# Given a gas under 600*"mm Hg" pressure, and at 303.15*K temperature, that occupies a volume of 2.02*L, what volume will it occupy at 273.15*K temperature, under 750*"mm Hg" pressure?

Aug 2, 2017

Well, the current definition of $\text{STP}$ is.... a temperature of $273.15 \cdot K$ and a pressure of exactly ${10}^{5} \cdot P a$ $\left(100 \cdot k P a , 1 \cdot \text{bar}\right)$.

#### Explanation:

And 1*"bar"-=(1*"bar")/(1.01325*atm*"bar"^-1)=0.987*atm

I do all this rigmarole BECAUSE I know that $1 \cdot a t m$ will support a column of mercury that is $760 \cdot m m$ high...........

And hence in terms of the length of a mercury column, $1 \cdot \text{bar} \equiv 760 \cdot m m \cdot H g \cdot a t {m}^{-} 1 \times 0.987 \cdot a t m = 750.0 \cdot m m \cdot H g$

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

We want ${V}_{2} = \frac{{P}_{1} {V}_{1}}{T} _ 1 \times {T}_{2} / {P}_{2}$

$\frac{600.0 \cdot m m \cdot H g \times 2.02 \cdot L}{303.15 \cdot K} \times \frac{273.15 \cdot K}{750 \cdot m m \cdot H g}$

$\cong 1.5 \cdot L$