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

$2.28$ $"mol"$

Use the ideal gas law equation

$PV=nRT$

where

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

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

$n=(PV)/(RT)$

$n=(5.60xx10.0)/(.082xx300)=2.28$ $"mol"$