# What is the pressure in mm*Hg if a 4.0*g mass of oxygen gas at 303*K was confined to a container whose volume of 3000*mL?

Jun 28, 2017

$P = 1.04 \cdot a t m \ldots \ldots .$

#### Explanation:

We use the Ideal Gas Equation, to give.....

$P = \frac{n R T}{V} = \frac{\frac{4.00 \cdot g}{32.00 \cdot g \cdot m o {l}^{-} 1} \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 303 \cdot K}{3000 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1}$

$= 1.04 \cdot a t m$

Now had the pressure been UNDER $1 \cdot a t m$ I could use the relationship $1 \cdot a t m \equiv 760 \cdot m m \cdot H g$; i.e. an atmosphere will support a column of mercury that is so high. You do not measure a pressure that is greater than $1 \cdot a t m$ in $m m \cdot H g$. Whoever set this problem was ignorant.........