How many moles of gas at 5.60 atm and 300 K will occupy a volume of 10.0 L?

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How many moles of gas at 5.60 atm and 300 K will occupy a volume of 10.0 L?

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Answer 1 · 2015-12-19T13:24:20

$2.28$ $2.28$ $mol$ $"mol"$

Explanation:

Use the ideal gas law equation

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

where

$P = pressure = 5.60$ $P="pressure"=5.60$ $atm$ $"atm"$
$V = volume = 10.0$ $V="volume"=10.0$ $L$ $"L"$
$n = moles$ $n="moles"$
$R = ideal gas constant = 0.082$ $R="ideal gas constant"=0.082$ $\frac{L atm}{K mol}$ $("L atm")/("K mol")$
$T = temperature = 300$ $T="temperature"=300$ $K$ $"K"$

Since we want to find the amount of moles, we can rearrange the equation by dividing by $R T$ $RT$ .

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

$n = \frac{5.60 \times 10.0}{.082 \times 300} = 2.28$ $n=(5.60xx10.0)/(.082xx300)=2.28$ $mol$ $"mol"$

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How many moles of gas at 5.60 atm and 300 K will occupy a volume of 10.0 L?

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