# How do you express 275 mmHg in atm. and Pascal?

Apr 17, 2016

$1$ $\text{atm}$ $=$ $760$ $m m \cdot H g$ $=$ $101.32$ $k P a$

#### Explanation:

Of course, pressure is not measured in millimetres of mercury, but in force per unit area. If you go thru the math, and I don't propose to do so, it turns out that a column of mercury $760$ $m m$ high exerts a force per unit area, a pressure, of $1$ $a t m$, or $101.32$ $k P a$.

Thus $275$ $m m$ $H g$ $=$ $\frac{275 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1}$ $=$ $\text{approx. 0.4 atm}$ or,

$\frac{275 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} \times 101.32 \cdot k P a \cdot a t {m}^{-} 1$ $=$ ??kPa

The use of a mercury barometer to measure pressures is of long standing. It was a very convenient means to measure small pressures when doing a distillation. Most laboratories would have an old (usually beautifully made) mercury barometer somewhere on their premises.

These days, the use of mercury is not so prevalent due to environmental concerns. Mercury in the lab tends to be fascinating, and sometimes students play with it, i.e. float coins on it, use it for electrical switches, dissolve alkali metals in it (I know I did). Inevitably, it all goes pear shaped, and you spill mercury on the floor or the desk, where it inhabits every crack. Water barometers are not so popular, because $1$ $a t m$ will support a column of water approx. $10$ $m$ high. $1$ $a t m$ will support a column of mercury $760$ $m m$ high - water is so much less dense than mercury.

Apr 17, 2016

275 mm Hg = 0.36 atm or 36477 pa

#### Explanation:

1 atm = 760 mm Hg

We can write the two conversion factors

1 atm / 760 mm Hg or 760 mm Hg / 1 atm

275 mm Hg x (1 atm \ 760 mm Hg)

0.36 atm

1 atm = 101325 pa

We can write two conversion factors

1 atm / 101325 pa or 101325 pa / 1 atm

0.36 atm x ( 101325 pa / 1 atm )

36477 pa