SOLUTIONS TOPIC 3 THERMAL
1. Temperature, Heat and the Atomic Model
Example : How many atoms are in 3.00 moles of C?
Example p. 5 of packet
= 1.81 x 1024 atoms
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Example : How many atoms are present in 3.5 g of N?
Example p. 6 of packet
2
Example p. 6 of packet
What is the ratio of
number of atoms of neon
?
number of atoms of helium
A.
0.4
B.
0.5
C.
2.0
D.
2.5
3
Tsokos p. 161 Q1 , p. 162 : 5 a, , 6 a
Examples p. 7 of packet
4
5
2. Specific Heat , Thermal Equilibrium
Q = mc∆t
Example: When a car brakes, an amount of thermal energy equal to 112500J is generated in
the brake drums. If the mass of the brake drums is 28kg and their specific heat capacity is
460.5Jkg-1K-1, what is the change in temperature?
Answer to Q1: 8.7K or 0C
Example p. 9 of packet
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Example 2 p. 9 of packet
Example2 : A piece of iron of mass 0.2kg and temperature 300ºC is dropped into 1.00kg of
water at temperature 20ºC. What will the temperature be at thermal equilibrium? (Take c
for iron as 470Jkg-1K-1 and for water c= 4200Jkg-1K-1.)
Answer = 26ºC
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Example p. 10 of packet
Example : What will be the final temperature (Tf) of 0.100kg of water at 20  C when
0.100 kg of iron nails at 40  C are submerged in the water? (Take c for iron as 470Jkg-1K1 and for water c= 4200Jkg-1K-1.)
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Tsokos p. 171 – 172 : 2
p. 11 of packet
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3. Thermal Properties : Change of State, Evaporation
Example p. 16 of packet
Example : How much energy is required to change 2.0 kg of ice at – 5 0C to boiling steam at
1100C ? ( c water = 4200Jkg-1K-1, LF water = 334 kJ kg-1 , LV water = 2257 kJ kg-1 )
Q of heating ice from – 5 0C to 00 C = mc∆t
+ Q melting ice = mLF ( change in state)
+ Q heating liqud water from 00C to 1000 C = mc∆t
+ Q boiling water = m Lv ( change in state)
+ Q heating steam water from 1000 C to 1100 C = mc∆t
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Tsokos p. 172 : 10, 11, p. 17 of packet
a) Q = mc∆t
cice = 2100
Q = ( 1.0) (2100)(10) = 2100J
b) Q = mL
Lwater = 334000
Q = (1.0) ( 334000) = 334 000 J or 334 kJ
c) Q = mc∆t
cwater = 4200
Q = ( 1.0) (4200)(10) = 42000J
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Q loss of water (200100 ) = mcΔt = (1.0 )(4200)(100 – 200) = - 42000J
Q gain ice + Q loss of water = 0
376000 mice + - 42000 J = 0
mice = 42000 = 0.112Kg = 112g
376000
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Deducing Information from Temperature-Thermal Energy Graph
p. 18 of packet
Figure 2.2 (Tsokos 2008; 166) shows how the temperature of 0.5kg of an unknown
substance changes as thermal energy is provided to it.
Figure 2.2 can be used to deduce the following:

The melting point is -10ºC and the boiling point is 130ºC.

The specific latent heat of fusion is:
Q
1
0
0
k
Jk

5
05
J0
k
J

1
L


1
0
0
k
J
.
k
g
f
m
0
.
5
k
g0
.
5
k
g

The specific heat capacity for the solid state is:
Q
5
0
k
J

0
k
J

1

1
c


1
.
4
k
J
.
K
g
K
(
s
)

m

T
0
.
5
k
g
(

1
0

8
0
)
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Calculate: Example p. 18 of packet
(a) the specific latent heat of vaporization
(b) The specific heat for the liquid phase
(c) The specific heat for the gas phase
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4. Thermal Properties : Kinetic Theory, Pressure
Examples p. 24 of packet
1. Which of the following is not an assumption on which the kinetic model of an ideal
gas is based? ( B)
A.
All molecules behave as if they are perfectly elastic spheres.
B.
The mean-square speed of the molecules is proportional to the kelvin
temperature.
C.
Unless in contact, the forces between molecules are negligible.
D.
The molecules are in continuous random motion.
2. When a gas in a cylinder is compressed at constant temperature by a piston, the
pressure of the gas increases. Consider the following three statements.
I.
The rate at which the molecules collide with the piston increases.
II.
The average speed of the molecules increases.
III.
The molecules collide with each other more often.
Which statement(s) correctly explain the increase in pressure?
A.
I only
B.
II only
C.
I and II only
D.
I and III only
A
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3. B
4. D
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Tsokos pp. 170- 171 Q8 – Q 10 p. 24 – 26 of packet
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5. Ideal Gases
Wilson / Buffa 31, 41, 43 p. 29 of packet
Tsokos p. 181 : 1 p. 31 of packet
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Wilson. Buffa : 43
Tsokos p. 181 : 2 , 3
p. 33 of packet
p. 33 - 34 of packet
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Tsokos p. 179 : Q 7 , Q 8 p. 37 of packet
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IB Questions:
1. In the drop tower shown, containers with experiments inside of them are fired
upwards inside a vertical tower.
The tower is 120 m high with an internal diameter 3.5 m. When most of the air has
beenremoved, the pressure in the tower is 0.96 Pa.
Determine the number of molecules of air in the tower when the temperature of the air
is 300 K:
Answer : 2.68 x 1023
Solution:
V = π r2 h
π (1.75) 2
3
PV = nRT
n = PV/RT
n = (0.96 )(1154.54)
(8.31) (300)
n = 0.44 = moles
n = N/NA
number of molecules N = n x NA = (0.44) (6.02 x 1023 ) = 2.68 x 1023
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2.
A container holds 20 g of Neon and 8 grams of helium. The molar mass of neon is 20
grams and that of helium is 4 grams. Calculate the ratio of the number of atoms of
neon to the number of atoms of helium.
3.
A fixed mass of an ideal gas is heated at constant volume. Sketch a graph to show
the variation with Celsius temperature T with pressure P of the gas.
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User what conditions does the equation of state for an ideal gas, PV = nRT , apply to
a real gas?
4.
5.



6.
7.
Explain the difference between an ideal gas and a real gas.
The ideal gas is assumed to be a low pressures making it easier to predict
The ideal gas can not be liquefied as a real gas can. This would be way too
complicated to understand and predict that a gas can actually be a liquid.
There are no intermolecular forces between the molecules so there is no potential
energy. All energy is considered to be kinetic. Hence the name : Kinetic model for a
gas.
Explain why the internal energy of an ideal gas comprises of kinetic energy only
 There are no intermolecular forces between the molecules so there is no potential
energy. All energy is considered to be kinetic. Hence the name : Kinetic model for a
gas.
 Remember : When matter changes state energy is needed to enable the molecules to
move more freely and thus molecules gain potential energy- a type of
intermolecular bonding energy
A fixed mass of an ideal gas has a volume of 870 cm 3 at a pressure of 1.00 ´10 5 Pa
and a temperature of
. The gas is heated at constant pressure to temperature of
. Calculate the change in volume of the gas:
V1 = V2
T1
T2
Convert C to K
Do not need to convert V
V2 = 872.97 cm3
Change in V = 2.97 cm3
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