# 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?

Nov 17, 2016

By ideal gas law we know

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

$\implies P M = \frac{w}{V} \times R T$

$\implies P M = D \times R T$

$\implies D = \frac{P M}{R T}$

Here

$D \to \text{Density of the gas}$

$P \to \text{Pressure of the gas} = 1.1 a t m$

$M \to \text{Molar mass of gas"=16.04" g/"mol}$

$T \to \text{Temperature of the gas} = 49.1 + 273 = 322.1 K$

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

So

$D = \frac{P M}{R T}$

$\implies D = \frac{1.1 \times 16.04}{0.082 \times 322.1} \frac{g}{L} = 0.67 \frac{g}{L}$