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Module G484.3
Thermal Physics
ANSWERS
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Lesson 31 questions – Solids Liquids Gases
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1
a)
Draw a diagram of each state of matter and describe each state in terms
of the spacing, ordering and motion of atoms or molecules.
A solid has definite shape and volume. Its particles are packed very close
together, so solids can't move, but they can vibrate. Solids are made up of
crystal shapes, which is the regular repeating order of particles (except for
amorphous solids).………………………………………………………… (3)
Liquids have definite volume but not definite shape, since it takes the shape of
its container. The particles that make up liquids are close together, but not
packed. Liquids have the ability to flow. Viscosity is the resistance of a liquid
to flow.…………………………………………………………………………… (3)
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Gases have neither a definite shape or volume. Their particles are spaced far
apart and they are able to move
freely.……………………………………………………………………………… (3)
SOME
Extension
The particles of plasma are very far apart. It also has very high energy, which
makes it quite dangerous. Although it is rare on Earth, it is the most common
phase in the universe. The sun and stars are made of plasma.
b)
Describe what is meant by a Kinetic Molecular Theory for Solids
Liquids and Gases.
Particles are in constant motion. In solids the particles are close together and
have limited motion. In a liquid some of the attraction between particles is
overcome which allows the particles more freedom of movement. In a gas
particles attraction between particles is minimized and the particles move
freely throughout the container.
…………………………………………………………………………………… (2)
SOME
2
A Student asks on a Physics blog “If all matter is either solid, liquid, gas or
plasma, what state of matter is a single atom or molecule?” What would
you write in answers to this question?
……There are more states of matter than just Solid Liquid and Gas. (1)
Solid, amorphous solid, liquid, gas, plasma, super-fluid, supersolid, degenerate
matter, neutronium, strongly symmetric matter, weakly symmetric matter, quarkgluon plasma, fermionic condensate, Bose-Einstein condensate and finally... Strange
matter
(1).……………………………………………………………………………………..
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…State of Matter" is not the same as "Matter" therefore a correct answer to this
question would be (1) All matter comes in many states, a single atom is an atom, a
single molecule is a molecule. (1)
The differing arrangements of atoms or molecules are what make up the various states
of matter it can fulfil. (1)
………………………………………………………………..
…………………………………………………………………………………… (4)
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Lesson 33 - Kinetic theory and Pressure
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1. (a)
What do you see when looking at Brownian motion in air?
……… smoke particles that appear as tiny points of light in random motion (1)
due to collisions with air molecules (1) ………………………………………………………… (2)
(b)
What changes do you see if the air is heated?
…… the random motion becomes more violent ……………………………………… (1)
2. Define pressure and give its units.
…………………………………………………………………………………………
……...
P (Pa) = F (N) / A (m2)
………………………………………………………………………………………………...
…………………………………………………………………………………………… (2)
3. List six assumptions that you have to make about molecules when deriving the
equation for the kinetic theory of gases.
1
A Gas consists of particles called molecules.
2
The molecules are in constant random motion. As many travelling in one
direction as any other. The centre of mass of the gas is at rest.
3
Intermolecular forces are negligible.
4
The duration of collisions between molecules is negligible.
5
Molecules move with constant velocity in between collisions.
6
The volume of gas molecules is negligible compared with the volume of the
gas.
7
All collisions are totally elastic.
8
Newtonian mechanics can be applied to the collisions.
……………………………………………………………6 of the above………… (6)
MOST
4.
(a)
a molecule of mass 3x10-26 kg moving at 4x102 ms-1 collides
elastically with a wall. What is the change of momentum of the molecule?
mv
= 3x10-26(4x102 – (- 4x102))
= 3x10-26x8x102
= 2.4x10-23 Ns
t
Change in momentum = ……2.4x10-23 ……………. Unit… Ns ……… (3)
(b)
How long would it take this molecule to travel a distance of 0.4 m?
= s/v
= 0.4/400
Time = ……0.01……….. s (2)
5.
(a)
What is meant by the root mean square speed of gas molecules?
……… the square root of the sum of the squares of the speeds of the molecules
divided by the number of molecules in the gas …………………………………...
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…………………………………………………………………………………………
……...
………………………………………………………………………………………………...……………………………
……………………………………………………………… (2)
(b)
what is the difference between the root mean square of a group
of gas molecules and their average speed?
Example
Consider four molecules with the following speeds :
200 ms-1 300 ms-1 400 ms-1 500 ms-1
R.M.S = [(2002+3002+4002+5002)/4]1/2 = 367 ms-1
Mean speed = [200 + 300 + 400 + 500]/4 = 350 ms-1
The r.m.s. speed is greater than the mean speed.…………………………………… (2)
6.
Calculate the root mean square speeds for the following gases at a
5
pressure of 10 Pa: USE P = 1/3 v2 to give: v2 = 3P/
(a)
air
density 1.29 kgm-3
(b)
root mean square speed = …482………….. ms-1 (2)
carbon dioxide
density 1.98 kgm-3
(c)
nitrogen
root mean square speed = …398………….. ms-1 (2)
density 1.25 kgm-3
(d)
root mean square speed = ……490 ……….. ms-1 (2)
chlorine
density 3.21 kgm-3
(e)
root mean square speed = …306 ………….. ms-1 (2)
hydrogen
density 0.09 kgm-3
root mean square speed = ……1826 ……….. ms-1 (2)
5. Calculate the pressure of three samples of air of density 1.29 kgm-3 with
different root mean square speeds:
USE P = 1/3 v2
(a) 4.0x102 ms-1
pressure = ……6.88x104 ………..Pa (2)
(b) 5.0x102 ms-1
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pressure = …1.075x105 …………..Pa (2)
(c) 6.0x102 ms-1
pressure = …1.548x105 …………..Pa (2)
9.
A sample of gas of volume 0.1 m3 and at a pressure of 2.0x105 Pa is
enclosed in a cylinder. If the root mean square speed of the gas molecules is
5.5x102 ms-1 and the mass of each molecule is 3.5x10-26 kg calculate the
number of gas molecules in the cylinder.
PV = 1/3 mNv2
So
N
= 3PV/mv2
= 3x2x105x0.1/[3.5x10-26x3.025x105]
And
N
= 6x104/1.059x10-20
= 5.67x1024
Number of gas molecules = ……5.67x1024………… (3)
10.
A sample of gas at a pressure of 1.5x105 Pa is contained in a cylinder
with a volume of 0.05 m3. If there are 1.2x1025 molecules of gas in the
cylinder with a root mean square speed of 350 ms-1 calculate the mass of one
gas molecule.
PV = 1/3 mNv2
So,
m
And
m
= 3PV/Nv2
= 3x1.5x105x0.05/[1.2x1025x1.225x105]
= 2.25x104/1.47x1030
= 1.53x10-26 kg
Mass of one gas molecule = ……1.53x10-26
…………kg (3)
SOME
11.
Suppose there are N molecules in a
rectangular box of dimensions a, b and l and
suppose that the molecule has a velocity v with
the components as shown.
vy
b
a)
What is the change in velocity
of the particle when it hits the shaded face if
the collision is totally elastic?
vx
vz
2mvx
a
(1)
L
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b)
i)
What is the time interval before the same molecule makes a 2nd
collision at the same face?
2Lvx.
(1)
ii)
An therefore what is the frequency of collisions?
vx/2L
(1)
c)
From the definition of Pressure and by finding the rate of change of
momentum for one molecule show that for N molecules:
P  Nmv2 / 3V .
Rate of change of momentum at the face = 2mvx x vx/2L= mvx2/L
(Since F=Δp/Δt)
(1)
Therefore the pressure exerted on the shaded face is by one molecule is:
P=F/A
(1)
P = (mvx2/L / (ab)
(1)
(ab = A, Area of shaded face)
If we sum N contributions, one from each particle in the box, each contribution
proportional to vx2 for that particle, the sum just gives us N times the average
value of vx2. That is to say,
P  F / A  Nmvx2 / LA  Nmvx2 / V
(1)
where there are N particles in a box of volume V. Next we note that the
particles are equally likely to be moving in any of 3 directions, so the average
value of vx2 must be the same as that of vy2 or vz2, and since v2 = vx2 + vy2 +
vz2, it follows that
(1)
P  Nmv2 / 3V .
(1)
(7)
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Lesson 34 questions – Internal Energy
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1
Describe the arrangement of atoms, the forces between the atoms and the
motion of the atoms in:
a)
a solid
The atoms in a solid are arranged in a three-dimensional structure. [1]
There are strong attractive forces between the atoms. [1]
The atoms vibrate about their equilibrium positions. [1]
…………………………………………………………………………………………
(3)
b)
a liquid
The atoms in a liquid are more disordered than those in a solid. [1]
There are still attractive electrical forces between the atoms but these are
weaker than those for similar atoms in a solid. [1]
The atoms in a liquid are free to move around and have kinetic energy. [1]
…………………………………………………………………………………………
(3)
c)
a gas.
The atoms in a gas move around randomly. [1]
There are virtually no forces between the atoms because they are further apart
than similar atoms in a liquid. [1]
The atoms of a gas move at high speeds (but no faster than those in a liquid at
the same temperature). [1]
…………………………………………………………………………………………
(3)
2
A small amount of gas is trapped inside a container. Describe the motion
of the gas atoms as the temperature of the gas within the container in
increased.
The atoms move faster [1]
because their kinetic energy increases as the temperature is increased. [1]
The atoms still have a random motion. [1]
…………………………………………………………………………………………
(3)
3
a)
Define the internal energy of a substance.
The internal energy of a substance is the sum (of the random distribution) of
the kinetic and potential energies of its particles (atoms or molecules). [1]
…………………………………………………………………………………………
(1)
b)
The temperature of an aluminium block increases when
it is placed in the flame of a Bunsen burner. Explain why this
causes an increase in its internal energy.
There is an increase in the average kinetic energy of the aluminium
atoms as
they vibrate with larger amplitudes about their equilibrium positions.
[1]
The potential energy remains the same because the mean
separation between
the atoms does not change significantly. [1]
Hence, the internal energy increases because there is an increase in
the kinetic
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energy of the atoms. [1]
…………………………………………………………………………………………
…………………………………………… (3)
SOME
c)
An ice cube is melting at a temperature of 0
°C. Explain whether its internal energy is increasing
or decreasing as it melts at 0°C.
As the ice melts, the mean separation between the
atoms increases. [1]
The increase in separation between the molecules
increases the potential
energy of the molecules. [1]
There is no change in the kinetic energy of the molecules because the
temperature remains the same. [1]
The internal energy of the molecules increases because there is an increase in
the potential energy of the molecules. [1]
…………………………………………………………………………………………
(4)
Complete the table below for each of the processes shown. Use the symbol ‘+’ for an
increase, the symbol ‘–’ for a decrease and ‘0’ for no change, as appropriate.
4
(3)
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Lesson 38 39 answers – Specific Heat Capacity
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1
What unit is used for all types of energy?
………Joule……………………………………………………………………………
(1)
2
State the equation for the energy needed to raise the temperature of a
particular material of known mass.
…………Energy = mass x specific heat capacity x change in temperature……………
…………………………………………………………………………………………
(1)
3
Use the following data to calculate the amounts of energy needed to change
the temperature of the questions below.
Material
Specific heat capacity (J/kg ˚C)
Copper
380
Water
4200
Aluminium
880
a)
2kg of water by 5˚C
…………2 x 380 x 5 = 3800 J…………………………………………………………
…………………………………………………………………………………………
(2)
b)
500g of water by 4˚C
………0.5 x 420 x 4 = 8400
J……………………………………………………………
…………………………………………………………………………………………
(2)
c)
100g of aluminium from 20˚C to 30˚C
………0.1 x 880 x 10 = 880 J…………………………………………………
…………………………………………………………………………………………
(2)
d)
200 g of copper from 60˚C to 10˚C
……….0.2 x 380 x –50 = -3800 J
(negative because it is the energy given out by the
copper)………………………………
……………………………………………………………………………………… (3)
4
A 2kg block of iron is given 10kJ of energy and its temperature rises by 10˚C.
What is the specific heat capacity of iron?
……… Energy = mass x specific heat capacity x change in temperature
………So, specific heat capacity = Energy / (mass x change in temperature)
…………… specific heat capacity = 10000 / (2 x 10)
…………… specific heat capacity = 500 J / kg˚C…………………………………(4)
5
Some cooks make toffee. Essentially, this is a process of boiling down a sugar
solution to concentrate it and then allowing the liquid to cool until it sets. Small
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children are usually warned not to touch the cooling toffee for a very long time –
much longer than the cooling for the same volume of pure water in the same vessel.
Why do you think that the cooling period so long?
…… There are a number of factors that determine the time it takes the toffee to cool sufficiently to eat:
The boiling point of the sugar solution is higher than that of water so the toffee is
cooling from a higher temperature.
The sugar solution has a higher specific heat capacity than pure water. So for an
equivalent temperature drop, more energy has to be lost.
[Perhaps most significant is the amount of energy that has to be lost for the liquid
toffee mixture to solidify, cooling and then changing from a liquid to a solid takes a
long time. This involves the idea of latent heat capacity, rather than specific heat
capacity] (2)
6
Show that the energy required to heat the air in your physics laboratory from a
chilly 10 °C to a more comfortable 20 °C is about 3 000 000 J if it has the
following dimensions: 3 m  10 m  10 m.
2
Assume the laboratory is 3 m  10 m  10 m, this leads to a volume of 300 m . Assume
further that the heating is just a question of warming up the air. The density of air is
–3
–1
–1
approximately 1 kg m so energy = 300 kg  1000 J kg C  10C, i.e. 3 MJ.
A reasonable heater might deliver this in 1000 s (about 20 minutes). Most people
would guess that the heating time would be much longer. This estimate ignores
heating the contents of the room, etc. Our perception of temperature is affected both
by the humidity of the air and by the cooling effect of any draughts. It would take
much longer in reality. (4)
7
This question is about the operation of an electrically operated shower.
a)
The water moves at constant speed through a pipe of cross section 7.5
2
x 10 m to a showerhead. See the diagram. The maximum mass of water which flows
per second is 0.090kgs-1.
i)
Show that the maximum speed of the water in the pipe is
-1
1.2ms .
Density of water = 1000kg m-3
-5
Density = m/V
Volume flowing in 1 second = 7.5 x 10-5m2 x 1.2ms-1
Therefore 1000 x 0.00009 = mass/second = 0.090kg/s QED
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(2)
ii)
The total cross-sectional area of the holes in the head is half
that of the pipe. Calculate the maximum speed of the water as it
leaves the shower head.
Same volume must flow therefore speed must be double
iii)
Speed = ……2.4………… ms-1 (1)
Calculate the magnitude of the force on the shower-head.
F = m (v-u)/t
F = 0.09 (2.4-1.2)
Force = ……0.11……….. N (3)
b)
The water enters the heater at the temperature of 15˚C. At the
maximum flow rate of 0.090kgs-3, the water leaves the shower head at a
temperature of 27˚C.
i) Calculate the rate at which energy is transferred to the water.
Give a suitable unit for your answer.
Specific heat capacity of water = 4200Jkg-1K-1
ΔQ=mcΔθ
= 0.09 x 4200 x (27-15)
Rate of energy transfer = ……4536………… unit ……W (Js-1)………..
(4)
ii) Suggest a reason that the power of the heater must be greater
than your answer to (b)(i)
……Energy (heat ) loss to
surroundings…………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (1)
iii)
Calculate the maximum possible temperature of the water at the
showerhead when the flow rate is half of the maximum.
4536 / 0.045 x 4200 = 24
If start temp still 15
So Δθ = 15 + 24
Temperature = ………39………….˚C (1)
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Lesson 40 questions – specific latent heat
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Question 1
Calculate the energy released when (a) 10 g water at 100 °C
and (b) 10 g of steam at 100 °C are each spilt on the hand.
Take the specific heat capacity of water to be 4200 J kg –1 K
–1 and the specific latent heat of vaporisation of water to be
2.2 MJ kg–1 . Assume that the temperature of the skin is 33
°C.
a)……… E = mcD0 = 0.01 × 4200 × (100 – 33) = 2814 J = 2.8 kJ (3)
b The latent heat given out in changing from steam at 100 °C to water at the same
temperature is
E = ml = 0.01 × 2.2 × 106 = 22 000 J
The heat given out when this condensed water drops in temperature from 100 °C to 33
°C is
E = mcq = 0.01 × 4200 × (100 – 33) = 2814 J
So, the total heat given out is = 25 kJ (3)
Question 2
When a falling hailstone is at a height of 2.00 km its
mass is 2.50 g. It has 49.05 J of gravitational potential
energy. Assuming that all of this potential energy is
converted to latent heat during the fall, calculate the
mass of the hailstone on reaching the ground. Take the
specific latent heat of fusion of ice to be 3.36 × 105 J kg–
1
.
The falling hailstone loses potential energy, and this is used to partly melt the
hailstone.
ml= 49.05
m × 3.36 × 105 = 49.05
m = 1.4598 × 10–4 kg (mass of hailstone that melted)
Total mass of hailstone = 2.50 g
=> Remaining mass that reaches the ground = 2.50 – 0.1458 g
= 2.354 g ……………………………………………………………… (3)
Question 3
0.30 kg of ice at 0 °C is added to 1.0 kg of water at 45 °C. What
is the final temperature, assuming no heat exchange with the
surroundings? Take the specific heat capacity of water to be 4200
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J kg -1 K -1 and the specific latent heat of fusion of ice to be 3.4 × 10 5 J kg -1 .
Let q be the final temperature.
Heat lost by water = heat gained in melting the ice + heat gained in warming the ice
water
mwcwDqw = micelice + micecwDqmelted ice
mwcw(45 – q) = micelice + micecwD0 melted ice
1 × 4200 × (45 – q) = (0.3 × 3.4 × 105 ) + (0.3 × 4200 × q)
4200 (45 – q) = 1.02 × 105 + 1260 q
1.89 × 105 – 1.02 × 105 = 1260 q + 4200 q
q = 16 oC (3)
4.
Why is a scald by steam at 100 oC much more painful than one by water at
100 oC?
Specific latent heat of fusion of ice = 335 000 J kg-1
Specific latent heat of evaporation of water 2.26 MJ kg-1
Because it has to condense first before cooling so giving out a large amount of latent heat.
(Better students will give a sample calculation, see the worked example)
(2)
5.
How long will it take a 50 W heater to melt 2 kg of ice at 0 oC?
Power x time = mL so time = mL/power = 2  335000 / 50 = 670000 / 50 = 13400 s = 223 min
= 3.7 hours = 3 hrs 43 minutes
(Assumes no heat lost to the surroundings and the water remains at 0 oC.
(3)
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Lesson 44 questions – Ideal Gas Equation
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1
What are the Avagadro Constant and the Mole and how are they linked?
……The Avagadro Constant NA is used to determine the number of particles in any
quantity of any substance.
One Mole of any substance contains 6.02x1023 (NA) number of atoms.
……………………………………………………………………………………… (2)
MOST
2
a)
The equation of state of an ideal gas is pV = nRT. Explain why the
temperature must be measured in Kelvin.
……when pressure (or Volume, or internal energy) tends to zero, the temperature
must tend to zero……………………………………………………………..
……The Kelvin scale with this zero of temp is the Kelvin scale………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
b)
A meteorological balloon rises through the atmosphere until it expands
to a volume of 1.0 x 106m3, where pressure is 1.0 x 103 Pa. The temperature also falls
from 17˚C to –43˚C.
The pressure of the atmosphere at the Earth’s surface = 1.0 x 105Pa.
Show that the volume of the balloon at take off is about 1.3 x 104m3.
P1V1/T1 = P2V2/T2
1.0 x 103 x 1.0 x 106/230 = 1.0 x 105 x V2/ 290
V2 = 12608.70 ~1.3 x 104m3 QED
(3)
c)
The balloon is filled with helium gas of molar mass 4.0 x 10-3kgmol-1
at 17˚C at a pressure of 1.0 x 105Pa. Calculate
i)
the number of moles of gas in the balloon
PV = nRT
n=PV/RT
n=1.0 x 105 x 1.3 x 104 /8.31 x 290
number of moles = …5.39 x 105…………. (2)
ii)
the mass of gas in the balloon.
n = mass of gas / molar mass of gas = n x molar mass of gas
Mass = ……2156…….kg (1)
d)
The internal energy of the helium gas is equal to the random kinetic
energy of all of its molecules. When the balloon is filled at ground level at a
temperature of 17˚C the internal energy is 1900 MJ. Estimate the internal energy of
the helium when the balloon has risen to a height where the temperature is –43˚C
Internal energy proportional to the temperature so
E = 1.9 x 230/290
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Internal energy = ……1500………MJ (2)
3
a)
Very high temperature, for example, the temperature of the solar
corona at half a million degrees, are often stated without a complete unit, i.e. degrees
Celsius of Kelvin.
Suggest why it is unnecessary to give degrees Celsius or Kelvin in this case.
……because there is only 273 degrees between 0 Kelvin and 0 Celsius when dealing
with thousands it becomes meaningless.……………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………… (2)
b)
Two students attempt the same experiment to find how air pressure
varies with temperature. They heat identical sealed glass flasks of air, to be
considered as an ideal gas, in an oil bath. The flasks are heated from 300k to 400k.
The pressure in flask A rises from atmospheric pressure, p0, as expected, but the
pressure in flask B remains at p0 because the rubber bung is defective and air leaks out
of the flask.
i)
Calculate the pressure in flask A at 400K in terms of p0.
P1V1/T1 = P2V2/T2 – V1=V2
p/400 = p0/300
ii)
to flask A at 400K.
f
Pressure = …1.3 p0………………… (2)
Calculate the fraction, f, of gas molecules in flask B compared
=
number of gas molecules in B at 400K
number of gas molecules in A at 400K
number of gas molecules is proportional to pressure (or 1/T) ( from PV = nRT)
f = ………0.75………………. (2)
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Lesson 45 questions – Boltzmann Constant
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Avogadro’s number (NA) = 6.02x1023
Ideal gas constant (R) = 8.31 Jmol-1K-1
mol-1
1.
Use the data given on this page to calculate the value of the Boltzmann
constant (k)
Boltzmann constant (k)
= R/NA
= 8.31/6.02x1023
k = ……1.38x1023 ……… units… JK-1………. (3)
2.
Calculate:
(a)
the kinetic energy of an individual gas molecule of mass 3.5x10-26 kg
moving at a speed of 600 ms-1.
2
½ mv = ½ x 3.5x10-26x6002
Kinetic energy = …6.3x10-21 ………….. J (3)
(b)
the kinetic energy of the gas molecules in a cylinder at a temperature
of 20oC
3/2 kT = 3/2x1.38x10-23x293
Kinetic energy = ……6.07x10-21 ……….. J (3)
3.
A cylinder contains 2 moles of a gas composed of molecules with a mass of
-26
2.5x10 kg moving with an r.m.s. speed of 500 ms-1. Calculate the temperature of the
gas.
RT
T
= 1/3 mNc2
= 1/3 [2xNAmc2/R]
= 1/3 x[2x6.02x1023x2.5x10-26x5002]
= 302 K
Temperature = …29……….. oC (4)
4.
A container of volume 1.5 m3 is full of gas with a density of 1.8 kgm-3 at a
pressure of 1.4x105 Pa.
(a)
Calculate the root mean square velocity of the molecules within the
gas.
2
1.4x105x1.5x3/1.8 = <c>2
<c> = 592 ms-1
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root mean square velocity = ……592………… ms-1 (3)
(b)
If the mass of an individual molecule of the gas is 3.0x10-27 kg
calculate the number of molecules in the container
Density (
N = V/m
= mN/V
= 1.8x1.5/3x10-27
Number of molecules = ……9x1026…………..(3)
5.
A deuterium plasma contains deuterium ions of mass 4.9x10-27 kg at a
temperature of 150x106 K.
Calculate:
(a) the r.m.s. speed and
RT = 1/3 mNc2
<c>2 = 3RT/mN
= [3x8.31x150x106]/[4.9x10-27x6.02x1023]
= 1.2677x1012
<c> = 1.13x106 ms-1
r.m.s. speed = ……1.13x106 ………….. ms-1 (3)
(b) the kinetic energy of a deuterium ion in the plasma
½ mv2 =½ (4.9x10-27)(1.13x106) 2
Kinetic energy = ……3.12 x10-15……………….. J (3)
6.
The mean kinetic energy of a gas molecule at an absolute temperature T is
given by:
kinetic energy = 3RT/2NA
where R is the molar gas constant and NA is the Avogadro constant.
a)
Calculate the mean kinetic energy of gas atoms at 0 °C.
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Kinetic energy = ………5.7x10-21…………….. J (2)
b)
Determine the speed of carbon dioxide molecules at 0 °C. The molar
mass
of carbon dioxide is 44 g.
Speed = ……390………. ms-1 (5)
Calculate the change in the internal energy of one mole of carbon
c)
dioxide
gas when its temperature increases from 0 °C to 100 °C.
change in the internal energy = ……1.2kJ………… (3)
SOME
7
from the kinetic theory, gas pressure is given by:
P = ⅓ρ<c>2
a)
State the ideal gas equation for n number of moles
PV = nRT
(1)
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RAB Plymstock School
b)
By using the equation for pressure given by the kinetic theory, the
ideal gas equation and the definition of kinetic energy show that E = 3/2kT for the
mean translational kinetic energy of an atom.
P = ⅓ρ<c>2
2
Is the same as P  Nmv / 3V .
(1)
PV = nRT
Therefore:
nRT = 1/3 Nm<c>2
(1)
Now,
KE = 1/2m v2
(1)
Rearranging
3nRT / N = m<c>2
(1)
So,
1/2 m<c>2= (3/2) nRT/N
(1)
We know that
NA = N/n so n/N = 1/NA
(1)
(N is the number of molecules in a volume of gas, n is the amount of gas in
moles, NA is Avagadro’s number).
Therefore:
E = 1/2 m<c>2= (3/2) RT / NA = (3/2)kT
(1)
(7)
What can we deduce about the relationship between the mean
translational kinetic energy of a molecule and the thermodynamic
temperature of a gas?
……… the mean kinetic energy of a molecule of an ideal gas is proportional to the
thermodynamic temperature
………………………………………………………………...
………………………………………………………………………………………(1)
b)
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RAB Plymstock School