Chemistry - A Molecular Science

Patm

P

h

Figure 7.2 A manometer A manometer is used to measure the pressure of a gas. In the above experiment, the pressure of the gas (P) is greater than barometric pressure (P

atm

) by h mm of Hg;

i.e

., P = P

atm

+ h.

Gas

285mm

Example 7.1c

pressure is sufficient to support a column of me

rcury, like that shown in Figure 7.1, to a

height of 760 mm. The following equalities re

late the units for measuring gas pressures.

14.7 lb/in

2 = 1 atm = 760 mm Hg = 760 torr = 1.01x10

5 Pa

Eq. 7.1

Pa is the pascal, the SI unit of pressure (1 Pa = 1 kg/m

(^2) s⋅
).
The device used to measure the pressure of a gas in the laboratory is called a
manometer

. One side of the manometer shown in

Figure 7.2 is open to the atmosphere,

while the other is attached to the gas whose

pressure is to be measured. Consequently, the

atmospheric pressure (P

atm

) is pushing on the right side and the unknown gas pressure (P)

is pushing on the left. When the two pressures are the same, the two mercury columns in the manometer are at the same height, and h = 0. However, when the two pressures are not the same, the side with the greater pressure

exerts the greater force, which causes the

mercury column on that side to drop and the column on the other side to rise.

The

difference in the pressures of th

e two sides is given by the

difference in the two mercury

levels,

h

. The gas in Figure 7.2 exerts a pressure greater than barometric pressure because

its mercury column is lower than the column on the atmosphere side, so P = P

atm

• 
  

h.

Example 7.1 a) What is the pressure of a gas, in atmosp

heres, if it can support a mercury column to

a height of 407 mm?

407 mm Hg

×

1 atm

760 mm Hg

= 0.536 atm

b) A pressure of 0.68 atm will support a column of mercury to what height?

0.68 atm

760 mm Hg×

1 atm

= 5.2

×^10

2 mm

c) What is the pressure of the gas shown in the margin if barometric pressure is 757

torr? Express the answer in torr and atmospheres. The column is lower on the side open to the atmosphere, which means that the pressure of the gas is less than barometric

(atmospheric) pressure. Thus,

P = P

atm

• h = 757 - 285 = 472 mm Hg = 472 torr.

We use the conversion factor taken from

Equation 7.1 to obtain atmospheres.

472 torr

×

1 atm760 torr

= 0.621 atm

Chapter 7 States of Matter and Changes in State

© by

North

Carolina

State

University