(a) Explain in terms of the kinetic theory why a tyre should not be overinflated. (b)The following results were obtained at room temperature in an experimen...
(a) Explain in terms of the kinetic theory why a tyre should not be overinflated.
(b)The following results were obtained at room temperature in an experiment to verify one of the gas laws using a glass syringe:
Pressure (P) of air in syringe (atm)
Volume (V) of air in syringe (cm\(^3\)
\(\frac{I}{V}\)
0.100
10.00
0.100
0.125
8.00
0.125
0.150
6.60
0.150
0.175
5.60
0.179
0.200
4.80
0.208
0.225
4.40
0.227
(i) Plot a graph of P against \(\frac{1}{v}\), using 1 cm to represent 0.01 atm on the vertical axis and 1cm to represent 0.02 unit on the horizontal axis.
(ii) Which of the gas laws is in agreement with the results?
(c) The flow chart below represents the stages involved in the manufacture of H\(_2\)SO\(_4\).
+x +Conc. H\(_2\)SO\(_4\) +H\(_2\)O
S + O\(_2\) \(\to\) SO\(_2\) \(\to\) SO\(_3\) \(\to\) Y \(\to\) Conc H\(_2\)SO\(_4\)
stage I stage II stage III stage IV
(i) Name the process represented by the chart.
(ii) Identify reactant X and product Y.
(iii) What are the operating temperature and pressure at stage II?
(iv) Mention the stage which requires a catalyst and state the catalyst used.
(v) Give the reason why the SO\(_3\) produced in stage II is not dissolved directly in water to form the acid
(d) When K\(_4\)Cr\(_2\)C\(_7\) dissolves in water, the following equilibrium is established:
(i) State the colour observed on adding a few drops of dilute H\(_2\)SO\(_4\) to the system.
(ii) Explain your answer in (d)(1).
(iii) What principle is applicable to this explanation?
(a) Why a tyre should not be overinflated (kinetic theory)
According to the kinetic theory, the air inside a tyre consists of tiny particles in continuous rapid random motion. These particles constantly collide with the inner walls of the tyre, and the total force of these collisions per unit area is the pressure the gas exerts. When the tyre is overinflated, more particles are forced into the same fixed volume, so the frequency of collisions with the walls increases and the pressure rises sharply. In addition, as the vehicle moves, friction warms the trapped air; the particles gain kinetic energy and move faster, striking the walls harder and more often, which raises the pressure still further. If the internal pressure exceeds the strength of the tyre material, the tyre bursts. The tyre should therefore not be overinflated.
(b) Verifying a gas law with a glass syringe
The recorded readings, with \(\tfrac{1}{V}\) computed to three significant figures, are:
Pressure, \(P\) (atm)
Volume, \(V\) (cm\(^3\))
\(\dfrac{1}{V}\) (cm\(^{-3}\))
\(P\times V\) (atm cm\(^3\))
0.100
10.00
0.100
1.00
0.125
8.00
0.125
1.00
0.150
6.60
0.152
0.99
0.175
5.60
0.179
0.98
0.200
4.80
0.208
0.96
0.225
4.40
0.227
0.99
(i) Plotting \(P\) (vertical axis) against \(\tfrac{1}{V}\) (horizontal axis) gives a straight line passing through the origin:
P plotted against 1/V gives a straight line through the origin; the constant gradient (~1.0 atm cm^3) confirms PV = constant, verifying Boyle's Law.
The graph is a straight line through the origin, which shows that \(P\) is directly proportional to \(\tfrac{1}{V}\). The gradient of the line,
which equals the constant product \(P\times V\) in the last column of the table. Hence \(PV=\text{constant}\).
(ii) The result that agrees with the data is Boyle's Law, which states that at constant temperature the volume of a fixed mass of gas is inversely proportional to its pressure, i.e. \(P\propto\tfrac{1}{V}\), so \(PV=\text{constant}\).
(c) Manufacture of H\(_2\)SO\(_4\)
(i) The process represented by the chart is the Contact Process.
(ii) Reactant X is oxygen, O\(_2\) (air); product Y is oleum (fuming sulphuric acid), H\(_2\)S\(_2\)O\(_7\).
(iii) At Stage II the operating temperature is about 450 °C (in the range 400-500 °C) and the pressure is about 1-2 atm (approximately atmospheric).
(iv) The stage that requires a catalyst is Stage II (the conversion of SO\(_2\) to SO\(_3\)), and the catalyst used is vanadium(V) oxide, V\(_2\)O\(_5\).
(v) The SO\(_3\) is not dissolved directly in water because the reaction is strongly exothermic; the large amount of heat released would vaporise the water and create a dense, choking mist (fog) of fine sulphuric acid droplets that is difficult to condense and would escape the plant. Instead SO\(_3\) is absorbed in concentrated H\(_2\)SO\(_4\) to form oleum, which is then safely diluted with water.
where \(\text{Cr}_2\text{O}_7^{2-}\) (dichromate) is orange and \(\text{CrO}_4^{2-}\) (chromate) is yellow.
(i) On adding a few drops of dilute H\(_2\)SO\(_4\), the colour changes from yellow to orange (the solution becomes orange).
(ii) Dilute H\(_2\)SO\(_4\) supplies H\(^+\) ions, increasing their concentration on the right-hand side of the equilibrium. The system responds by shifting the position of equilibrium to the left so as to remove some of the added H\(^+\). This produces more of the orange dichromate ion, \(\text{Cr}_2\text{O}_7^{2-}\), and less of the yellow chromate ion, so the solution turns orange.
(iii) The principle applicable to this explanation is Le Chatelier's Principle.
(a) Why a tyre should not be overinflated (kinetic theory)
According to the kinetic theory, the air inside a tyre consists of tiny particles in continuous rapid random motion. These particles constantly collide with the inner walls of the tyre, and the total force of these collisions per unit area is the pressure the gas exerts. When the tyre is overinflated, more particles are forced into the same fixed volume, so the frequency of collisions with the walls increases and the pressure rises sharply. In addition, as the vehicle moves, friction warms the trapped air; the particles gain kinetic energy and move faster, striking the walls harder and more often, which raises the pressure still further. If the internal pressure exceeds the strength of the tyre material, the tyre bursts. The tyre should therefore not be overinflated.
(b) Verifying a gas law with a glass syringe
The recorded readings, with \(\tfrac{1}{V}\) computed to three significant figures, are:
Pressure, \(P\) (atm)
Volume, \(V\) (cm\(^3\))
\(\dfrac{1}{V}\) (cm\(^{-3}\))
\(P\times V\) (atm cm\(^3\))
0.100
10.00
0.100
1.00
0.125
8.00
0.125
1.00
0.150
6.60
0.152
0.99
0.175
5.60
0.179
0.98
0.200
4.80
0.208
0.96
0.225
4.40
0.227
0.99
(i) Plotting \(P\) (vertical axis) against \(\tfrac{1}{V}\) (horizontal axis) gives a straight line passing through the origin:
P plotted against 1/V gives a straight line through the origin; the constant gradient (~1.0 atm cm^3) confirms PV = constant, verifying Boyle's Law.
The graph is a straight line through the origin, which shows that \(P\) is directly proportional to \(\tfrac{1}{V}\). The gradient of the line,
which equals the constant product \(P\times V\) in the last column of the table. Hence \(PV=\text{constant}\).
(ii) The result that agrees with the data is Boyle's Law, which states that at constant temperature the volume of a fixed mass of gas is inversely proportional to its pressure, i.e. \(P\propto\tfrac{1}{V}\), so \(PV=\text{constant}\).
(c) Manufacture of H\(_2\)SO\(_4\)
(i) The process represented by the chart is the Contact Process.
(ii) Reactant X is oxygen, O\(_2\) (air); product Y is oleum (fuming sulphuric acid), H\(_2\)S\(_2\)O\(_7\).
(iii) At Stage II the operating temperature is about 450 °C (in the range 400-500 °C) and the pressure is about 1-2 atm (approximately atmospheric).
(iv) The stage that requires a catalyst is Stage II (the conversion of SO\(_2\) to SO\(_3\)), and the catalyst used is vanadium(V) oxide, V\(_2\)O\(_5\).
(v) The SO\(_3\) is not dissolved directly in water because the reaction is strongly exothermic; the large amount of heat released would vaporise the water and create a dense, choking mist (fog) of fine sulphuric acid droplets that is difficult to condense and would escape the plant. Instead SO\(_3\) is absorbed in concentrated H\(_2\)SO\(_4\) to form oleum, which is then safely diluted with water.
where \(\text{Cr}_2\text{O}_7^{2-}\) (dichromate) is orange and \(\text{CrO}_4^{2-}\) (chromate) is yellow.
(i) On adding a few drops of dilute H\(_2\)SO\(_4\), the colour changes from yellow to orange (the solution becomes orange).
(ii) Dilute H\(_2\)SO\(_4\) supplies H\(^+\) ions, increasing their concentration on the right-hand side of the equilibrium. The system responds by shifting the position of equilibrium to the left so as to remove some of the added H\(^+\). This produces more of the orange dichromate ion, \(\text{Cr}_2\text{O}_7^{2-}\), and less of the yellow chromate ion, so the solution turns orange.
(iii) The principle applicable to this explanation is Le Chatelier's Principle.
