Question

What is the pressure (in millimeters of mercury) inside a container of gas connected to a mercury-filled open-end manometer of the sort shown in Figure 9.4 in the textbook when the level in the arm connected to the container is 18.0 cm lower than the level in the arm open to the atmosphere, and the atmospheric pressure reading outside the apparatus is 765.3 mm Hg ?

Answer 1

...........P atm
.................I
.................I
................V
...............I....I
......__......I__I<--- P1
.......I........I....I
.....Δh.......I....I.....__________
.......I........I....I....I...._______... P gas
P2 left --->I.....I....I__I<--- P2 right
...............IHg.I....I....I
...............I.....I....IHgI
...............I.....I__I....I
...............I___Hg___I

Given point 1 and point 2 above,
we have the following pressures:

P1 = P atm = 765.3 mm Hg

P2 left = P1 + P Hg

---------
Finding P Hg:

Density Hg in kg/m^3 = (g/mL) * 1,000
13.534 g/mL (wikipedia) * 1,000
= 13,534 kg/m^3

P Hg = ρ g Δh
where,
ρ = density of Hg = 13,534 kg/m^3
g = 9.8 m/s^2
Δh = 18.0 cm x (1 m / 100 cm) = 0.18 m

Substituting numbers:
P Hg = (13,534 kg/m^3) x (9.8 m/s^2) x (0.176 m)
Units: [(kg/m^3) x (m/s^2) x (m)] = [kg·m/s^2 / m^2] = [N / m^2] = [Pa]
= (13,534) x (9.8) x (0.18) Pa
= 23,873.52 Pa

P Hg = 23,873.52 Pa x (760 mm Hg / 1.01325 x 10^5 Pa)

==> P Hg = 179.07 mm Hg

---------
P2 left = P1 + P Hg
= 765.3 mm Hg + 179.07 mm Hg
= 944.37 mm Hg