# How do you find the number of grams of CO_2 that exert a pressure of 785 mmHg at a volume of 325 L and a temperature of 32°C?

Nov 30, 2016

Approx. $600 \cdot g$ of carbon dioxide gas.

#### Explanation:

We know (or should know) that $n = \frac{P V}{R T}$, and further that $1 \cdot a t m \equiv 760 \cdot m m \cdot H g$.

And thus $n = \frac{\frac{785 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} \times 325 \cdot L}{0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 305 \cdot K} \cong 13 \cdot m o l$.

And $13 \cdot m o l \times 44.0 \cdot g \cdot m o {l}^{-} 1 \cong 600 \cdot g$.

From where did I get the figure $44.0 \cdot g \cdot m o {l}^{-} 1$?

My own feeling is that you should never use a column of mercury to measure pressure ABOVE atmospheric pressure. You will end up getting mercury EVERYWHERE. A column of mercury is suitable for measuring LOW pressures. These days, mercury is rarely present in a laboratory due to safety concerns.