# How many grams of fluorine gas will exert a pressure of 134 atm in a 3.2-liter container at 40°C?

May 11, 2017

It will have to be a stout container.

#### Explanation:

We use the Ideal Gas equation:

$\text{Mass} = \frac{P V}{R T} \times 38.00 \cdot g \cdot m o {l}^{-} 1$

$= \frac{134 \cdot a t m \times 3.2 \cdot L}{0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 313 \cdot K} = 16.7 \cdot m o l \times 38.00 \cdot g \cdot m o {l}^{-} 1 = 635 \cdot g$

Why did I use $38.00 \cdot g \cdot m o {l}^{-} 1$ for fluorine gas and not $19.00 \cdot g \cdot m o {l}^{-} 1$?