A sample of chlorine gas is confined in a 3.93-L container at 407 torr and 21.0 degrees Celsius. How many moles of gas are in the sample?

Jul 6, 2017

Well, here we measure pressure in $\text{Torr}$, and of course there is assumed knowledge........I make $P = 0.09 \cdot a t m$

Explanation:

We use the old Ideal Gas equation.....

$P V = n R T$, and now all the given measurements are kosher except for the PRESSURE measurement. Pressure, $\text{force per unit area}$ is something that is hard and non-intuitive to measure. Physical scientists, however, have long known that $\text{1 atmosphere}$ of pressure will support a column of mercury that is $760 \cdot m m$ high (and $\text{1 Torr"-="1 mm Hg}$). And thus we can use a measurement of length to give the pressure in atmospheres. Are you with me.....?

So here $P = \frac{407 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} = 0.536 \cdot a t m$.

The problem is almost done now.........

$P V = n R T$; and so $n = \frac{P V}{R T}$

$n = \frac{0.536 \cdot a t m \times 3.93 \cdot L}{294.15 \cdot K \times 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1}$

=??*atm

Note the dimensional consistency of the answer.....

$n = \frac{0.536 \cdot \cancel{a t m} \times 3.93 \cdot \cancel{L}}{294.15 \cdot \cancel{K} \times 0.0821 \cdot \cancel{L} \cdot \cancel{a t m \cdot {K}^{-} 1} \cdot m o {l}^{-} 1}$

$= \frac{1}{m o {l}^{-} 1} = \frac{1}{\frac{1}{m o l}} = m o l$ as required...............this is what we want, and the consistent unit is an internal check on our calculation......