Properties of Gases
15. Define the problem: Convert a series of pressure quantities into other pressure units.
Develop a plan: Use Table 10.2 to design conversion factors to achieve the conversions.
Execute the plan:
1 atm
= 0.95 atm
760 mmHg
(a)
720 mmHg ×
(b)
1.25 atm×
(c)
542 mmHg ×
760 torr
= 542 torr
760 mmHg
(d)
740 mmHg ×
1 atm
101.325 kPa
×
= 99 kPa
760 mmHg
1 atm
(e)
700 kPa ×
760 mmHg
= 950. mmHg
1 atm
1 atm
= 7 atm
101.325 kPa
Check your answers: The unit of atm represents a lot more pressure than the units of kPa, torr, and
mmHg, so it makes sense that the numbers of atm are always much smaller than the pressure expressed
in these other units. The unit of kPa represents more pressure than the units of torr and mmHg, so it
makes sense that the numbers of kPa are always smaller than the pressure expressed in the other units
of torr and mmHg. Torr and mmHg are the same size, so their quantities should be identical.
17. Define the problem: Given the density of mercury, the density of an oil used to construct a barometer,
and the atmospheric pressure, determine the height in meters of the oil column in the oil barometer.
Develop a plan: Use Table 10.2 to convert the pressure into mmHg. Pressure on the liquid pushes the
liquid up the barometer until its mass exerts the same force per unit area as the air pressure: Pliquid =
2
Force/Area = mg /Area. Since g is a constant, the force per unit area of the liquid counteracting the air
pressure is proportional to just the mass per unit area. Relate the mass per unit area of mercury to the
mass per unit area of the oil for 1 atm pressure. Relate the mass of each liquid to its respective density
and volume. Relate the volume of each liquid to the dimensions of its respective barometers, including
the height. Relate the height of mercury in the mercury barometer to the height of the oil in the oil
barometer.
Execute the plan: According to Table 10.2, at 1.0 atm the height of a column of mercury in a mercury
barometer is 760 mm. Let m = mass of the liquid, A = cylindrical area of the barometer’s column, d =
density of the liquid, V = volume of the liquid, and h = the height of the liquid in the barometer. Now,
equate the mass per unit area of each barometer and derives an equation relating their heights:
mHg/AHg = moil/Aoil
dHgVHg/AHg = doilVoil/Aoil
h
V = (Area) × (height) = Ah
dHgAHghHg/AHg = doilAoilhoil/Aoil
dHghHg = doilhoil
A
hoil =
d HghHg
doil
(
)
13.596 g / cm 3  760 mm
1m


=
×
= 14 m
1000 mm
0.75 g / cm 3 


Check your answer: A larger mass of oil will be needed to counterbalance the atmospheric pressure
because it is less dense than mercury. A higher column of oil makes sense.
18. If a perfect vacuum is established, atmospheric pressure will push the liquid mercury up the tube 760
millimeters. To whatever extent the vacuum is imperfect, the height will be less than that.
19. With a perfect vacuum at the top of the well, this system would resemble a water barometer. Using the
equation derived in the answer to Question 12:
(
)
13.596 g/ cm3  760 mm
1m
3.281 feet


=
×
×
= 34 feet
hwater =
1000 mm
1m
d water
1.00 g/ cm3 


d Hgh Hg
That means atmospheric pressure can only push water up to about 34 feet. So, the well cannot be
deeper than that, not even using a high quality vacuum pump.
20. According to Question 19, each 34 feet of water creates 1 atm of pressure.
60 feet×
1 atm
= 2 atm
34 feet
(1.77 atm rounded to 1 sig. fig.)
At sea level the pressure of the air bubbles will be atmospheric pressure, or 1 atm.