A cylinder of argon gas contains 50.0 L of $Ar$ at 18.4 atm and 127 °C. How many moles of argon are in the cylinder?

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Answer 1 · 2016-02-19T20:21:15

The cylinder contains $"28.0 mol Ar"$ .

$PV=nRT$ , where $n$ is moles and $R$ is the gas constant.

Known
$P="18.4 atm"$
$V="50.0 L"$
$T=127^@"C+273.15=400 K"$
$R="L atm K"^(-1) "mol"^(-1)$
https://en.m.wikipedia.org/wiki/Gas_constant

Unknown
$n$

Solution
Rearrange the equation to isolate $n$ . Substitute the known values into the equation and solve for $n$ .

$PV=nRT$

$n=(PV)/(RT)$

$n=(18.4cancel"atm" * 50.0cancel"L")/(0.082057338 cancel"L" * cancel"atm" * cancel"K"^(-1) * "mol"^(-1) * 400"K")="28.0 mol Ar"$

A cylinder of argon gas contains 50.0 L of ArAr at 18.4 atm and 127 °C. How many moles of argon are in the cylinder?