1. Kinetic theory mainly explains the behaviour of a gas by considering
ⓐ. only the shape of the container
ⓑ. molecular motion and collisions
ⓒ. colour changes during heating
ⓓ. weight of the container walls
Correct Answer: molecular motion and collisions
Explanation: Kinetic theory connects the visible properties of a gas with the microscopic motion of its molecules. A gas may be described macroscopically by quantities such as pressure \(P\), volume \(V\), and temperature \(T\), but kinetic theory looks at the molecules that produce those effects. The molecules are taken to be in continuous motion and to collide with one another and with the container walls. This molecular picture helps explain why a gas fills its container and why heating changes its pressure or volume. The theory does not begin with colour, container shape alone, or the weight of the walls; it begins with molecular motion as the cause behind gas behaviour.
2. Which description best represents molecular motion in an ordinary gas?
ⓐ. Molecules remain fixed at regular points
ⓑ. Molecules move only upward inside the container
ⓒ. Molecules move randomly in all directions
ⓓ. Molecules move in circular paths around the centre
Correct Answer: Molecules move randomly in all directions
Explanation: Gas molecules are not fixed like particles in an ideal crystal. They move continuously and change direction after collisions with other molecules and with the walls. The motion is random, so no single direction is preferred for all molecules in a gas at equilibrium. This randomness is important because it allows the gas to exert pressure on all walls of the container. The average velocity direction of many molecules may cancel out, but their speeds and kinetic energies are still meaningful.
3. A sealed container filled with air is kept on a table. The pressure exerted by the air on the side walls is mainly due to
ⓐ. the downward weight of air molecules only
ⓑ. momentum transfer in wall collisions
ⓒ. attraction between the walls and the table
ⓓ. molecules becoming motionless near the wall
Correct Answer: momentum transfer in wall collisions
Explanation: Gas pressure in kinetic theory is explained through molecular collisions with the container walls. When a molecule strikes a wall and rebounds, its momentum changes, so the wall receives an impulse. A very large number of such collisions per second produces a steady force on the wall. Pressure is this force per unit area, so it can act on side walls as well as the bottom wall. Treating gas pressure as only the weight of gas misses the microscopic collision mechanism that works in every direction.
4. Heating a gas without changing its chemical nature most directly increases the molecules'
ⓐ. total electric charge
ⓑ. molecular identity
ⓒ. number of protons in each molecule
ⓓ. average molecular kinetic energy
Correct Answer: average molecular kinetic energy
Explanation: In kinetic theory, absolute temperature is connected with the average translational kinetic energy of gas molecules. When the temperature of a gas increases, the molecules move faster on average. This does not mean every molecule has exactly the same speed, because a gas contains a distribution of molecular speeds. Heating does not change the number of protons or the molecular identity in ordinary physical conditions. The useful link is between \(T\) and average molecular kinetic energy, not between \(T\) and chemical composition.
5. An ideal-gas model treats gas molecules as particles for which molecular size and intermolecular attraction are
ⓐ. dominant at all distances
ⓑ. zero only during heating
ⓒ. neglected except during collisions
ⓓ. larger than the size of the container
Correct Answer: neglected except during collisions
Explanation: The ideal-gas model simplifies a real gas by assuming that molecular size is negligible compared with the volume of the container. It also neglects intermolecular forces except during collisions. These assumptions make it possible to connect \(P\), \(V\), and \(T\) using simple relations. The model does not say that molecules have no mass or that collisions do not matter. The phrase “ideal gas” means a simplified gas model, not a gas whose molecules are absent or motionless.
6. Match the symbols with their usual meanings in kinetic theory.
| Symbol | Meaning |
|---|
| P. \(P\) | 1. Absolute temperature |
| Q. \(V\) | 2. Pressure |
| R. \(T\) | 3. Volume |
| S. \(N\) | 4. Number of molecules |
ⓐ. P-3, Q-2, R-1, S-4
ⓑ. P-2, Q-1, R-3, S-4
ⓒ. P-4, Q-3, R-1, S-2
ⓓ. P-2, Q-3, R-1, S-4
Correct Answer: P-2, Q-3, R-1, S-4
Explanation: In kinetic theory and the ideal-gas equation, \(P\) denotes pressure and \(V\) denotes volume. The symbol \(T\) represents absolute temperature, which must be measured in \(\text{K}\) when used in gas relations. The symbol \(N\) is used for the number of molecules, not the number of moles. This distinction matters because \(N\) counts particles, while \(n\) counts amount of substance in \(\text{mol}\). Confusing \(N\) with \(n\) leads to mixing the molecule-scale equation with the mole-scale equation.
7. Which row contains a fully suitable SI unit pairing for the listed gas quantities?
| Row | Pressure \(P\) | Volume \(V\) | Temperature \(T\) | Speed \(v\) |
|---|
| P | \(\text{Pa}\) | \(\text{m}^3\) | \(\text{K}\) | \(\text{m s}^{-1}\) |
| Q | \(\text{J}\) | \(\text{m}^3\) | \(\text{K}\) | \(\text{m s}^{-1}\) |
| R | \(\text{Pa}\) | \(\text{m}\) | \(\degree\text{C}\) | \(\text{N}\) |
| S | \(\text{mol}\) | \(\text{Pa}\) | \(\text{K}\) | \(\text{J}\) |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row P
Explanation: Pressure \(P\) is measured in pascal, written as \(\text{Pa}\). Volume \(V\) is measured in \(\text{m}^3\), not \(\text{m}\), because volume is three-dimensional space. Absolute temperature \(T\) is measured in \(\text{K}\), especially in gas equations and kinetic-energy relations. Speed \(v\) has the unit \(\text{m s}^{-1}\). The row with \(\text{Pa}\), \(\text{m}^3\), \(\text{K}\), and \(\text{m s}^{-1}\) keeps all four quantities in their proper SI form.
8. A gas pushes normally on a wall with a total force of \(120\,\text{N}\) over an area of \(0.40\,\text{m}^2\). What pressure does the wall experience?
ⓐ. \(48\,\text{Pa}\)
ⓑ. \(120\,\text{Pa}\)
ⓒ. \(300\,\text{Pa}\)
ⓓ. \(480\,\text{Pa}\)
Correct Answer: \(300\,\text{Pa}\)
Explanation: \( \textbf{Given:} \) Force on the wall \(F=120\,\text{N}\).
\( \textbf{Given area:} \) \(A=0.40\,\text{m}^2\).
\( \textbf{Required:} \) Pressure \(P\) on the wall.
\( \textbf{Definition of pressure:} \)
\[
P=\frac{F}{A}
\]
\( \textbf{Why this relation applies:} \) Pressure is normal force per unit area, and the force is spread over the wall area.
\( \textbf{Substitution:} \)
\[
P=\frac{120}{0.40}
\]
\( \textbf{Calculation:} \)
\[
P=300\,\text{N m}^{-2}
\]
\( \textbf{Unit conversion:} \) \(1\,\text{N m}^{-2}=1\,\text{Pa}\).
\( \textbf{Final answer:} \) The pressure is \(300\,\text{Pa}\), so the area must be divided into the force rather than multiplied by it.
9. In gas equations, the symbol \(n\) is best interpreted as
ⓐ. number of molecules in the gas sample
ⓑ. amount of gas measured in \(\text{mol}\)
ⓒ. mass of one molecule in \(\text{kg}\)
ⓓ. molecular speed in \(\text{m s}^{-1}\)
Correct Answer: amount of gas measured in \(\text{mol}\)
Explanation: The symbol \(n\) represents the number of moles of a gas. It is different from \(N\), which represents the actual number of molecules. The two are connected by \(N=nN_A\), where \(N_A\) is Avogadro's constant. In the equation \(PV=nRT\), the gas amount is measured in \(\text{mol}\). In the equation \(PV=Nk_BT\), the same gas is described using the number of molecules, so the symbol choice tells us which scale is being used.
10. The relation connecting the gas constant \(R\), Avogadro's constant \(N_A\), and Boltzmann constant \(k_B\) is
ⓐ. \(R=\frac{k_B}{N_A}\)
ⓑ. \(R=N_Ak_B\)
ⓒ. \(k_B=N_AR\)
ⓓ. \(N_A=Rk_B\)
Correct Answer: \(R=N_Ak_B\)
Explanation: The gas constant \(R\) belongs to the mole-scale form of the ideal-gas equation, \(PV=nRT\). The Boltzmann constant \(k_B\) belongs to the molecule-scale form, \(PV=Nk_BT\). Since \(N=nN_A\), the two descriptions must give the same value of \(PV\) for the same gas. Substituting \(N=nN_A\) into \(PV=Nk_BT\) gives \(PV=nN_Ak_BT\). Comparing this with \(PV=nRT\) gives \(R=N_Ak_B\), so \(R\) is the per-mole form of the molecular constant \(k_B\).
11. A gas sample contains \(2.0\,\text{mol}\) of molecules. Using \(N_A=6.0\times10^{23}\,\text{mol}^{-1}\), how many molecules are present?
ⓐ. \(3.0\times10^{23}\)
ⓑ. \(6.0\times10^{23}\)
ⓒ. \(1.2\times10^{24}\)
ⓓ. \(2.0\times10^{24}\)
Correct Answer: \(1.2\times10^{24}\)
Explanation: \( \textbf{Given:} \) Amount of gas \(n=2.0\,\text{mol}\).
\( \textbf{Given constant:} \) \(N_A=6.0\times10^{23}\,\text{mol}^{-1}\).
\( \textbf{Required:} \) Number of molecules \(N\).
\( \textbf{Relation between moles and molecules:} \)
\[
N=nN_A
\]
\( \textbf{Why this relation applies:} \) Each mole contains \(N_A\) molecules, so \(n\) moles contain \(n\) times that number.
\( \textbf{Substitution:} \)
\[
N=(2.0)(6.0\times10^{23})
\]
\( \textbf{Calculation:} \)
\[
N=12.0\times10^{23}
\]
\( \textbf{Scientific notation:} \)
\[
12.0\times10^{23}=1.2\times10^{24}
\]
\( \textbf{Final answer:} \) The gas contains \(1.2\times10^{24}\) molecules; the unit \(\text{mol}^{-1}\) of \(N_A\) cancels with \(\text{mol}\).
12. A formula sheet shows the two relations \(PV=nRT\) and \(PV=Nk_BT\). The safest way to choose between them is to check whether the data gives
ⓐ. pressure in \(\text{Pa}\) or pressure in \(\text{N}\)
ⓑ. volume in \(\text{m}^3\) or volume in \(\text{cm}\)
ⓒ. amount in \(\text{mol}\) or number of molecules
ⓓ. temperature in \(\text{K}\) or temperature in \(\degree\text{C}\)
Correct Answer: amount in \(\text{mol}\) or number of molecules
Explanation: Both ideal-gas equations describe the same gas, but they use different counting scales. The relation \(PV=nRT\) uses \(n\), the amount of gas in \(\text{mol}\). The relation \(PV=Nk_BT\) uses \(N\), the actual number of molecules. Temperature must still be in \(\text{K}\) for both equations, so Kelvin use does not decide between \(R\) and \(k_B\). The key distinction is whether the gas quantity is given as moles or as particles.
13. The root mean square speed \(v_{\text{rms}}\) of gas molecules means
ⓐ. the square root of mean squared speed
ⓑ. the ordinary arithmetic mean of speeds
ⓒ. the speed of every molecule in the gas
ⓓ. the speed of the container wall
Correct Answer: the square root of mean squared speed
Explanation: The rms speed is defined as \(v_{\text{rms}}=\sqrt{\overline{v^2}}\). This means that the speeds are squared first, then their mean is taken, and finally the square root is taken. It is not the same as the ordinary average speed. It also does not mean that every molecule moves with exactly \(v_{\text{rms}}\), because gas molecules have a range of speeds. The rms value is especially useful because translational kinetic energy depends on \(v^2\), not directly on \(v\).
14. For kinetic-theory formulas involving temperature, a gas at \(27\degree\text{C}\) should be entered as ______.
ⓐ. \(27\,\text{K}\)
ⓑ. \(100\,\text{K}\)
ⓒ. \(273\,\text{K}\)
ⓓ. \(300\,\text{K}\)
Correct Answer: \(300\,\text{K}\)
Explanation: Kinetic-theory relations use absolute temperature, so the temperature must be expressed in \(\text{K}\). The conversion from Celsius to Kelvin is \(T_{\text{K}}=T_{\degree\text{C}}+273\) for ordinary calculations. For \(27\degree\text{C}\), this gives \(T=27+273=300\,\text{K}\). Using \(27\) directly would make molecular kinetic energy far too small. The Kelvin scale is essential because molecular kinetic energy is proportional to absolute temperature.
15. Read the situation below and choose the best microscopic explanation.
A closed bicycle tyre is left in sunlight for some time. Its volume changes very little, but the tyre becomes harder and the air pressure inside increases.
ⓐ. The air molecules become larger in size and press on the wall
ⓑ. Molecules hit the tyre wall more often and more strongly
ⓒ. The air molecules stop moving and settle at the bottom
ⓓ. The tyre wall creates new air molecules from heat
Correct Answer: Molecules hit the tyre wall more often and more strongly
Explanation: Sunlight warms the air inside the tyre, increasing its absolute temperature \(T\). In kinetic theory, higher \(T\) means greater average molecular kinetic energy. When the tyre volume changes very little, the molecules remain confined in nearly the same space. Faster molecular motion causes more effective momentum transfer to the wall during collisions, so the pressure rises. The gas pressure increase is not due to molecules becoming larger or to heat creating new air molecules in the tyre.
16. Assertion: Gas pressure on a container wall is produced by molecular impacts on that wall.
Reason: During a wall collision, a molecule changes momentum and transfers impulse to the wall.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The assertion is true because kinetic theory explains gas pressure through molecular collisions with the container walls. When a molecule rebounds from a wall, its momentum component perpendicular to the wall changes. This change in momentum means an impulse is delivered to the wall. Many such impulses per second produce an average force, and pressure is force per unit area. The reason directly gives the microscopic cause of the pressure described in the assertion. This is why pressure can act on vertical walls even though the weight of the gas acts downward.
17. A data table lists some constants used in kinetic theory. Identify the row that needs correction.
| Row | Symbol | Meaning | Usual role |
|---|
| P | \(N_A\) | Avogadro's constant | Connects \(\text{mol}\) to number of molecules |
| Q | \(k_B\) | Boltzmann constant | Appears in molecule-scale energy and gas equations |
| R | \(R\) | Ideal gas constant | Appears in \(PV=nRT\) |
| S | \(n\) | Number of molecules | Appears in \(PV=Nk_BT\) |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: Row P is suitable because \(N_A\) connects the mole scale to the molecule count scale. Row Q is suitable because \(k_B\) is the constant used for a single-molecule scale of thermal energy and in \(PV=Nk_BT\). Row R is also suitable because \(R\) appears in the mole-scale equation \(PV=nRT\). Row S is not suitable because \(n\) represents the number of moles, not the number of molecules. The number of molecules is \(N\), so the molecule-scale equation uses \(N\) with \(k_B\).
18. A gas sample is described once by \(PV=nRT\) and once by \(PV=Nk_BT\). If the sample has \(n=3\,\text{mol}\), then the molecule count \(N\) used in the second equation is
ⓐ. \(\frac{3}{N_A}\)
ⓑ. \(3N_A\)
ⓒ. \(N_A+3\)
ⓓ. \(3k_B\)
Correct Answer: \(3N_A\)
Explanation: \( \textbf{Given:} \) Amount of gas \(n=3\,\text{mol}\).
\( \textbf{Required:} \) Number of molecules \(N\) for the equation \(PV=Nk_BT\).
\( \textbf{Link between the two descriptions:} \)
\[
N=nN_A
\]
\( \textbf{Why this relation applies:} \) One mole contains \(N_A\) molecules, so \(3\,\text{mol}\) contains three times \(N_A\).
\( \textbf{Substitution:} \)
\[
N=3N_A
\]
\( \textbf{Meaning of result:} \) The answer is symbolic because the question asks for \(N\) in terms of \(N_A\).
\( \textbf{Scale check:} \) A multi-mole gas sample should contain more molecules than \(N_A\), not less.
\( \textbf{Final answer:} \) The molecule count is \(3N_A\), and using \(3k_B\) would confuse a counting constant with an energy-temperature constant.
19. A gas spreads through the whole available space because its molecules
ⓐ. are fixed at one point until heated
ⓑ. have no mass inside a container
ⓒ. move continuously and randomly
ⓓ. attract only the upper wall
Correct Answer: move continuously and randomly
Explanation: Gas molecules are in continuous motion and their directions keep changing because of collisions. Since the motion is random, molecules do not stay collected near one corner of the container. They eventually reach all parts of the available space, so the gas fills the container. This does not require the molecules to lose mass or to be pulled specially toward one wall. The spreading of a gas is a direct everyday sign that molecular motion is not ordered like the motion of a rigid body.
20. When the smell of perfume slowly reaches the other side of a room, the observation supports the idea that
ⓐ. perfume molecules become heavier than air molecules
ⓑ. the room temperature must become zero
ⓒ. gas molecules can move only in straight vertical lines
ⓓ. perfume and air molecules move randomly
Correct Answer: perfume and air molecules move randomly
Explanation: The spreading of smell through air is explained by molecular motion. Perfume molecules mix with air molecules and move from one region to another through repeated random motion and collisions. The motion is not restricted to vertical lines, so the smell can spread sideways as well. The observation does not need a change in molecular weight or a fall of temperature to \(0\,\text{K}\). This example links an ordinary macroscopic observation with the microscopic motion assumed in kinetic theory.
