# If 1.93 * 10^-3 mol of argon occupies a 65.8-mL container at 20 degrees Celsius, what is the pressure (in torr)?

Jul 6, 2017

P~=0.706*atm-=??"Torr"

#### Explanation:

From the Ideal Gas equation, which you must be getting sick of.......

$P = \frac{n R T}{V} = \frac{1.93 \times {10}^{-} 3 \cdot m o l \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 293 \cdot K}{65.8 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1}$

$\equiv 0.706 \cdot a t m$

And here the trick is to know that $1000 \cdot m L \equiv 1 \cdot L$, i.e. $1 \cdot m L \equiv {10}^{-} 3 \cdot L$.

Anyway, you know the relationship between $\text{atmospheres}$ and $m m \cdot H g$ so have at it.............