equation so that $V$ $V$ is isolated and solve.

$V = \frac{n R T}{P}$ $V=(nRT)/P$

$V=(0.33378cancel"mol"xx8.3144598 "L" cancel"kPa" cancel("K"^(-1)) cancel("mol"^(-1))xx273.15cancel"K")/(100 cancel"kPa")="7.58 L"$

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Question #73fe8

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