__Chapter 14:__

__Expansion of Gases:__

The thermal expansion of a gas involves 3 variables:
volume, temperature, and pressure. Pressure
of a gas, in a closed container, is the result of
the collision of its molecules on the walls of that container. It is
important to note that the kinetic energy of each gas molecule
depends on its temperature only. Recall the definition of
temperature**:** " the temperature of an object is a result of the vibrations
of its atoms and molecules**.** For a gas, molecules are free to move and
bounce repeatedly against each other and their container's walls**.** **In
each collision, a gas molecule transfers some momentum to its container's wall.
Gas pressure is the result of such momentum transfers.** The faster they move, the
greater impulse per collision they impart to the container's walls causing a
higher pressure**.** For a fixed
volume, if the temperature of a gas increases (by heating), its pressure
increases as well. This is simply because of increased
kinetic energy of gas molecules that cause more number of collisions per second
and therefore increased pressure**.**

One important formula to know is the formula for the **
average** kinetic energy of a number of gas molecules that are at a given
temperature.

The average K.E. of gas molecules is a function of
temperature only**.** The formula is

(K.E.)avg**.** =
(3/2) kT

where T is the
absolute temperature in Kelvin scale and
k is the " Boltzman's
constant " with a value of k
= 1**.**38x10^{-23} J /K**.**

By a number of gas molecules, we do not
mean 1000 or even 1000,000 molecules**.** Most often we mean much more
than 10^{24} molecules**.**

Note that kinetic energy on the other hand
is K.E. = (1/2)MV^{2} **, **where
V is the average
speed of the gas molecules**.**

**Since at a given temperature, the average kinetic energy of
gas molecules is constant, a gas molecule that has a greater mass oscillates
slower, and a gas molecule that has a smaller mass oscillate faster.**
The following example clarifies this concept**.**

Example 1: Calculate
the average K**.**E**.** of air molecules at 27**.**0^{o}C**.**
Also, calculate the average speed of its constituents**:** oxygen molecules and nitrogen** **
molecules**.**
Note that O_{2}
= 32**.**0 grams / mole, and N_{2} = 28**.**0 grams / mole.

Solution: K.E. =
(3/2) k T
;
K.E. = (3/2) (1**.**38x10^{-23}
J/K)(27+273)K
= 6**.**21x10^{-21} J/molecule.

This means that **any gas molecule**, on the average,
has this energy**.**

For an oxygen molecule**:** K**.**E**.**
= (1/2)MV^{2}
**;**

6**.**21x10^{-21} J/molecule =
(1/2) ( 32**.**0x10^{-3} kg
/ 6**.**02x10^{23}molecule)V^{2}.

483m/s = V.

For a nitrogen molecule**:** K**.**E**.**
= (1/2)MV^{2}
**;**

6**.**21x10^{-21} J/molecule =
(1/2) ( 28.0x10^{-3} kg
/ 6**.**02x10^{23}molecule)V^{2}.

V = 517m/s.

Nitrogen is lighter; therefore, its average speed
is
higher**.** Oxygen is heavier than nitrogen; therefore, its average speed
is lower than that of nitrogen at the same temperature**.** **Note that in SI**, grams must be converted to kilograms**.**
Avogadro number (6**.**02x10^{23}
molecules of) oxygen has a mass of 32**.**0 grams**.** 32**.**0grams
means 32**.**0x10^{-3} kg**.**

__Expansion of Gases: Perfect Gas Law:__

If a gas fulfills two conditions, it is called a " perfect gas"
or an " ideal gas " and its expansion follows the perfect gas law**:**

PV = nRT

where P is the gas absolute pressure (pressure with respect to
vacuum), V is its
volume (the volume of its container), n
is the number of moles of gas in the container,
R is the Universal
gas constant, R = 8**.**314
[ J / (mole ^{
o}K)]**, **and
T is the gas
absolute temperature in Kelvin scale**. **

The two conditions for a gas to
follow this equation are**:**

** 1) **The gas
pressure should not exceed about 10 atmospheres**.**

** 2)** The gas
must be superheated (**gas temperature sufficiently
above its boiling point**) at the operating pressure and volume**.**

__The Unit of " PV ":__

Note that the product " PV " has dimensionally the unit of
"energy**.**" In SI, the unit of "P" is [ N / m^{2} ] and the
unit of volume " V " is [ m^{3} ]**.** On this basis,
the unit of the product " PV
" becomes [ Nm ] or [ Joule
]**.** The " Joule " that appears in R = 8**.**314 J
/(mole K) is for
this reason**.**

Example 2: A 0**.**400m^{3}
tank contains nitrogen at 27^{o}C**.**
The pressure gauge on it reads 3**.**75 atmosphere**.** Find (a) the
number of moles of gas in the tank, and (b) the gas mass in kg**.**

Solution: PV
= nRT ; n = (PV) / (RT) ; Use horiz**.**
fraction bars when solving**.**

n = [(4**.**75x101000Pa)(0**.**400m^{3})]
**/ **[(8**.**314
J / (mole K))( 27 + 273)K].

(a) n = 76.9 moles.

(b)
M = (76**.**9 moles)(28**.**0 grams /mole) = 2150 grams = 2**.**15
kg.

Note:
P_{absolute} = P_{gauge} + 1atmosphere
= 4**.**75 atmosphere**.**

T_{absolute} = ^{
o}C
+ 273 or, K
= ^{o}C
+ 273.

Example 3: A 0**.**770m^{3}
hydrogen tank contains 0**.**446 kg of hydrogen at 127 ^{
o}C**.** The pressure gage on it
is not working**.** What pressure should the gauge show? Note that 6**.**02x10^{23 }
molecules of H_{2 }amount to 2**.**00
grams**.**

Solution: n
= (0**.**446x10^{3} grams) /
(2**.**00 grams / mole) = 223 moles**.**

PV = nRT ; P = (nRT) /
V ; Use horiz.
fraction bars when solving**.**

P = (223 moles)**(** 8**.**314 (J / mole K)**)** (127 + 273)K
**/ **( 0**.**770 m^{3} )**.**

P_{abs}
= 1,105,000 Pascals**.**

P_{gauge} = P_{abs} - 1 atm =
1,105,000 Pascals
- 101,000Pascals = 1,004,000 Pa ( about
10 atmospheres)**.**

__Equation of State:__

Equation PV = nRT is also called the
"equation of state**.**" The reason is that for a certain amount of a gas,
i.e., a fixed mass, the number of moles is fixed**.**
A change in any of the variables**:** P ,
V , or T, or any two of them, results in a change
in one or the other two**.** Regardless of the changes, PV = nRT
holds true for any state that the gas is in**,** as long as the two
conditions of a perfect gas are maintained**.** That's why it is called the
equation of state**.** A gas is considered to be ideal if its
temperature is quite above its boiling point and its pressure is under 80
atmospheres**.** These two conditions must be met in any state
that the gas is in, in order for this
equation to be valid**.**

Now suppose that a fixed mass of a gas is in **state 1**: P_{1}, V_{1},
and T_{1}**.** We can write P_{1}V_{1} = nRT_{1}**.**
If the gas goes through a certain change and ends up in state 2:
P_{2}, V_{2}, and T_{2}, the equation of state for it becomes P_{2}V_{2} = nRT_{2}**.**

Dividing the 2nd equation by the
1st one, side by side, results in (P_{2}V_{2} / P_{1}V_{1})
= ( nRT_{2} / nRT_{1} )**.** Simplifying
yields**:**

(P_{2}V_{2}
/ P_{1}V_{1}) = ( T_{2} / T_{1} ).

This equation simplifies the solution to many problems**.
**Besides its general form shown above, it finds 3 other forms**:**
one for constant pressure, one for constant temperature, and one for constant volume.

Example 4: 1632
grams of oxygen is at 2**.**80 atm**.** of gauge pressure and a temperature
of 127^{o}C**.**
Find (a) its volume**.** It is then compressed to 6**.**60 atm**.** of
gauge pressure while cooled down to 27^{o}C**.**
Find (b) its new volume**.**

Solution: n = (1632 / 32**.**0) moles
= 51.0 moles ;
(a) PV = nRT ; V = nRT/p ;

V = (51 moles)**(** 8**.**314 (J / mole K)**)**(127+273)K
**/** [3**.**80x101,000]Pa = 0.442m^{3}.

**(b)** P_{2}V_{2}
/ P_{1}V_{1} = T_{2 }/ T_{1}
;
(7**.**6atm)(V_{2})
**/** [(3**.**8atm)(0**.**442m^{3})]
**=** 300.K**/**400.K (Use horiz**.** frac**.** bars)**.**

V_{2} = 0**.**166m^{3}.

__Constant Pressure (Isobar)
Processes:__

A process in which the pressure of an ideal gas
does not change is called an " isobar
process."
Constant pressure means P_{2}=P_{1}**.**
This simplifies the equation P_{2}V_{2} / P_{1}V_{1}
= T_{2}/ T_{1} to equation: V_{2} / V_{1} = T_{2 }/ T_{1}.

Example 5: A
piston-cylinder mechanism as shown below may be used to keep a constant
pressure**. **The pressure on the gas under the piston is
0 gauge plus the extra pressure that the weight generates.
Let the piston's radius be 10**.**0 cm and the weight 475N, and suppose that
the position of the piston at 77^{o}C is 25**.**0cm
from the bottom of the cylinder**.** Find its position when the system is
heated and the temperature is 127^{o}C**.**

Solution: V V P T T Using
V ( 10 h |

__Constant Temperature (Isothermal)
Processes:__

A process in which the temperature of an ideal gas
does not change is called an **" isothermal process."** Constant temperature
means T_{2}=T_{1}.
This simplifies the equation P_{2}V_{2} / P_{1}V_{1}
= T_{2}/ T_{1} to equation: P_{2}V_{2} / P_{1}V_{1} =
1. Cross-multiplication results in:
P_{2}V_{2}
**=**
P_{1}V_{1}.

Example 6: A
piston cylinder system has an initial volume of 420 cm^{3} and the
air in it is at a pressure of 3**.**00
atmospheres as its gauge shows**.** The gas is compressed to a volume of 140cm^{3}
by pushing the piston**.** The generated heat is removed by enough cooling
such that the temperature remains constant**.** Find the final pressure of the
gas**.**

Solution: Since T = constant,
therefore, P_{2}V_{2} **=**
P_{1}V_{1 }; P_{2}(140cm^{3})
= (4**.**00atm)(420cm^{3}) ;

(P_{2})_{absolute} =
12.0 atm
;
(P_{2})_{gauge} = 11.0 atm**.**

__Constant Volume (Isometric)
Processes:__

A process in which the volume of an ideal gas
does not change is called an " isometric
process."
Constant volume means V_{2}=V_{1}**.**
This simplifies the equation P_{2}V_{2} / P_{1}V_{1}
= T_{2}/ T_{1} to equation: P_{2} / P_{1} = T_{2
}/ T_{1}. Gas
cylinders have constant volumes**.**

Example 7: A 15**.**0
liter gas cylinder contains helium at 7^{o}C
and 11.0atm
of gauge pressure.
It is warmed up to 147^{o}C.
Find its new gauge pressure**.**

Solution: P_{2}
/ P_{1} = T_{2 }/ T_{1} ; P_{2}
= P_{1 }(T_{2 }/ T_{1 }) ; (P_{2})_{gauge} = 17.0 atm.

Note: If your solution resulted in 16**.**5atm and
you rounded it to 17 atmospheres, it is wrong**. **The answer is
exactly 17**.**0 atm**.** without rounding**.**

Chapter 14 Test Yourself 1:

1) The **temperature of a gas** is the result of the (a) average K**.**E**.** of
its atoms or molecules (b) average P**.**E**.** of its atoms and molecules
(c) net momentum its atoms and molecules have**.**
click here

2) The net momentum of gas molecules in a container is (a) equal to the net
K**.**E**.** of the atoms or molecules (b) zero (c) neither a nor b**.**

3) The **pressure of a gas** in its container (a) is the result of the
average momentum transfer of its molecules to the walls of the container
(b) depends on the temperature of the gas that in turn depends on the average K**.**E**.** of the gas molecules (c) both a & b**.**
click here

4) In a head-on collision of two equal and rigid masses, (a) the same-mass
molecules exchange velocities (b) if one molecule is initially at rest,
it will move at the velocity of the colliding molecule, and the colliding molecule
comes to stop**.** (c) both a & b**.**

5) According to the "Kinetic Theory of Gases", the average K**.**E**.** of gas
molecules is a function of (a) pressure only (b) volume only (c)
temperature only**.**
click here

6) The average K**.**E**.** of gas molecules in a container that is at
temperature T is equal to (a) kT (b)
1/2 kT (c) 3/2
kT where K is the Boltzman's constant**.**

7) The Boltzman's constant (k) has a value of (a) 1**.**38x10^{-23}
/K (b) 1**.**38x10^{-23} J (c) 1**.**38x10^{-23
}J/ K**.**
click here

8) The formula (K**.**E**.**)_{avg.} = (3/2) kT is applicable to (a)
all gases (b) monoatomic gases only (c) diatomic
gases only**.**

9) The energy of a gas in a container due to its temperature is because of
the energy it constantly receives from (a) the surroundings via heat transfer
(b) the pressure from its container (c) the gravitational field of the
Earth**.**
click here

10) The ratio of the mass of an oxygen molecule to that of a hydrogen
molecule is (a) 32 (b) 16 (c) 8**.**

11) The (K**.**E**.**)_{avg.} of a gas molecule is on one hand 3/2kT and on
the other hand is (a) Mv^{2} (b)
1/2 Mv^{2} (c) Mv.

12) At a given temperature, if the average speed of oxygen molecules is
480m/s, the average speed of hydrogen molecules is (a)120m/s (b)1920m/s (c)960m/s**.**
click here

13) A gas is treated as a **perfect gas** if its** pressure and
temperature** are respectively**:** (a) less than 80atm, under boiling
point (b) more than 80atm, at boiling point (c) less then 80atm,
above boiling point**.**

14) When the temperature of a gas is quite above its boiling point, the molecules
are quite energetic and bounce around, and therefore do not stick together to
condensate into liquid phase**.** This is a good reason for a gas to
be a perfect gas and follow the perfect gas formula. (a) True
(b) False**.**
click here

15) When the pressure of a gas is under 80atm, the pressure is not too high to keep
the molecules closer together to where the molecular attraction forces between
the molecules become significant**.** Since no provision is made in the
perfect gas law (PV=nRT) for molecular attraction;
therefore, the formula is valid for pressures under 80
atmospheres**.** (a) True (b) False

16) In the formula PV = nRT, (a) P is absolute pressure only (b)
T is absolute temperature only (c) both P and T are absolute
quantities**.**
click here

17) In the formula PV = nRT, the value of R is 8**.**314 J/(mole K) in (a)
SI units only (b) all systems (c) American systems
only**.**

18) The product PV has units if (a) force (b) energy (c) power**.**

19) In SI, since R = 8**.**314 J/(mole K), the product PV is in (a) lb-ft
(b) Joules (c) watts**.**
click here

20) The gauge on a gas tank shows the pressure as 2**.**5atm**.**
The absolute pressure is (a) 1**.**5atm (b) 3**.**5atm (c)
also 2**.**5atm**.**

21) The gas pressure in a tank is (340kPa)_{gauge}**.** The
absolute pressure is (a) 440kPa (b) 240kPa
(c) also 140kPa**.**

22) The absolute pressure of a gas tank is 44**.**7psi**.** Its gauge
should show (a) 59**.**4psi**.** (b) 43**.**7psi**.** (c)
30**.**0psi**.**

23) The Avogadro number is (a) 6**.**02x10^{23} molecules
(b) 6**.**02x10^{23} molecules/mole (c) 6**.**02x10^{23}
grams**.**

24) 6**.**02x10^{23} molecules of O_{2} have a mass of (a)
32**.**0grams (b) 16**.**0grams (c)
8**.**0grams**.**
click here

25) 6**.**02x10^{23} molecules of N_{2} have a mass of (a)
34**.**0grams (b) 18**.**0grams (c)
28**.**0grams**.**

26)
One mole of N_{2} has a mass of (a) 34**.**0grams
(b) 18**.**0grams (c) 28**.**0grams**.**

27) One mole of O_{2} has a mass of (a) 32**.**0grams
(b) 16**.**0grams (c) 8**.**0grams**.**
click here

Problem: Let us select **1 mole of a perfect gas**, any
perfect gas, hydrogen, nitrogen, helium, etc**.**, and
put it in a container that can have a variable volume. Also, let us create STP
(Standard Pressure and Temperature) for it, that is 1atm of absolute pressure and 0ºC**.**
In Metric units, the standard pressure and temperature are**:** 101,000Pa and
273K**.** Answer the following**:**

28) Plugging these values**:** n = 1 mole, P_{abs} = 101,000Pa,
T = 273K in the perfect gas formula PV = nRT , and solving for volume, yields**:**
(a) V = 0**.**0224m^{3} (a) V = 22**.**4 Liter
(c) both a & b**.** ** Note that: **1m^{3} = 1000 liter**.**

29) We may say that: **one mole of any perfect gas** at STP ( 1atm of absolute
pressure, or 0 gauge pressure, and 0^{o}C, or 273^{o}K) occupies the same volume
of 22**.**4
liter**.** (a) True (b) False
click here

30) In an isothermal process, (a) T_{2} = T_{1}
(b) P_{2}V_{2} = P_{1}V_{1} (c)
both a & b**.**

31) In an isometric process, (a) P_{2} = P_{1}
(b) V_{2} = V_{1} (c) P_{2}V_{2}
= P_{1}V_{1}**.**

32) In an isometric process, (a) V_{2} = V_{1}
(b) P_{2 } / P_{1} = T_{2 }/ T_{1}
(c) both a & b**.**

33) In an isobar process, (a) P_{2} = P_{1}
(b) V_{2} = V_{1} (c) P_{2}V_{2}
= P_{1}V_{1}**.**

34) In an isobar process, (a) P_{2} = P_{1}**.**
(b) V_{2 } / V_{1} = T_{2 }/ T_{1}**.**
(c) both a & b**.**
click here

35) At constant pressure, if the absolute temperature of a gas is doubled,
its volume (a) doubles (b) triples (c) becomes half of what it
was**.**

36) At constant volume, if the absolute temperature of a gas is doubled, its
pressure (a) doubles (b) triples (c) becomes half of what it was**.**
click here

37) At constant temperature, if the volume of a gas tripled by expansion, its
pressure (a) doubles (b) triples (c) becomes 1/3 of what it was**.**

38) At constant temperature, if the pressure of a gas quadrupled by
compression, its volume (a) doubles (b) triples (c) becomes 1/4 of
what it was**.**

39) At constant temperature, when the pressure of a gas increases, its volume
(a) increases (b) decreases (c) does not change**.**
click here

40) The trick to create a constant pressure for a gas is to (a) place it in a
tank with fixed boundaries
(b) place it under a piston that can easily move up and down a cylinder without allowing
the gas to escape (c) neither a nor b**.**

41) To keep a constant temperature for a gas (a) its volume must be kept
constant regardless of its pressure (b) its pressure must be kept
constant regardless of its volume (c) heat must be supplied to the gas or
removed from it to keep it at the desired temperature**.**

42) Constant volume for a gas means keeping it (a) in a rigid closed
cylinder (b) keep it under a piston-cylinder system such that the piston
cannot move (c) both a & b**.**
click here

43) A certain amount of gas in a closed rigid cylinder loses pressure if it
is (a) cooled down (b) warmed up (c) both a & b**.**

44) If there is a certain amount of gas under a piston-cylinder system and
the piston is pulled up such that the gas volume is increased, the gas
temperature (a) goes up (b) goes down (c) remains
unchanged (d) insufficient information**.**
click here

Problems:

1) Calculate (a) the mass (in kg) of each
CO2 molecule as well as each He molecule knowing that their molar weights are 44**.**0
gr/mole and 4**.**0 gr/mole, respectively**.** Find (b) the average K**.**E**.** of
each at 57**.**0^{o}C**.**
Find (c) the average speed of each at 57**.**0^{o}C**.**
The Avogadro Number is 6**.**02x10^{23} atoms/mole.

2) A 0**.**200m^{3}
tank contains CO_{2} at 77^{o}C**.**
The pressure gauge on it reads 7**.**75 atmospheres**.** Find (a) the
number of moles of gas in the tank, and (b) the mass in kg**.** Each mole of
CO_{2} has a mass of 44**.**0grams**.** R = 8**.**314 J/(mole K)**.**

3) A 555-liter tank contains 1**.**2 kg of
helium at -73 ^{
o}C**.** What pressure should the gauge
on it show (a) in kPa and (b) in psi? Each mole of He_{ }is 4**.**00
grams**.** 1m^{3} = 1000 liter**.**

4) A tank contains 2**.**800 kg of
nitrogen at a pressure of 3**.**80 atmosphere as its gauge shows**.** It is kept
at a temperature of 27^{o}C**.**
Find (a) its volume in liters**.** It is then kept in another room
for several hours that has a temperature of 77^{o}C**.**
Find (b) its new pressure**. **Neglect the very small change in the
volume of the tank due to thermal expansion and treat the volume as a constant**.**

5) A piston cylinder system has an
initial volume of 960 cm^{3} and the
air in it is at zero gauge pressure**.** (a) If it is taken to outer space
while keeping its volume and temperature constant,
what pressure will its gauge show? Here on Earth, if the gas is compressed to a volume of 160cm^{3}
by pushing the piston while keeping its temperature constant by cooling, find (b) its final
gauge pressure**.**

6) A 36**.**0
liter metal cylinder contains nitrogen at 127^{o}C
and 15.0atm
of gauge pressure.
It is warmed up to 177^{o}C.
Find (a) its new gauge pressure. (b) how much
gas does it contain?

Answers:

1) 7**.**31x10^{-26}kg ,
6**.**64x10^{-27}kg ,
6**.**83x10^{-21}J, 432m/s,
1430m/s

2) 60**.**7moles, 2**.**67kg 3)
798kPa, 116psi 4) 513
liters, 4**.**6 atm.

5) 1**.**0atm, 5**.**0atm 6) 17atm,
491grams