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So we solve for $P_2$, and thus $P_2=(P_1V_1)/T_1xxT_2/V_2$, which clearly has the required units of pressure; why?

$P_2=(500*cancel(cm^3)xx700*"Torr")/(700*cancel(cm^3))=500*"Torr"$

The beauty of using these , is that (excluding temperature) given the proportionality, we can use whatever outlandish units we want, i.e. $"pounds per square inch, pints, furlongs"^3$.

And for part (b), we must use units of $"absolute temperature"$, $T_1=298*K; T_2=308*K$:

$P_2=(700*"Torr"*500*cm^3)/(298*K)xx(308*K)/(700*cm^3)$,

$=517*"mm Hg"$

What are these pressure in $"atmospheres"$?

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