# What is the volume expressed by a 74.3*g mass of carbon dioxide at 294.9*K enclosed in a piston at a pressure of 439*"mm Hg"?

Mar 23, 2017

$V = \frac{n R T}{P} \cong 70 \cdot L$

#### Explanation:

The key to this question is to express the pressure, which is given in $\text{mm Hg}$, to atmospheres. We know (or should know) that $1 \cdot a t m$ will support a column of mercury that is $760 \cdot m m$ high. Mercury columns (which are fast disappearing in laboratories these days) provides a highly visual and accessible measurement of pressures BELOW atmospheric.

And thus $P = \left(\text{439 mm Hg")/("760 mm Hg} \cdot a t {m}^{-} 1\right) = 0.578 \cdot a t m$

And thus $V = \frac{\frac{74.3 \cdot g}{44.0 \cdot g \cdot m o {l}^{-} 1} \times 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 294.9 K}{0.578 \cdot a t m}$

$\cong 70 \cdot L$

Do the units in the calculation cancel out to give an answer in $L$? If not they should!