OCR (A) specifications: 5.4.10c,d,e,i,j,k
Chapter 11
Ideal gases
Worksheet
Worked examples
Practical: Determining absolute zero of
temperature from the ‘pressure law’
End-of-chapter test
Marking scheme: Worksheet
Marking scheme: End-of-chapter test
Worksheet
Avogadro constant NA = 6.02 × 1023 mol–1
universal gas constant R = 8.31 J mol–1 K–1
Intermediate level
1
Determine the number of atoms or molecules in each of the following:
a
1.0 mole of carbon;
[1]
b
3.6 moles of water;
[1]
c
0.26 moles of helium.
[1]
2
The molar mass of helium is 4.0 g. Determine the mass of a single atom of helium
in kilograms.
[2]
3
The molar mass of uranium is 238 g.
4
5
a
Calculate the mass of one atom of uranium.
b
A small rock contains 0.12 g of uranium. For this rock, calculate the number of:
i
moles of uranium;
[2]
ii
atoms of uranium.
[1]
Explain what is meant by the absolute zero of temperature.
a
b
[3]
Change the following temperatures from degrees Celsius into kelvin.
i
0 °C
ii
80 °C
iii –120 °C
[3]
Change the following temperatures from kelvin into degrees Celsius.
i
6
[2]
400 K
ii
272 K
iii 3 K
[3]
a
Write the ideal gas equation in words.
[1]
b
One mole of an ideal gas is trapped inside a rigid container of volume
0.020 m3. What pressure is exerted by the gas when the temperature
within the container is 293 K?
[3]
Higher level
7
A fixed amount of an ideal gas is trapped in a container of volume V. The pressure
exerted by the gas is P and its absolute temperature is T.
a
b
8
Using a sketch of PV against T, explain how you can determine the number
of moles of gas within the container.
[4]
Sketch a graph of PV against P when the gas is kept at a constant temperature.
Explain the shape of the graph.
[3]
A rigid cylinder of volume 0.030 m3 holds 4.0 g of air. The molar mass of air is 29 g.
a
b
Calculate the pressure exerted by the gas when the temperature within the
cylinder is 34 °C.
[4]
What is the temperature of the gas in degrees Celsius when the pressure is
twice your value from part a?
[4]
11 Ideal gases
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9 The diagram shows two insulated containers holding gas. The containers are
connected together by tubes of negligible volume.
valve
A
B
The internal volume of each container is 2.0 × 10–2 m3. The temperature within each
container is –13 °C. The gas in container A exerts a pressure of 180 kPa and the gas
in container B exerts a pressure of 300 kPa.
a
Show that the amount of gas within the two containers is about 4.4 moles.
[3]
b
The valve connecting the containers is slowly opened and the gases are allowed
to mix. The temperature within the containers remains the same. What is the
new pressure exerted by the gas within the containers?
[3]
10 The diagram shows a cylinder containing air at a
cylinder
temperature of 5.0 °C.
A force of 400 N is applied normally to the piston of
cross-sectional area 1.6 × 10–3 m2. The volume occupied
by the compressed air is 2.4 × 10–4 m3. The molar mass
of air is 29 g. Calculate:
400 N
a
the pressure exerted by the compressed air;
[2]
b
the number of moles of air inside the cylinder.
[3]
c
Use your answer to b to determine:
i
the mass of air inside the cylinder;
[1]
ii
the density of the air inside the cylinder.
[2]
piston cross-sectional
area = 1.6× 10–3 m2
11 The mean speed of a helium atom at a temperature of 0 °C is 1.3 km s–1.
What is the mean speed of helium atoms on the surface of a star where the
temperature is 10 000 K?
[6]
Extension
12 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.
[2]
b
Determine the speed of carbon dioxide molecules at 0 °C. The molar mass
of carbon dioxide is 44 g.
[5]
Calculate the change in the internal energy of one mole of carbon dioxide
gas when its temperature increases from 0 °C to 100 °C.
[3]
c
Total: ––– Score:
68
128
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%
11 Ideal gases
Worked examples
Example 1
A sealed container holds air at a pressure of 100 kPa
at 80 °C. The container is cooled to a temperature of
–10 °C. Assuming that there is negligible change in the
volume of the container, calculate the new pressure
exerted by the air in the container.
Initial
Final
Assuming that the air behaves as an ideal gas, we have:
PV = nRT
Hence:
P = 100 kPa
θ = 80 °C
PV
= nR
T
P=?
θ = –10°C
In the relation above, the terms n, R and V are all constant. Therefore:
P
= constant
T
or
P1 P2
=
T1 T2
The new pressure P is given by:
100 × 103
P
(273 + 80) = (273 – 10)
P=
It is very important to change the temperature
into kelvin. To change temperature from °C to
kelvin, just add 273.
263 × 100 × 103
= 7.45 × 104 Pa ≈ 75 kPa
353
Example 2
A particular filament lamp contains argon gas at a low pressure. When the lamp is
switched on, the temperature within the lamp increases from 20 °C to 80 °C. Calculate
the ratio:
speed of argon atom at 80 °C
speed of argon atom at 20 °C
The mean kinetic energy of an atom is proportional to the absolute temperature:
1 2
mv ∝ T
2
or
v2 ∝
2T
m
Since the mass m of the atom is constant, we have:
v∝
T
and so
v1
T1
v2 = T 2
If the speed at 80 °C is v80 and the speed at 20 °C is v20, then:
ratio
v80
273 + 80
=
=
v20
273 + 20
353
= 1.2 ≈ 1.1
293
Once again, do not forget to convert the temperature into kelvin.
Tip
353
= 1.2
293
The mean kinetic energy of an atom is directly proportional to the absolute
temperature. Therefore the mean kinetic energy also increases by the same factor
of 1.2. For a given atom, the kinetic energy is proportional to the speed2. Hence the
speed must increase by a factor of 1.2 ≈ 1.1 .
The temperature in kelvin increases by a factor of:
11 Ideal gases
© Cambridge University Press 2005
129
Practical
Determining absolute zero of temperature from the
‘pressure law’
Safety
Be careful when working around the heater and ensure that the hot beaker is stable
when stirring the water. Teachers and technicians should follow their school and
departmental safety policies and should ensure that the employer’s risk assessment has
been carried out before undertaking any practical work.
Apparatus
•
•
•
•
electrical heater (hot plate)
water bath
conical flask
pressure gauge or sensor
• thermometer
• crushed ice
• stirrer
Introduction
The details of the ideal gas equation are given on page 117 of Physics 2. In this
experiment you will investigate how the pressure exerted by a fixed amount of trapped
air depends on its temperature. You can use your results to estimate the value of
absolute zero of temperature. The apparatus is shown in the diagram.
pressure gauge
thin glass pipe
thermometer
rubber bung
water bath
flask
dry air
heater
Procedure
1
2
3
4
5
6
130
Put some water and crushed ice into the water bath and leave the water to cool.
Fully immerse the conical flask in the cold water.
Measure and record the temperature θ of the water in degrees Celsius and the
pressure P as measured by the pressure gauge.
Warm the water on the heater.
Carefully stir the water and measure the pressure for every 10 °C change in
temperature.
Continue taking readings until the water boils.
© Cambridge University Press 2005
11 Ideal gases
7
Plot a graph of pressure P against temperature θ on axes in the range –300 °C
to 100 °C.
extrapolation
P
absolute zero
0
–300
0
100
θ (°C)
8
Draw a straight line of best fit and extrapolate it until it crosses the temperature
axis. What is your experimental value for the absolute zero of temperature?
9
What is the relationship between the pressure P and the thermodynamic (absolute)
temperature T of the air?
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131
End-of-chapter test
Answer all questions.
Avogadro constant NA = 6.02 × 1023 mol–1
universal gas constant R = 8.31 J mol–1 K–1
1
2
3
The molar mass of water is 18 g. Calculate:
a
the number of molecules in 2.0 moles of water;
[1]
b
the number of molecules of water in a cup containing 200 g of water.
[3]
The temperature of solid A is 210 K and the temperature of solid B is –70 °C.
Which of these two solids is at a higher temperature?
a
[2]
Use the ideal gas equation to explain why for a fixed amount of gas:
PV
= constant
T
where P is the pressure exerted by the gas, V is the volume occupied by the gas
and T is the thermodynamic (absolute) temperature.
[2]
b
4
At a research station in the Arctic a weather balloon is filled with helium gas.
The volume of the balloon is 0.56 m3. At a temperature of –20 °C the pressure
exerted by the helium gas is 180 kPa.
i
Calculate the number of moles of helium.
[3]
ii
When the balloon is released it reaches a stable height above the ground.
The pressure within the balloon drops to 120 kPa. Assuming there is no
change in the temperature, calculate the new volume of the balloon.
[3]
At a temperature of 30 °C, the mean kinetic energy of a proton is 6.3 × 10–21 J. The
mass of a proton is 1.7 × 10–27 kg. For a single proton, calculate:
a
its speed at a temperature of 30 °C;
[3]
b
its mean kinetic energy at 600 °C.
[3]
Total: ––– Score:
20
132
© Cambridge University Press 2005
%
11 Ideal gases
Marking scheme
Worksheet
1
a
Number of atoms = number of moles × NA
number of atoms = 1.0 × 6.02 × 1023 ≈ 6.0 × 1023 [1]
2
b
Number of molecules = 3.6 × 6.02 × 1023 ≈ 2.2 × 1024 [1]
c
Number of atoms = 0.26 × 6.02 × 1023 ≈ 1.6 × 1023 [1]
There are 6.02 × 1023 atoms in 4.0 g of helium. [1]
mass of atom =
3
a
0.004
= 6.645 × 10–27 kg ≈ 6.6 × 10–27 kg [1]
6.02 × 1023
There are 6.02 × 1023 atoms in 0.238 kg of uranium. [1]
mass of atom =
b
i
ii
0.238
= 3.95 × 10–25 kg ≈ 4.0 × 10–25 kg [1]
6.02 × 1023
Number of moles =
mass of uranium
[1]
molar mass of uranium
number of moles =
0.12 g
= 5.11 × 10–4 ≈ 5.1 × 10–4 [1]
235 g
Number of atoms = number of moles × NA
number of atoms = 5.11 × 10–4 × 6.02 × 1023 ≈ 3.1 × 1020 [1]
4
The absolute zero of temperature is –273 °C or 0 K. [1]
This is the lowest temperature any substance can have. [1]
At absolute zero of temperature, the substance has minimum internal energy. [1]
5
a
i
T = 273 + 0 = 273 K [1]
ii
T = 273 + 80 = 353 K [1]
iii T = 273 – 120 = 153 K [1]
b
i
θ = 400 – 273 = 127 °C [1]
ii
θ = 272 – 273 = –1 °C [1]
iii θ = 3 – 273 = –270 °C [1]
6
a
Pressure × volume
= number of moles × universal gas constant × thermodynamic temperature [1]
b
PV = nRT [1]
P=
nRT 1.0 × 8.31 × 293
=
[1]
V
0.020
P ≈ 1.2 × 105 Pa (120 kPa) [1]
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7
a
PV = nRT [1]
Comparing this equation with y = mx, we have:
y = PV,
x = T,
gradient, m = nR [1]
A graph of PV against T is a straight line through the origin.
PV
Correct graph. [1]
n=
gradient
R
(R = molar gas constant) [1]
gradient = nR
0
0
b
T
PV = nRT [1]
At a constant temperature, the product PV is a constant. [1]
Hence a graph of PV against P is a straight horizontal line (see below). [1]
PV
0
0
8
a
P
PV = nRT [1]
n=
4.0
= 0.138 [1]
29
P=
nRT 0.138 × 8.31 × (273 + 34)
=
[1]
V
0.030
P = 1.17 × 104 Pa ≈ 1.2 × 104 Pa (12 kPa) [1]
b
P
= constant when the volume of the gas is constant. [1]
T
The pressure is doubled, hence the absolute temperature of the gas is also
doubled. [1]
Therefore:
temperature = 2 × (273 + 34) = 614 K [1]
temperature in °C = 614 – 273 = 341 °C ≈ 340 °C [1]
134
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11 Ideal gases
9
a
n=
PV
[1]
RT
n=
300 × 10 × 2.0 × 10
× 2.0 × 10
+ [ 8.31 × (273 – 13) ]
[ 1808.31× 10× (273
– 13) ]
3
3
–2
–2
[1]
Hence, n = 4.44 ≈ 4.4 moles [1]
b
Total volume, V = 4.0 × 10–2 m3,
P=
nRT
[1];
V
P=
T = 273 – 13 = 260 K
4.44 × 8.31 × 260
[1]
4.0 × 10–2
P ≈ 2.4 × 105 Pa (240 kPa) [1]
10 a P = AF = 400 –3 [1]
1.6 × 10
P = 2.5 × 105 Pa [1]
b
n=
PV
[1]
RT
n=
2.5 × 105 × 2.4 ×10–4
[1]
8.31 × (273 + 5.0)
n = 2.6 × 10–2 moles [1]
c
i
Mass = number of moles × molar mass
mass = 2.6 × 10–2 × 29 = 0.754 g ≈ 0.75 g [1]
ii
Density =
mass
volume
density =
0.754 × 10–3
[1]
2.4 × 10–4
density = 3.14 kg m–3 ≈ 3.1 kg m–3 [1]
11 Mean kinetic energy of atom ∝ absolute temperature [1]
1 2
2 mv ∝ T [1]
or
v2 ∝
2T
m
Since the mass m of the atom is constant, we have: v ∝ T [1]
The temperature of 0 °C in kelvin is: T = 273 K
The absolute temperature increases by a factor of:
10 000
(= 36.6) [1]
273
Hence the speed will increase by a factor of:
10 000
= 6.05 [1]
273
The speed of the atoms at 10 000 K is:
speed = 1.3 × 6.05 ≈ 7.9 km s –1 [1]
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12 a Mean kinetic energy = 3RT = 3 × 8.31 × 273
23 [1]
2NA
2 × 6.02 × 10
mean kinetic energy = 5.65 × 10–21 J ≈ 5.7 × 10–21 J [1]
b
There are 6.02 × 1023 atoms in 44 g of carbon dioxide. [1]
mass of molecule = m =
0.044
= 7.31 × 10–26 kg [1]
6.02 × 1023
1 2
mv = 5.65 × 10–21 [1]
2
v=
2 × 5.65 × 10–21
[1]
7.31 × 10–26
v ≈ 390 m s –1 [1]
c
Total kinetic energy of 1 mole of gas =
3RT
3RT
× NA=
[1]
2NA
2
The change in the internal energy of the gas is almost entirely kinetic energy.
3
3
change in internal energy = 2 R(373 – 273) = ×8.31 × 100 [1]
2
change in internal energy ≈ 1.2 kJ [1]
136
© Cambridge University Press 2005
11 Ideal gases
Marking scheme
End-of-chapter test
1
a
Number of molecules = number of moles × NA
number of molecules = 2.0 × 6.02 × 1023 ≈ 1.2 × 1024 [1]
b
Number of moles =
mass of water
[1]
molar mass of water
number of moles =
200
[1]
18
number of molecules = number of moles ×NA
number of molecules =
2
200
× 6.02 × 1023 = 6.69 × 1024 ≈ 6.7 × 1024 [1]
18
Temperature of B in kelvin is: T = 273 – 70 = 203 K [1]
Solid A has a higher temperature of 210 K. [1]
3
a
The ideal gas equation is:
PV = nRT [1]
For a fixed amount of gas, the number n of moles is a constant and R is a
constant for all ideal gases. [1]
Hence:
b
i
PV
= nR = constant.
T
n=
PV
[1]
RT
n=
180 × 103 × 0.56
[1]
8.31 × (273 – 20)
n = 47.9 moles ≈ 48 moles [1]
ii
At constant temperature, PV = constant [1]
180 × 103 × 0.56 = 120 × 103 × V [1]
V=
4
a
180 × 103 × 0.56
= 0.84 m3 [1]
120 × 103
1 2
mv = 6.3 × 10–21 [1]
2
v=
2 × 6.3 × 10–21
[1];
1.7 × 10–27
v ≈ 2.7 × 103 m s–1 [1]
b
Mean kinetic energy ∝ absolute temperature [1]
The temperature increases by a factor of:
273 + 600
= 2.88 [1]
273 + 30
The mean kinetic energy increases by the same factor. Hence:
mean kinetic energy = 2.88 × 6.3 × 10–21 J ≈ 1.8 × 10–20 J [1]
11 Ideal gases
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