‫المحاضرة الخامسة‬
Boyle's Law
Ideal Gas Equation
When gas is kept at constant temperature
its pressure is inversely proportional to the
volume.
Charle's Law
When the pressure of the gas kept
constant the volume directly proportional
to the temperature.
V a T at constant pressure.
result can be summarized in one equation
called the equation of state for an ideal gas.
Where n is the number of moles, R is a constant for a
specific gas, which can be determined experimentally,
and T is the absolute temperature in Kelvin
When the pressure goes to zero then the quantity PV/nT
become the same value of R for all gasses, therefore R
called the universal gas constant
R = 8.31 J/mole.K
The ideal gas law can be expressed in terms of the total
number of molecules N where N = nNA
where NA is the Avogadro's number = 6.022´1023molecules/mole
where K is called Boltzmann's constant, which has the
value R/NA
K = R/NA = 1.38´10-23J/K
¯One mole of substance is that mass of the substance
that contains Avogadro's number of molecules
Example (1)
An ideal gas occupies a volume of 100cm3 at
20oC and a pressure of 100Pa. Determine the
number of moles of gas in the container.
Solution
PV = nRT
What is the number of molecules in the
container?
Example (2)
Pure helium gas is admitted into a tank containing
a movable piston. The initial volume, pressure and
temperature of the gas are 15´10-3m3, 200kPa and
300K respectively. If the volume is decreased to
12´10-3m3 and the pressure is increased to
350KPa, find the final temperature of the gas.
Solution
Since the gas can not escape from the tank then
the number of moles is constant, therefore, PV =
nRT
at the initial and final points of the
process
Exercise
Pure helium gas is admitted into a leak-proof
cylinder containing a movable piston. The initial
volume, pressure, and temperature of the gas are 15
L, 2.0 atm, and 300 K. If the volume is decreased to
12 L and the pressure increased to 3.5 atm, find the
final temperature of the gas. (Assume helium
behaves as an ideal gas.)
T=420K
Exercise
An ideal gas undergoes the process shown in the
figure. Find V2, Ta, Td and Tb. where n=44kmole and
R=8.314J/mole.K
Heat and the first law of thermodynamics
The word of ''heat flow'' is an energy
transfer that take place as a consequence of
temperature difference only.
temperature is defined as the derivative of the internal
energy with respect to the entropy.
Unit of Heat
The unit of heat is ''calorie'' which is defined as the amount of heat
(energy) required to raise the temperature of 1g of water from 14.5oC to
15.5oC.
1cal = 4.186J
or
1J = 0.2389cal
Temperature is a physical property of matter that
quantitatively expresses the common notions of hot and cold.
Objects of low temperature are cold, while various degrees of
higher temperatures are referred to as warm or hot. Heat
spontaneously flows from bodies of a higher temperature to
bodies of lower temperature, at a rate that increases with the
temperature difference and the thermal conductivity. No heat
will be exchanged between bodies of the same temperature;
such bodies are said to be in "thermal equilibrium".
The temperature of a substance typically varies with the
average speed of the particles that it contains, raised to the
second power; that is, it is proportional to the mean kinetic
energy of its constituent particles. Formally,
Example
A student eats a dinner rated at 2000 (food) Calories. He wishes to do
an equivalent amount of work in the gymnasium by lifting 50Kg
mass. How many times must he raise the weight to expend this much
energy? Assume that he raises the weight a distance of 2m each time
and no work is done when the weight is dropped to the floor.
Solution
Since 1 (food) Calories = 1000 cal then the work required
is 2x106cal.
Converting this to joule, then the work required is
W = 2x106cal x 4.186J/cal = 8.37x106J
‫والشغل الكلي لرفع‬mgh ‫يساوي‬h ‫الشغل المبذول لرفع االثقال لمسافة‬
‫وعليه فإن عدد مرات رفع االثقال‬nmgh ‫األثقال عدة مرات يعطي بالعالقة‬
‫هو‬
W = nmgh = 8.37x106J
Since m = 50 Kg, and h = 2m
n = 8.54x103 times
Heat capacity and specific heat
The heat capacity is defined as the
amount of heat energy needed to raise
the temperature of a sample by 1 degree
Celsius.
J/Co
The specific heat capacity is defined
as the amount of heat energy needed to
raise 1kg of sample by 1 degree Celsius.
where c is called the specific heat capacity or specific heat.
Al
900J/kg.Co
wood
1700J/kg.Co
Cu
387J/kg.Co
glass
837J/kg.Co
Ag
129J/kg.Co
water
4186J/kg.Co
Pb
128J/kg.Co
ice
2090J/kg.Co
Example
A 0.05kg of metal is heated to 200oC and then
dropped into a beaker containing 0.4kg of water
initially at 20oC. If the final equilibrium
temperature of the mixed system is 22.4oC find the
specific heat of the metal. What is the total heat
transferred to water in cooling the metal?
Solution
Heat lost by the metal = heat gained by water
mx cx (Ti-Tf) = mw cw (Tf-Ti)
(0.05Kg) cx (200oC-22.4oC) = (0.4kg)(4186J/kg.Co)(22.4oC-20oC)
cx = 453J/kg.Co
(b) Q = m c (Ti-Tf) = 0.05 ´ 453 ´ (200-22.4) = 4020J
Example
A man fires a silver bullet of mass 2g
with a velocity of 200m/sec into a
wall. What is the temperature change of
the bullet?
Solution
The kinetic energy of the bullet Ek = 1/2 m v2 =
40J
Q = m c DT
where c for silver is 234Jkg.Co
DT = Q/mc = 85.5Co
Latent heat
The heat or energy required to change the phase ( ‫تغير‬
‫ )المادة حالة‬of a given mass m of a substance is given by
where L is called the latent heat (hidden heat) of substance
Solid ---- Liquid
Liquid ------ gas
melting
boiling
Latent heat of fusion Lf
melting (Solid º> Liquid)
Latent heat of vaporization Lv boiling) Liquid º>
gas)
Lv > Lf
Work and heat in thermodynamic processes
Process
Isothermal process
Isobaric process
Isochoric process
Adiabatic process
dF = P dA
dF dy = P dA dy
dW = P dV
dV=0
)Isochoric(
W=
0
The first law of thermodynamics
change in the internal energy of a closed system is equal to
the amount of heat supplied to the system, minus the amount
of work performed by the system on its surroundings. The
law can also be stated: The energy of an isolated system is
constant.
If a system (gas) in its initial state at Pi, Vi
change to Pf, Vf by adding quantities of heat
and applying work on the system. If the
quantity Q-W is measured for various paths
from i to f.
we find
Q-W is always constant.
Q-W is called the change in the internal
energy of the system DU
Q and W depend on the path but Q-W is
independent of the path
DU = Uf - Ui = Q - W The first law of thermodynamics
for small changes
dU = dQ - dW
Special cases
In isolated system there is no heat
flow and work is zero the change in
internal energy is zero, i.e. DU=0
If the process is done on a system
taken through a cycle, the change in
the internal energy is zero, i.e. DU=0
and Q = W
Example
A thermodynamic
process is shown in
Figure. In process ab,
600J of heat are added,
and in process bd 200J
of heat are added. Find
1) the internal energy
change in process ab
2) the internal energy
change in process abd
3) the total heat added
in process acd
Solution
(a) in ab W = 0 and DU = Q = 600J
(b) in bd pressure is constant
W = P (V2 - V1) = 8x104 pa (5x10-3 - 2x10-3)
= 240J
W abd = 240 + 0
Q
abd = 800J
DU = 800 - 240 = 560J
(c) in acd DU = 560J as well
W = 3x104 pa (5x10-3 - 2x10-3) = 90J
Q = DU + W = 560 + 90 = 650J
Example
One gram of water occupies a volume of 1cm3 at
atmospheric pressure. When this amount of water is boiled,
it becomes 1671cm3 of steam. Calculate the change in
internal energy for this process.
Where the latent energy of water is (2.26x106J/kg)
Solution
DU = Q - W
Q = m Lv = (1x10-3kg) x (2.26x106J/kg) = 2260J
W = P (Vf - Vi) = (1.013x105) x [(1671-1)10-6] = 169J
DU = Q - W = 2260J - 169J = 2091J
Example
A 1kg bar of copper is heated at atmospheric
pressure(1.013x105N/m2) . If its temperature increases from
20oC to 50oC, (a) find the work done by the copper. (b) What
quantity of heat is transferred to the copper? (c) What is the
increase in internal energy of the copper?
Where
intial volume of copper is (1kg/8.92x103kg/m3)
the volumetric expansion coefficient β= 5.1x10-5Co-1
(a) the work done by the copper
DV = [5.1x10-5Co-1](50oC20oC)x(1kg/8.92x103kg/m3)
DV = 1.7x10-7 m3
W = P DV = (1.013x105N/m2) x (1.7x10-7 m3) =
1.9x10-2J
(b) What quantity of heat is transferred to the copper?
Q = m C DT = (1kg) x (387J/kg.Co) x (30Co) = 1.16 x 104J
(c) What is the increase in internal energy of the copper?
DU = Q - W = 1.16 x 104J- 1.9x10-2J