An unknown gas at 49.1 °C and 1.10 atm has a molar mass of 16.04 g/mol. Assuming ideal behavior, what is the density of the gas?

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Answer 1 · 2016-11-17T13:46:55

$P V = \frac{w}{M} \times R T$ $PV=w/MxxRT$

$\Rightarrow P M = \frac{w}{V} \times R T$ $=>PM=w/VxxRT$

$\Rightarrow P M = D \times R T$ $=>PM=DxxRT$

$\Rightarrow D = \frac{P M}{R T}$ $=>D=(PM)/(RT)$

Here

$D \to Density of the gas$ $D->"Density of the gas"$

$P \to Pressure of the gas = 1.1 a t m$ $P->"Pressure of the gas"=1.1atm$

$M \to Molar mass of gas = 16.04 g/ m o l$ $M->"Molar mass of gas"=16.04" g/"mol"$

$T \to Temperature of the gas = 49.1 + 273 = 322.1 K$ $T->"Temperature of the gas"=49.1+273=322.1K$

$R \to Ideal gas constant = 0.082 L a t m m o l^{- 1} K^{- 1}$ $R->"Ideal gas constant"=0.082Latmmol^-1K^-1$

So

$D = \frac{P M}{R T}$ $D=(PM)/(RT)$

$\Rightarrow D = \frac{1.1 \times 16.04}{0.082 \times 322.1} \frac{g}{L} = 0.67 \frac{g}{L}$ $=>D=(1.1xx16.04)/(0.082xx322.1)g/L=0.67g/L$