What is the density of a dioxygen gas at 840*"mm Hg" pressure at a temperature of 291*K?

May 5, 2017

$\rho < 0.15 \cdot g \cdot m {L}^{-} 1$

Explanation:

$P V = n R T$; i.e. $P = \frac{n R T}{V} = \text{mass"/Vxx(RT)/"molar mass}$.

And thus $\rho = \text{mass"/V=P/(RT)xx"Molar mass}$.

The signal problem with this question is that you do not measure pressure OVER $1 \cdot a t m$ in $m m \cdot H g$. A mercury column is good for measuring pressures $\le 1 \cdot a t m$.

We know that $1 \cdot a t m \equiv 760 \cdot m m \cdot H g$, i.e. an atmosphere will support a column of mercury that is $760 \cdot m m$ high......A mercury column is thus suitable to measure atmospheric pressure, or for pressures LESS than one atmosphere.

And thus, here, $P = \frac{840 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} = 1.11 \cdot a t m .$

And so, $\rho = \frac{1.11 \cdot \cancel{a t m} \times 32.0 \cdot g \cdot \cancel{m o {l}^{-} 1}}{0.0821 \cdot L \cdot \cancel{a t m} \cdot \cancel{{K}^{-} 1} \cdot \cancel{m o {l}^{-} 1} \times 291 \cdot \cancel{K}}$

$\cong 1.5 \cdot g \cdot {L}^{-} 1. \ldots \ldots . .$