A non volatile oil has a density of 0.9 g / ml what height of a coloum of oil will represent an atmospheric pressure value of 740 torr (of Hg) density of Hg = 13.6 g/ml and temperature is 298K?

Question

anonymous · Answer

@Callisto

callisto · Answer

I'm sorry that I haven't learnt that... :(

anonymous · Answer

OK!

unklerhaukus · Answer

$$1[\text{Torr}]=1[\text{ml}]\cdot[\text{Hg}]=1[\text{cm}^3]\cdot[\text{Hg}]$$
$$[\text{Hg}] =13.6 \left[\frac {\text{g}}{{\text{cm}^3}}\right]$$$$[\text {Oil}]=0.9 \left[\frac {\text{g}}{{\text{cm}^3}}\right]$$
$$P=740[\text{Torr}]$$$$=740[\text{cm}^3]\cdot1[\text{Hg}]$$$$=h[\text{cm}]\times 1[\text{cm}^2] \cdot[\text {Oil}]$$

unklerhaukus · Answer

So you are given a pressure, with units that depend on the density of mercury

Mercury ;$\text{Hg} $ can be found in a $1[\text{cm}^2] $ cylinder i a barometer, |dw:1343106244830:dw|