Section 11.7: Atmospheric Pressure
Tutorial 1 Practice, page 543
1. (a) Given: P = 203 kPa
Required: pressure in mm Hg
Solution:
Step 1. Find the relationship between kPa and mm Hg.
760 mm Hg = 101.325 kPa
Step 2. Write the relationship as a fraction, with the unit you want to find as the numerator.
760 mm Hg
101.325 kPa
Step 3. Multiply the given value by the conversion factor developed in Step 2.
760 mm Hg
203 kPa !
= 1520 mm Hg
101.325 kPa
Statement: The pressure is 1520 mm Hg.
(b) Given: P = 40.0 kPa
Required: pressure in Torr
Solution:
Step 1. Find the relationships that can be used to relate kPa and Torr.
760 mm Hg = 101.325 kPa; 1 Torr = 1 mm Hg
Step 2. Write the relationships as fractions, with the unit you want to find as the numerator.
760 mm Hg
1 Torr
,
101.325 kPa 1 mm Hg
Step 3. Multiply the given value by the conversion factors developed in Step 2.
40.0 kPa !
760 mm Hg
101.325 kPa
!
1 Torr
1 mm Hg
= 3.00 ! 102 mm Hg
Statement: The pressure is 3.00 × 102 Torr.
(c) Given: P = 717 mm Hg
Required: pressure in Pa
Solution:
Step 1. Find the relationships that can be used to relate mm Hg and Pa.
101.325 kPa = 760 mm Hg; 1000 Pa = 1 kPa
Step 2. Write the relationships as fractions, with the unit you want to find as the numerator.
101.325 kPa 1000 Pa
,
760 mm Hg
1 kPa
Step 3. Multiply the given value by the conversion factor developed in Step 2.
101.325 kPa 1000 Pa
717 mm Hg !
!
= 9.56 ! 104 Pa
760 mm Hg
1 kPa
Statement: The pressure is 9.56 × 104 Pa.
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Chapter 11: The Gas State and Gas Laws 11.7-1
Mini Investigation: How Strong Is Your Pop Can?, page 545
Answers may vary. Sample answers:
A. Heating the water causes it to vapourize inside of the can, and some of the air in the can will
be displaced.
B. The water in the pail will rapidly cool the inside of the can.
C. The can collapses when it is placed in the pail. Before the can is placed in the pail, the
atmospheric pressure exerts an inward force on the can. The water vapour and air mixture inside
the can exerts a force outward. When water vapour/air mixture inside the can is rapidly cooled,
the water vapour becomes a liquid. Since liquids take up much less volume than gases, this
greatly reduces the pressure inside the can. At this point, the pressure of the atmosphere acting
inward is greater than the pressure exerted outward from inside the can. These conditions cause
the can to be crushed by the atmospheric pressure.
Section 11.7 Questions, page 546
1. (a) Given: P = 58 kPa
Required: pressure in mm Hg
Solution:
Step 1. Find the relationship between kPa and mm Hg.
760 mm Hg = 101.325 kPa
Step 2. Write the relationship as a fraction, with the unit you want to find as the numerator.
760 mm Hg
101.325 kPa
Step 3. Multiply the given value by the conversion factor developed in Step 2.
760 mm Hg
58 kPa !
= 440 mm Hg
101.325 kPa
Statement: The pressure is 440 mm Hg.
(b) Given: P = 125 mm Hg
Required: pressure in kPa
Solution:
Step 1. Find the relationships that can be used to relate mm Hg and kPa.
101.325 kPa = 760 mm Hg
Step 2. Write the relationships as fractions, with the unit you want to find as the numerator.
101.325 kPa
760 mm Hg
Step 3. Multiply the given value by the conversion factor developed in Step 2.
101.325 kPa
125 mm Hg !
= 16.7 kPa
760 mm Hg
Statement: The pressure is 16.7 kPa.
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Chapter 11: The Gas State and Gas Laws 11.7-2
(c) Given: P = 130 kPa
Required: pressure in mm Hg
Solution:
Step 1. Find the relationship between kPa and mm Hg.
760 mm Hg = 101.325 kPa
Step 2. Write the relationship as a fraction, with the unit you want to find as the numerator.
760 mm Hg
101.325 kPa
Step 3. Multiply the given value by the conversion factor developed in Step 2.
760 mm Hg
130 kPa !
= 980 mm Hg
101.325 kPa
Statement: The pressure is 980 mm Hg.
(d) Given: P = 950 mm Hg
Required: pressure in kPa
Solution:
Step 1. Find the relationships that can be used to relate mm Hg and kPa.
101.325 kPa = 760 mm Hg
Step 2. Write the relationships as fractions, with the unit you want to find as the numerator.
101.325 kPa
760 mm Hg
Step 3. Multiply the given value by the conversion factor developed in Step 2.
101.325 kPa
950 mm Hg !
= 130 kPa
760 mm Hg
Statement: The pressure is 130 kPa.
Table 3 Equivalent Values of Pressure in Different Units
Pressure (kPa)
Pressure (mm Hg)
(a)
58
440
(b)
16.7
125
(c)
130
980
(d)
130
950
2. (a) Standard temperature and pressure (STP) is defined as conditions of 0 °C and 101.325 kPa.
Standard ambient temperature and pressure (SATP) is defined as conditions of 25 °C and
100 kPa.
(b) Scientists defined the standard temperature and pressure (STP) conditions to allow them to
compare results under identical conditions. They then defined a second set of standard
conditions, the standard ambient temperature and pressure (SATP) conditions, because they
more closely resembled the conditions in a laboratory.
(c) SATP is used more often because it is more convenient to use 25 °C rather than 0 °C in a
laboratory setting.
3. As altitude increases the density of gases in the atmosphere decreases. This is because there
are fewer entities of air at higher altitudes. At elevations greater than 4572 m there is insufficient
oxygen to breathe, so skydivers at these altitudes require oxygen to be safe.
4. Low pressure tents are designed to simulate the conditions of high altitude training. Ideally
long term use of low pressure tents should cause athletes to produce more red blood cells. More
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Chapter 11: The Gas State and Gas Laws 11.7-3
red blood cells should cause greater oxygen distribution and better performance. One of the
disadvantages of high altitude training is that during the adjustment period to the high altitude
athletes cannot train as rigorously. The use of the tent is intended to eliminate the adjustment
period. In theory, the use of low pressure tents should be advantageous to athletes.
5. (a) Pressurized cabins have compressed air pumped into the cabin to ensure that the air
pressure inside the cabin is greater than outside the plane. At high altitudes, the density of
atmospheric gases is low, and this causes the external pressure to be low. Pressurized cabins
ensure that passengers and crew have enough oxygen to breathe and are comfortable during the
flight.
(b) If the pressurization fails, the conditions inside the plane will become similar to the external
conditions. At high altitudes there are fewer molecules of oxygen available. Oxygen masks will
be deployed to ensure passengers and crew have enough oxygen to breathe. Without
pressurization passengers and crew can experience hypoxia, altitude sickness, decompression
sickness, and barotrauma. Hypoxia is reduced oxygen intake. Symptoms include impaired mental
function, dimming vision, and, in extreme cases, death. Altitude sickness causes an increase of
blood pH because of problems expelling carbon dioxide. Passengers can experience nausea,
fatigue, and headache, among other symptoms. Decompression sickness can cause dissolved
gases in the blood like nitrogen to form bubbles in the bloodstream. This can result in the bends;
symptoms include tiredness, and headache. Barotrauma is caused by rapid changes in altitude,
and typical symptoms include acute pain in the middle ear, gastrointestinal tract, or teeth.
6. At high elevations there is decreased air pressure. Athletes not accustomed to low air pressure
performed poorly. However, low air pressure means that air resistance is reduced, so some
events where air resistance is a significant obstacle had athletes performing better.
7. (a)
(b) As altitude increases, the air pressure decreases. As the altitude increases, there are fewer
entities of air. With fewer entities, there are fewer collisions per unit area causing decreased
atmospheric pressure compared to lower altitudes.
(c) Answers may vary. Sample answer: I live in Sudbury, Ontario. The altitude is 273 m above
sea level and the typical air pressure is 99 kPa. This follows the trend in the graph in part (b).
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Chapter 11: The Gas State and Gas Laws 11.7-4
8. Answers may vary. Student’s answers should include some of the following information:
Causes: Lowered oxygen and pressure levels. At high altitudes the body requires more breaths to
take in adequate oxygen. If the body does not take in adequate oxygen, this can impair mental
and physical function.
Occurrence: Acute altitude sickness can occur as low as 2500 m, with the onset of systems
beginning anytime from 8-96 hours after ascent.
Symptoms: Headache, nausea, vomiting, fatigue, shortness of breath, swelling of the extremities,
and disturbed sleep are common symptoms. More extreme cases can produce a build up of fluid
on the lungs or brain that can be life threatening.
People at Increased Risk: Those who quickly ascend to a high altitude, people who have
experienced altitude sickness previously, people who reside at less than 3000 feet, people who
are not well hydrated, and those under 50 years of age are at increased risk.
To Combat Altitude Sickness: Ascend to an altitude over a period of several days, limit physical
activity for the first few days, and stay hydrated.
9. In a pneumatic nail gun, the force to drive nails comes from compressed air. An air
compressor has one or more pistons that draw air in from the atmosphere. This air is then pushed
out into the gun mechanism to drive nails. This process supplies compressed air to the air
reservoir in the nail gun. A piston in the gun slides to provide the nailing mechanism. This piston
is driven by differences in air pressure. When the pressure above the piston is greater than below
it, the piston is forced downward, which is the mechanism that hammers the nail. As the nail is
hammered, the pressure inside the piston reverses and the piston retracts.
10. High-pressure injectors use high-pressure jets of liquid rather than a needle to inject medicine
or vaccines under the skin. They are powered by compressed air or other gas that is released
from a gas cylinder or a built-in gas cartridge.
Benefits: Jet injectors are designed to minimize pain and damage to surrounding tissue. Jet
injectors can be used in areas where needles do not work as well, for example, inside the mouth,
nose, ears, and in open wounds. They are precise methods of delivery. The dosage of medication
can be easily customized, they are low cost, and very easy to operate. They can be used
repeatedly unlike syringes, which should only be used once. With jet injectors there are no sharp
syringes can be an issue for disposal.
Disadvantages: They are more expensive compared to syringes. Bruising and localized pain
have been reported. It does require time to prepare and set up the injector. They are not used as
widely so finding parts can be difficult (especially during travel). Sterilization had been a
problem in early models.
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Chapter 11: The Gas State and Gas Laws 11.7-5