What is the volume, in liters, of 5.0 mol of Ne gas at STP?

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Answer 1 · 2018-03-24T00:24:17

$V_{N e}$ $V_(Ne)$ $=$ $=$ $22.41$ $22.41$ $L$ $L$

$We use the ideal gas law,$ $"We use the ideal gas law,"$

$P V = n R T$ $color(orange)(PV=nRT)$

$We know the conditions for STP are the following,$ $"We know the conditions for STP are the following,"$

$P = 101.325$ $P=101.325$ $k P a$ $kPa$

$V = ?$ $V=?$

$n = 1.0$ $n = 1.0$ $m o l$ $mol$

$R = 8.314$ $R=8.314$

$T = 273.15$ $T=273.15$ $K$ $K$

Plug in your values.

$V_{N e} = \frac{n R T}{P}$ $color(orange)(V_(Ne)=(nRT)/(P)$

$= \frac{(1.0 m o l) (8.314) (273.15)}{101.325}$ $=((1.0mol)(8.314)(273.15))/(101.325)$

$= \frac{2270.9691}{101.325}$ $=(2270.9691)/(101.325)$

$V_{N e}$ $V_(Ne)$ $=$ $=$ $22.41$ $22.41$ $L$ $L$

$This is true because the volume of an ideal gas is 22.41 L/mol at STP.$ $"This is true because the volume of an ideal gas is 22.41 L/mol at STP."$