Atmospheric Pressure & the Height of Mercury in a Tube [closed]

Question

I wanted to find out what the height of some mercury reaching 1m in a tube when inverted and placed in a beaker, again of Mercury, would be?

I know that:

• Atmospheric Pressure ($P$) $ \approx 10^5 Pa = 100,000 \frac{N}{m^2} = 10 \frac{N}{cm^2}$
• Density of Hg ($\varrho$) $ \approx 13.6 \frac{g}{cm^3} $
• Gravity ($g$) $ \approx 9.8 \frac{m}{s^2} = 980 \frac{m}{s^2} $
• $ P = \varrho h g $

But I get: $$ h = \frac{P}{\varrho \cdot g} = \frac{10 N cm^3 s^2}{13.6 g \cdot 980 cm \cdot cm^2} = \frac{10N s^2}{13.328 g} \approx 1.33 s^2 g^{-1} $$

I wanted a value in cm, what went wrong?

Answer 1

If you want everything in centimeters, you should convert the Newton (which has units $\text{kg}\cdot\text{m}\over\text{s}^2$) to dynes. Converting $1\,\text N$ to dynes gives $1\,{\text N}= 10^5{\text{g}\cdot \text{cm}\over \text{s}^2}$. Note your units for the gravitational constant are off, you should have $g\approx 980\,{\text{cm}/\text{s}^2}$.

When doing calculations and checking units, I find it easier to separate the "unit part" and the "number part": $$\eqalign{ h={P\over\rho\cdot g} &= {10{\text{N}\over\text{cm}^2}\over 13.6{\text{g}\over\text{cm}^3}\cdot 980{\text{cm}\over s^2} } \cr &= {10\cdot10^5{\text{g}\cdot\text{cm}\over \text{s}^2} \cdot{1\over\text{cm}^2}\over 13.6{\text{g}\over\text{cm}^3}\cdot 980{\text{cm}\over \text{s}^2 } }\cr &= {10\cdot10^5 \over 13.6 \cdot 980 } {\text{g}\cdot\text{cm}\over \text{s}^2}\cdot{1\over\text{cm}^2}\cdot {\rm{s}^2\over\text{cm} }\cdot {\rm{cm}^3\over\text{g}}\cr &={10^6\over13.6\cdot980 }\,\text{cm}\cr &\approx 75\text{cm}. } $$

Note this is slightly off as we used somewhat crude approxiations for $g$ and $\rho$.