# What is the pressure in millimeters of mercury of 0.0130 mol of helium gas with a volume of 208 mL at a temperature of 30 C?

Dec 5, 2016

We use the ideal gas equation to give an approximate pressure of $130 \cdot m m \cdot H g$

#### Explanation:

From the Ideal Gas equation $P = \frac{n R T}{V}$

$= \frac{0.0130 \cdot \cancel{m o l} \times 0.0821 \cdot \cancel{L} \cdot a t m \cdot \cancel{{K}^{-} 1 \cdot m o {l}^{-} 1} \times 303 \cdot \cancel{K}}{208 \cdot \cancel{m L} \times {10}^{-} 3 \cancel{L} \cdot \cancel{m {L}^{-} 1}}$

$=$ $0.17 \cdot a t m$

Now we were asked to quote the pressure in $m m \cdot H g$. It is a fact that $1$ $\text{atmosphere}$ will support a column of mercury $760 \cdot m m$ high. This property is a handy laboratory measurement because you can measure small pressure differences by the difference in heights of mercury columns, relative to the daily atmospheric pressure.

"Pressure "mm*Hg-=0.17*atmxx760*mm*Hg*atm^-1=??mm*Hg.