Gases are subject to three closely interrelated factors—temperature, pressure, and
volume. As the kinetic theory of gases points out, a change in one of these factors
must result in some measurable change in the other factors. Further, the theory
indicates that the kinetic behavior of any one gas is the same for all gases or
mixtures of gases. Consequently, basic laws have been established to help predict
the changes that will be reflected in one factor as the conditions of one or both of
the other factors change. A diver needs to know how changing pressure will effect
the air in his suit and lungs as he moves up and down in the water. He must be able
to determine whether an air compressor can deliver an adequate supply of air to a
proposed operating depth. He also needs to be able to interpret the reading on the
pressure gauge of his tanks under varying conditions of temperature and pressure.
The answers to such questions are calculated using a set of rules called the gas
laws. This section explains the gas laws of direct concern to divers.
Boyle’s law states that at constant temperature, the absolute pressure
and the volume of gas are inversely proportional. As pressure increases the
gas volume is reduced; as the pressure is reduced the gas volume increases.
Boyle’s law is important to divers because it relates to change in the volume of a gas caused by the change in pressure, due to depth, which defines the relationship
of pressure and volume in breathing gas supplies.
The formula for Boyle’s law is: C = P X V
| where: |
|
|
|
|
| C |
|
= |
|
a constant |
| P |
|
= |
|
absolute pressure |
| V |
|
= |
|
Volume |
Boyle’s law can also be expressed as: P1V1 = P2V2
| where: |
|
|
|
|
| P1 |
|
= |
|
initial pressure |
| V1 |
|
= |
|
initial volume |
| P2 |
|
= |
|
final pressure |
| V2 |
|
= |
|
final volume |
When working with Boyle’s law, pressure may be measured in atmospheres absolute.
To calculate pressure using atmospheres absolute:
| |
|
|
|
Depth fsw + 33 fsw |
|
|
|
|
|
|
|
psig + 14.7psi |
| Pata |
|
= |
|
----------------------------- |
|
or |
|
Pata |
|
= |
|
---------------------------- |
| |
|
|
|
33fsw- |
|
|
|
|
|
|
|
14.7psi |
Sample Problem 1. An open diving bell with a volume of 24 cubic feet is to be
lowered into the sea from a support craft. No air is supplied to or lost from the bell.
Calculate the volume of the air in the bell at 99 fsw.
-
Rearrange the formula for Boyle’s law to find the final volume (V2):
| |
|
|
|
P1V2 |
| V2 |
|
= |
|
--------------------------------- |
| |
|
|
|
P2 |
-
Calculate the final pressure (P2) at 99 fsw:
| |
|
|
|
99 fsw + 33 fsw |
| P2 |
|
= |
|
--------------------------------- |
| |
|
|
|
33 fsw |
| |
|
|
|
|
| |
|
= |
|
4ata |
-
Substitute known values to find the final volume:
| |
|
|
|
1ata X 24ft3 |
| V2 |
|
= |
|
--------------------------------- |
| |
|
|
|
4 ata |
| |
|
|
|
|
| |
|
= |
|
6ft3 |
The volume of air in the open bell has been compressed to 6 ft.3 at 99 fsw.
When working with Boyle’s law, the temperature of
the gas is a constant value. However, temperature significantly affects the pressure
and volume of a gas. Charles’/Gay-Lussac’s law describes the physical relationships
of temperature upon volume and pressure. Charles’/Gay-Lussac’s law states
that at a constant pressure, the volume of a gas is directly proportional to the
change in the absolute temperature. If the pressure is kept constant and the absolute
temperature is doubled, the volume will double. If the temperature decreases,
volume decreases. If volume instead of pressure is kept constant (i.e., heating in a
rigid container), then the absolute pressure will change in proportion to the absolute
temperature.
The formulas for expressing Charles’/Gay-Lussac’s law are as follows.
For the relationship between volume and temperature:
| V1 |
|
|
|
T1 |
| -------- |
|
= |
|
-------- |
| V2 |
|
|
|
T2 |
| where: |
|
|
|
|
| T 1 |
|
= |
|
intial temperature (absolute) |
| T 2 |
|
= |
|
final temperature (absolute) |
| V1 |
|
= |
|
initial volume |
| V2 |
|
= |
|
final Volume |
And, for the relationship between pressure and temperature:
| P1 |
|
|
|
P2 |
| -------- |
|
= |
|
-------- |
| T1 |
|
|
|
T2 |
| where: |
|
|
|
volume is constant |
| P 1 |
|
= |
|
intial pressure (absolute) |
| P2 |
|
= |
|
final pressure (absolute) |
| T1 |
|
= |
|
initial temperature |
| T2 |
|
= |
|
final temperature |
Sample Problem 1. An open diving bell of 24 cubic feet capacity is lowered into
the ocean to a depth of 99 fsw. The surface temperature is 80° ;F, and the
temperature at depth is 45° ;F. From the sample problem illustrating Boyle’s law,
we know that the volume of the gas was compressed to 6 cubic feet when the bell
was lowered to 99 fsw. Apply Charles’/Gay-Lussac’s law to determine the volume
when it is effected by temperature.
1. Convert Fahrenheit temperatures to absolute temperatures (Rankine):
| ° ;R |
|
= |
|
° ;F |
|
+ |
|
460 |
| T1 |
|
= |
|
80° ;F |
|
+ |
|
460 |
| |
|
= |
|
505° ;R |
|
|
|
|
| |
|
|
|
|
|
|
|
|
| T2 |
|
= |
|
45° ;F |
|
+ |
|
460 |
| |
|
= |
|
505° ;R |
|
|
|
|
2. Transpose the formula for Charles’/Gay-Lussac’s law to solve for the final
volume (V2):
| |
|
|
|
V1 T2 |
| V2 |
|
= |
|
-------------- |
|
| |
|
|
|
T1
|
3. Substitute known values to solve for the final volume (V2):
| |
|
|
|
6 ft.3 X 505 |
| V2 |
|
= |
|
-------------- |
|
| |
|
|
|
540
|
| |
|
|
|
|
| |
|
= |
|
5.61 ft3 |
The volume of the gas at 99 fsw is 5.61 ft.
Sample Problem 2. A 6-cubic foot flask is charged to 3000 psig and the
temperature in the flask room is 72 °F. A fire in an adjoining space causes the
temperature in the flask room to reach 170 °F. What will happen to the pressure in
the flask?
1. Convert gauge pressure unit to atmospheric pressure unit:
| P1 |
|
= |
|
3000 psig |
|
+ |
|
14.7 psi |
| |
|
= |
|
3014.7 psia |
|
|
|
|
2. Convert Fahrenheit temperatures to absolute temperatures (Rankine):
| ° ;R |
|
= |
|
° ;F |
|
+ |
|
460 |
| T1 |
|
= |
|
72° ;F |
|
+ |
|
460 |
| |
|
= |
|
532° ;R |
|
|
|
|
| |
|
|
|
|
|
|
|
|
| T2 |
|
= |
|
170° ;F |
|
+ |
|
460 |
| |
|
= |
|
630° ;R |
|
|
|
|
3. Transpose the formula for Gay-Lussac’s law to solve for the final pressure
(P2):
| |
|
|
|
P1 T2 |
| P2 |
|
= |
|
-------------- |
|
| |
|
|
|
T1
|
4. Substitute known values and solve for the final pressure (P2):
| |
|
|
|
3014.7 X630 |
| P2 |
|
= |
|
-------------- |
|
| |
|
|
|
532
|
| |
|
|
|
|
| |
|
|
|
1,899,261 |
| |
|
= |
|
-------------- |
|
| |
|
|
|
532
|
| |
|
|
|
|
| |
|
= |
|
3570.03 psia - 14.7 |
| |
|
|
|
|
| |
|
= |
|
3555.33 psig |
The pressure in the flask increased from 3000 psig to 3555.33 psig. Note that the
pressure increased even though the flask’s volume and the volume of the gas
remained the same.
This example also shows what would happen to a scuba cylinder that was filled to
capacity and left unattended in the trunk of an automobile or lying in direct
sunlight on a hot day.
Boyle, Charles, and Gay-Lussac demonstrated that
temperature, volume, and pressure affect a gas in such a way that a change in one
factor must be balanced by corresponding change in one or both of the others.
Boyle’s law describes the relationship between pressure and volume, Charles’/
Gay-Lussac’s law describes the relationship between temperature and volume and
the relationship between temperature and pressure. The general gas law combines
the laws to predict the behavior of a given quantity of gas when any of the factors
change.
The formulas for expressing Charles’/Gay-Lussac’s law are as follows.
The formula for expressing the general gas law is:
| P1V1 |
|
= |
|
P2V2 |
| -------- |
|
= |
|
-------- |
| T1 |
|
|
|
T2 |
| where: |
|
|
|
|
| P 1 |
|
= |
|
initial pressure (absolute) |
| V 1 |
|
= |
|
initial volume |
| T1 |
|
= |
|
initial temperature (absolute) |
| P2 |
|
= |
|
final pressure (absolute) |
| V2 |
|
= |
|
final volume |
| T2 |
|
= |
|
final temperature (absolute) |
Two simple rules must be kept in mind when working with the general gas law:
- There can be only one unknown value.
- The equation can be simplified if it is known that a value remains unchanged
(such as the volume of an air cylinder) or that the change in one of the
variables is of little consequence. In either case, cancel the value out of both
sides of the equation to simplify the computations.
Sample Problem 1.
Your ship has been assigned to salvage a sunken LCM
landing craft located in 130 fsw. An exploratory dive, using scuba, is planned to survey the wreckage. The scuba cylinders are charged to 2,250 psig, which raises
the temperature in the tanks to 140° ;F. From experience in these waters, you know
that the temperature at the operating depth will be about 40° ;F. Apply the general
gas law to find what the gauge reading will be when you first reach the bottom.
(Assume no loss of air due to breathing.)
1. Simplify the equation by eliminating the variables that will not change. The
volume of the tank will not change, so V1 and V2 can be eliminated from the
formula in this problem:
| P1 |
|
|
|
P2 |
| -------- |
|
= |
|
-------- |
| T1 |
|
|
|
T2 |
2. Calculate the initial pressure by converting the gauge pressure unit to the
atmospheric pressure unit:
| P |
|
= |
|
2,250 psig + 14.7 |
| |
|
= |
|
2,264.7 psia |
3. Convert Fahrenheit temperatures to Rankine (absolute) temperatures:
Conversion formula: ° ;R = ° ;F + 460
| T1 |
|
= |
|
140 ° F |
+ |
460 |
| |
|
|
|
|
|
|
| |
|
= |
|
600 ° R |
|
|
| |
|
|
|
|
|
|
| T2 |
|
= |
|
40 ° F |
+ |
460 |
| |
|
|
|
|
|
|
| |
|
= |
|
500 ° R |
|
|
4. Rearrange the formula to solve for the final pressure (P2):
5. Fill in known values:
| |
|
|
|
2,264.7 psia X 500 ° ;R |
| P2 |
|
= |
|
----------- |
| |
|
|
|
600° ;R |
| |
|
|
|
|
| |
|
= |
|
1887.25 psia |
6. Convert final pressure (P2) to gauge pressure:
| P2 |
|
= |
|
1,887.25 psia - 14.7 |
| |
|
|
|
|
| |
|
= |
|
1,872.55 psig |
The gauge reading when you reach bottom will be 1,872.55 psig.
Sample Problem 2. During the survey dive for the operation outlined in Sample
Problem 1, the divers determined that the damage will require a simple patch. The Diving Supervisor elects to use surface-supplied MK 21 equipment. The
compressor discharge capacity is 60 cubic feet per minute, and the air temperature
on the deck of the ship is 80° F.
Apply the general gas law to determine whether the compressor can deliver the
proper volume of air to both the working diver and the standby diver at the operating
depth and temperature.
1. Calculate the absolute pressure at depth (P2):
| |
|
|
|
130 fsw + 33FSW |
| P2 |
|
= |
|
------------------------- |
| |
|
|
|
33FSW |
| |
|
|
|
|
| |
|
= |
|
4.93 ata |
2. Convert Fahrenheit temperatures to Rankine (absolute) temperatures:
Conversion formula:
| ° ;R |
|
= |
|
° ;F |
|
+ |
|
460 |
| T1 |
|
= |
|
80° ;F |
|
+ |
|
460 |
| |
|
= |
|
540° ;R |
|
|
|
|
| |
|
|
|
|
|
|
|
|
| T2 |
|
= |
|
40° ;F |
|
+ |
|
460 |
| |
|
= |
|
500° ;R |
|
|
|
|
3. Rearrange the general gas law formula to solve for the volume of air at depth
(V2):
| |
|
|
|
P1V1T2 |
| V2 |
|
= |
|
------------ |
| |
|
|
|
P2T1 |
4. Substitute known values and solve:
| |
|
|
|
1ata X 60cfm X 500 ° R |
| V2 |
|
= |
|
------------------------------- |
| |
|
|
|
4.93ata X 540 ° ;R |
4. Based upon an actual volume (displacement) flow requirement of 1.4 acfm for a
deep-sea diver, the compressor capacity is sufficient to support the working and
standby divers at 130 fsw.
Sample Problem 3. Find the actual cubic feet of air contained in a 700-cubic inch
internal volume cylinder pressurized to 3,000 psi.
1. Simplify the equation by eliminating the variables that will not change. The
temperature of the tank will not change so T1 and T2 can be eliminated from
the formula in this problem:
P1V1 = P2V2
2. Rearrange the formula to solve for the initial volume:
Where:
| P1 |
|
= |
|
14.7 PSI |
|
| P2 |
|
= |
|
3,000 psi + 14.7 PSI |
|
| V2 |
|
= |
|
700 in3 |
|
3. Fill in the known values and solve for V1:
| |
|
|
|
3014.7 psia X 700 in 3 |
| V1 |
|
= |
|
------------------------------ |
| |
|
|
|
14.7 psia |
| |
|
|
|
|
| |
|
= |
|
143,557.14 in3 |
| |
|
|
|
|
4. Convert V 1to cubic feet:
| |
|
|
|
143 557.14 in 3 |
| V1 |
|
= |
|
------------------ |
| |
|
|
|
1728 in3 |
(1728 in3 = 1 ft3 |
| |
|
|
|
|
| |
|
= |
|
83.07 scf |
| |
|
|
|
|
|
