A sample of carbon monoxide occupies 2.44 L at 295.0 K and 771 torr. Find its volume at -47 degrees Celsius and 350 torr?

Jul 5, 2017

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$; ${V}_{2} \cong 4 \cdot L$

Explanation:

The key to solving this problem is to realize that a mercury column is used as a highly visual measure of pressure. One atmosphere of pressure will support a column of mercury that is $760 \cdot m m$ high.

i.e. $1 \cdot a t m \equiv 760 \cdot m m \cdot H g$

....And thus we can use the length measurement to represent pressure...We also realize that we must use $\text{Absolute temperature}$, where ${\text{Absolute Temperature"-=}}^{\circ} C + 273.15 \cdot K$.

So we solve for ${V}_{2} = \frac{{P}_{1} {V}_{1}}{T} _ 1 \times {T}_{2} / {P}_{2}$, and clearly the RHS gives an answer with units of volume.......

And so........

$\frac{771 \cdot m m \cdot H g \times 2.44 \cdot L}{295 \cdot K} \times \frac{226.1 \cdot K}{350 \cdot m m \cdot H g}$....

$\cong 4 \cdot L$

Of course, I could have reduced each pressure measurement to units of $a t m$, but the increase in volume is consistent with a decrease in PRESSURE, and the moderate decrease in temperature.

Do you see how the units cancel in the expression to give an answer in $\text{litres}$, i.e. as required for a $\text{volume}$?

Jul 5, 2017

$4.18 \text{L}$

Explanation:

As there are a constant number of moles:
("P"_1"V"_1)/"T"_1=("P"_2"V"_2)/"T"_2

$- {47}^{\circ} C$=226^circK

$2.44 {\text{L"=0.00244"m}}^{3}$

$771 \text{torr"=102791.546"Pa}$
$350 \text{torr"=46662.8288"Pa}$

If ("P"_1"V"_1)/"T"_1=("P"_2"V"_2)/"T"_2#, then $\frac{102791.546 \cdot 0.00244}{295} = \frac{46662.8288 \cdot {\text{V}}_{2}}{226}$.

$\text{V"_2=(102791.546*0.00244*226)/(295*46662.8288)=0.00411777473"m"^3~~0.00418"m"^3=4.18"L}$'