# If a manometer reads "681.7 torr", and the mercury is higher on the left half of the sidearm (where the right side is open to the atmosphere) by "9.96 cm", what is the pressure of the gas inside in "bars"?

## Jan 24, 2017

$\text{0.776 bars}$.

The idea is that the pressure of the gas in the bulb balances out the pressure of the air (the atmospheric pressure) to some extent.

It is implied that mercury is in the manometer, and the height $\Delta h$ tells you how many $\boldsymbol{\text{mm Hg}}$ by which the two pressures differ.

Since the manometer reads $\text{681.7 torr}$, or $\text{681.7 mm Hg}$, and the difference in height is

$\Delta h = \text{9.96 cm Hg" = "99.6 mm Hg" = "99.6 torr}$,

it means that there is a difference in pressure of $\text{99.6 torr}$ relative to atmospheric pressure, in some direction (less/greater than). To figure out which direction...

Since the mercury is higher on the left half of the sidearm, it means the pressure of the gas inside is weaker than the atmospheric pressure.

Therefore, ${P}_{\text{gas" < "681.7 torr}}$, and:

$\textcolor{g r e e n}{{P}_{\text{gas") = "681.7 torr" - "99.6 torr" = color(green)("582.1 torr}}}$

In $\text{bar}$ then:

$\textcolor{b l u e}{{P}_{\text{gas") = 582.1 cancel"torr" xx cancel"1 atm"/(760 cancel"torr") xx (1.01325xx10^5 cancel"Pa")/cancel"1 atm" xx "1 bar"/(10^5 cancel"Pa}}}$

$=$ $\textcolor{b l u e}{\text{0.776 bars}}$

CHALLENGE: What would the pressure of the gas inside be in $t o r r$ if the mercury level was higher on the right half of the sidearm, with the same $\Delta h$?

Answer: $\textcolor{w h i t e}{\text{681.7 torr + 99.6 torr " = " 781.3 torr}}$
(highlight to see)