Gases and Gas Laws.

The first substances to be produced and studied in high purity were gases. Gases
are more difficult to handle and manipulate than solids and liquids, since any
minor mistakes generally results in the gas escaping to the atmosphere. However,
the ability to produce gases in very high purity made the additional difficulty
acceptable. The most common way of producing a gas was by some sort of
chemical reaction, and the gas was collected by liquid displacement (either water
or mercury). Figure 8.1 shows the general process of collecting gas by liquid
displacement.

Characteristics of Gases
Unlike liquids and solids, gases:
- Expand spontaneously to fill their containers
- Are highly compressible
- Have extremely low densities
- Change volume dramatically with changing temperature.
- Can diffuse and mix rapidly with other gases in the same container (different
gases in a mixture do not separate upon standing)
- The ideal gas law is the quantitative relationship between pressures (P),
volume (V), moles gas present (n), and the absolute temperature (T).
- R is the universal gas constant.
- R = 0.08206 L atm mol-1 K-1: used in most gas equations
- R = 8.314 J mol-1 K-1 : used in equations involving energy
-

PV=nRT

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 - STP (Standard Pressure and Temperature = 1 atm & 0oC)
- NC (Normal Conditions = 1 atm & 25oC)

Pressure
- Pressure is force per unit area.
- Atmospheric pressure is the force attributed to the weight of air molecules
attracted to Earth by gravity.
- As altitude increases, atmospheric pressure decreases.
- Pressure results from molecular collisions between gas molecules and
container walls.
- Each collision imparts a small amount of force.
- Summation of the forces of all molecular collisions produces the
macroscopic property of pressure.
History and Application of the Gas Law
• Gases change significantly when the conditions in which they are found are
altered.
• These changes are determined empirically using gas laws.
• Charles’s Law: relationship between T and V
• Boyle’s Law: relationship between P and V
• Gay-Lussac’s law: relationship between P and T
• The empirical gas laws led to the ideal gas law

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 Laws of Perfect Gases

1. Charle's Law
The volume of a given mass of perfect gas varies directly as its absolute
temperature when the pressure remains constant.
𝑉 ∝ 𝑇 ( 𝑃 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 )

V1T1= nR/P= constant =V2T2

𝑉1 𝑉2 𝑉3
= = … = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑇1 𝑇2 𝑇3

2. Boyl's Law
It state, the absolute pressure at a given mass of a perfect gas varies in verily
as its volumes, when the temperature remains constant.
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𝑃 ∝ ( 𝑇=𝐶)
𝑉
𝑃𝑉 = 𝐶 ⟸ ‫حاصل الضرب يساوي دائما كمية ثابتة‬

𝑃1 𝑉1 = 𝑃2 𝑉2 = 𝑃3 𝑉3 … = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑊ℎ𝑒𝑟𝑒:

𝑃 ∶ 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑁 𝑚2

𝑉 ∶ 𝑣𝑜𝑙𝑢𝑚𝑒 (𝑚3 )

𝑵𝒐𝒕𝒆 𝒕 𝒇𝒐𝒓 °𝑪 , 𝑻 𝒇𝒐𝒓 °𝑲

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 3. Gay-lussac Law
The absolute pressure of a given mass of a perfect gas varies directly as its
absolute temperature when the volume remains constant.
𝑃 ∝ 𝑇 (𝑉 = 𝐶)
𝑃
∴ = 𝐶
𝑇
𝑃1 𝑃2 𝑃3
= = … = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝐶)
𝑇1 𝑇2 𝑇3

General Gas Equation

𝐴𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 𝑎 𝑏𝑜𝑙𝑦′𝑠 𝐿𝑎𝑤
1
𝑃 ∝ (𝑇 =𝐶)
𝑉
1
𝑉 ∝
𝑃
𝐴𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 𝑎 𝑐ℎ𝑎𝑟𝑙𝑒′𝑠 𝐿𝑎𝑤

𝑉∝𝑇 (𝑃 =𝐶)
1 𝑇
𝑠𝑜, (𝑉 ∝ 𝑎𝑛𝑑 𝑇 𝑏𝑜𝑡ℎ ) 𝑜𝑟 𝑉 =
𝑃 𝑃
𝑃𝑉 ∝ 𝑇 𝑜𝑟 𝑃𝑉 = 𝐶𝑇

𝑃𝑉 ( ‫) المعادلة العامة للغازات‬

= 𝐶 (‫) المعادلة االساسية‬
𝑇
𝐶
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑤ℎ𝑒𝑟𝑒 𝑖𝑡𝑠 𝑣𝑎𝑙𝑢𝑒 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑚𝑎𝑠𝑠 𝑎𝑛𝑑 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑖𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑠

𝑃1 𝑉1 𝑃2 𝑉2 𝑃3 𝑉3
∴ = = … = 𝐶
𝑇1 𝑇2 𝑇3

𝑃𝑉 = 𝑚𝑅𝑇
R = the constant of proportionality is known as R, the gas constant.

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Example / find the mass of the gas which occupies (5.6 m3) at (7 N/m2) and (200
ºC). (Assume R = 0.289 KJ/ (Kg ºK.)

Solution/ V=5.6m3 ,P=7N/m2,K=200+273=473,R=0.289*1000J/(KG/K)

PV=MRT

7N/m2*5.6m3= M*289*473= M= 0.0002868Kg

Example: A 2.5 L container has a gas pressure of 4.6 atm. If the volume is
decreased to 1.6 L, what will be the new pressure inside the container?
Solution: V1 = 2.5 L, V2 = 1.6 L, p1 = 4.6 atm and p2 =? atm
(4.6 atm)(2.5 L) = p2(1.6 L) p2 = 7.19 atm

Example: The pressure of a gas in a rigid container is 125 kPa at 300 K. What is
the new pressure if the temperature increases to 900 K?
Solution: T1 = 300 K, T2 = 900 K, p1 = 125 kPa and p2 =?
p1/ T1 = p2/ T2
125 kPa/300 K = p2/ 900 K
p2= 375 kPa

Specific heat of gas (C)

It may be defined as the amount of heat required to raise the temperature of
the unit mass through 1 ºC
Specific heat (C)
Specific heat of constant volume (𝐶𝑣 )
Specific heat of constant pressure (𝐶𝑝 )
Specific heat of constant volume (𝐶𝑣 )
It is the amount of heat required to raise the temperature of the unit mass
through 1 Cº when it is heated at constant volume

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Specific heat of constant pressure (𝐶𝑝 )
It is the amount of heat required to raise the temperature of the unit mass
through 1 Cº when it is heated to a constant pressure.

𝑄 = 𝑚 𝐶 ∆𝑇

𝑄 = 𝐻𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑

𝑚 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑠

𝐶 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑠

∆𝑇 = 𝑇1 − 𝑇2

𝑇1 ∶ 𝑖𝑛𝑖𝑡𝑖𝑎𝑡 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑠𝑒𝑠

𝑇2 ∶ 𝑓𝑖𝑛𝑎𝑙 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑠

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The first law of thermodynamics
A thermodynamic system is a collection of objects we can regard as a unit, which
can exchange energy with its surroundings. We can now think about energy
transfers into and out of these systems; through
– Heat transfer Q and
– Work W
Sign convention
We need to be careful about signs:
Qin is positive W done by the system is positive
Qout is negative W done on the system is negative

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Work W
Let’s look at work done during volume changes. Pressure exerts a force on the
piston, which moves from x1 to x2:

W > 0 energy removed from system (work is done by the system against its
surroundings) (expansion)
W < 0 energy added to system (work is done on the system) (compression)
So the work done equals the area under a pV curve.

There can be many different paths from one thermodynamic state to another, so the
work done by a system during a transition between two states depends on the path
chosen.

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Example: Isothermal expansion an ideal gas undergoes a constant-temperature
Expansion at temperature T, so its volume changes from V1 to V2. How much work
does the gas do?
Heat transfer Q
A system can interact with its surroundings by doing work. It can also interact with
its surroundings by means of heat transfer.

How much heat is transferred also depends on the path taken.

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Heat and work
So both heat and work are only recognized as they cross the boundary of a system.
• They are associated with a process, not a state.
• They are both path-dependent functions.
• A system in general does not possess heat or work.
[Wrong description: ‘work in a body; ‘heat in a body’]
Internal energy U
We define the internal energy of a system to be the sum of
– The kinetic energy of all its particles
– The potential energy of interactions between particles
Kinetic energy: translation, vibration, rotation

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 U =ΣKE +ΣPE
KE = Random chaotic motion
PE = Interaction between atoms & molecules
The kinetic and potential energies associated with the random motion of molecules
constitute the internal energy U. Value of U not important, ΔU during a thermal
process is what matters:

The first law of thermodynamics

The first law of thermodynamics: The total change in internal energy of a system is
the sum of the heat added to it and the work done on it.

∆𝑈 = 𝑄 − 𝑊
(Conservation of energy)
Remember:
Qin is positive W done by the system is positive
Qout is negative W done on the system is negative

While Q and W depend on the path, ΔU = Q – W does not. The change in internal
energy of a system during any thermodynamic process depends only on the initial
and final states, not on the path leading from one to the other. I.e.
U is an intrinsic property of the system.

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Equivalence of heat and work
The equivalence of heat and work was demonstrated by Joule in the 1840s. His
apparatus converted the potential energy of falling weights into work done on the
water by a rotating paddle.

U for an ideal gas

Recall that the internal energy is

U =ΣKE +ΣPE
KE = Random chaotic motion
PE = Interaction between atoms & molecules
We showed that for an ideal gas.

Where f is the number of degrees of freedom.

For an ideal gas, there are no forces between molecules, so PE = 0.
Hence for an ideal gas,

So the internal energy for an ideal gas depends only on its temperature.
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Heat capacity of an ideal gas
Consider adding heat to a fixed volume of an ideal gas. Since the volume is fixed,
W = 0, so
ΔU = Q.
The temperature of the gas changes when we add heat:

But our definition of heat capacity was

(Where C is the molar heat capacity) and so we must have

So any ideal gas has the same heat capacity (with the same number of degrees of
freedom).

Change in internal energy:

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U for an ideal gas, again
We calculated U for a process at constant volume. But since the internal energy of
an ideal gas only depends on temperature, the change in internal energy during any
process must be determined only by the temperature change.
For an ideal gas, the internal energy change in any process is given by

Whether the volume is constant or not

Heat capacity of an ideal gas
Now consider adding heat to an ideal gas at constant pressure.

It takes greater heat input to raise the temperature of a gas a given amount at
constant pressure than constant volume
Ratio of heat capacities
Look at the ratio of these heat capacities: we have

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