Kinetic Theory MCQs with Answers – Part 2 (Class 11 Physics)

Explanation: The molecule-scale form of the ideal-gas equation is \(PV=Nk_BT\). Here \(N\) is the number of molecules and \(k_B\) is Boltzmann constant. The mole-scale form \(PV=nRT\) uses \(n\) and \(R\). The two forms are consistent because \(N=nN_A\) and \(R=N_Ak_B\). The constant must match the counting scale: \(R\) goes with moles, while \(k_B\) goes with molecules.

Explanation: \( \textbf{Given pressure:} \) \(P=1.0\times10^5\,\text{Pa}\). \( \textbf{Given volume:} \) \(V=0.020\,\text{m}^3\). \( \textbf{Given temperature:} \) \(T=300\,\text{K}\). \( \textbf{Gas constant:} \) \(R=8.3\,\text{J mol}^{-1}\text{K}^{-1}\). \( \textbf{Required:} \) Amount of gas \(n\). \( \textbf{Ideal-gas equation:} \) \[ PV=nRT \] \( \textbf{Rearrangement:} \) \[ n=\frac{PV}{RT} \] \( \textbf{Substitution:} \) \[ n=\frac{(1.0\times10^5)(0.020)}{(8.3)(300)} \] \( \textbf{Numerator:} \) \[ PV=2.0\times10^3\,\text{J} \] \( \textbf{Denominator:} \) \[ RT=2490\,\text{J mol}^{-1} \] \( \textbf{Calculation:} \) \[ n=\frac{2000}{2490}\approx0.80\,\text{mol} \] \( \textbf{Final answer:} \) The amount of gas is approximately \(0.80\,\text{mol}\).

Explanation: For an ideal gas, \(PV=nRT\). In a closed rigid vessel, \(n\) and \(V\) remain constant. Therefore, \(P\) is directly proportional to \(T\). If the absolute temperature is doubled from \(300\,\text{K}\) to \(600\,\text{K}\), the pressure also doubles. The amount \(2.0\,\text{mol}\) does not change during the heating, so it does not weaken the proportionality. The result depends on doubling Kelvin temperature, not Celsius reading.

Explanation: \( \textbf{Initial data:} \) \(P_1=1.0\times10^5\,\text{Pa}\), \(V_1=2.0\,\text{L}\), and \(T_1=300\,\text{K}\). \( \textbf{Final data:} \) \(V_2=3.0\,\text{L}\) and \(T_2=450\,\text{K}\). \( \textbf{Condition:} \) Amount of gas is fixed. \( \textbf{Combined gas relation:} \) \[ \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \] \( \textbf{Reason for relation:} \) It follows from \(PV=nRT\) when \(n\) is constant. \( \textbf{Rearrangement:} \) \[ P_2=P_1\frac{V_1}{V_2}\frac{T_2}{T_1} \] \( \textbf{Substitution:} \) \[ P_2=(1.0\times10^5)\left(\frac{2.0}{3.0}\right)\left(\frac{450}{300}\right) \] \( \textbf{Temperature ratio:} \) \[ \frac{450}{300}=1.5 \] \( \textbf{Volume and temperature factors:} \) \[ \left(\frac{2}{3}\right)(1.5)=1 \] \( \textbf{Final answer:} \) \(P_2=1.0\times10^5\,\text{Pa}\), because the temperature increase and volume increase balance in this case.

Explanation: From \(PV=nRT\), the product \(PV\) has the unit \(\text{Pa m}^3\). Since \(1\,\text{Pa}=1\,\text{N m}^{-2}\), \(1\,\text{Pa m}^3=1\,\text{N m}=1\,\text{J}\). Therefore, \(R=\frac{PV}{nT}\) has unit \(\frac{\text{J}}{\text{mol K}}\). This is written as \(\text{J mol}^{-1}\text{K}^{-1}\). The unit \(\text{J K}^{-1}\) would miss the per-mole part required for \(R\).

Explanation: Row P is consistent because the mole-scale form of the ideal-gas equation is \(PV=nRT\). Row Q is also consistent because the molecule-scale form is \(PV=Nk_BT\). Row R mixes moles with Boltzmann constant, so it uses the wrong constant for the counting scale. Row S mixes molecule count with the gas constant \(R\), which belongs to the mole scale. The constants \(R\) and \(k_B\) are connected, but they cannot be swapped without also changing \(n\) and \(N\).

Explanation: For the molecule-scale description of an ideal gas, \[ PV=Nk_BT \] \( \textbf{Fixed quantities:} \) \(N\), \(k_B\), and \(T\) remain constant. \( \textbf{Therefore:} \) \[ PV=\text{constant} \] \( \textbf{Initial state:} \) \[ P_1V_1=\text{constant} \] \( \textbf{Final volume:} \) \[ V_2=2V_1 \] \( \textbf{Using } P_1V_1=P_2V_2 \textbf{:} \) \[ P_1V_1=P_2(2V_1) \] \( \textbf{Simplification:} \) \[ P_2=\frac{P_1}{2} \] \( \textbf{Final answer:} \) The pressure becomes \(\frac{P}{2}\), because the same molecules at the same temperature are spread through twice the volume.

Explanation: In the molecule-scale ideal-gas equation, \(PV=Nk_BT\). The product \(PV\) has the unit \(\text{Pa m}^3=\text{J}\). The symbol \(N\) is a pure molecule count, so it has no unit like \(\text{mol}\). Therefore, \(k_BT\) must have the unit of energy. Since \(T\) is measured in \(\text{K}\), \(k_B\) has the unit \(\text{J K}^{-1}\). The unit \(\text{J mol}^{-1}\text{K}^{-1}\) belongs to \(R\), because \(R\) is used with \(n\) in \(\text{mol}\).

Explanation: \( \textbf{Given molecule count:} \) \(N=4.0\times10^{23}\). \( \textbf{Temperature:} \) \(T=300\,\text{K}\). \( \textbf{Volume:} \) \(V=1.0\times10^{-2}\,\text{m}^3\). \( \textbf{Boltzmann constant:} \) \(k_B=1.38\times10^{-23}\,\text{J K}^{-1}\). \( \textbf{Required:} \) Pressure \(P\). \( \textbf{Molecule-scale ideal-gas equation:} \) \[ PV=Nk_BT \] \( \textbf{Rearrangement:} \) \[ P=\frac{Nk_BT}{V} \] \( \textbf{Substitution:} \) \[ P=\frac{(4.0\times10^{23})(1.38\times10^{-23})(300)}{1.0\times10^{-2}} \] \( \textbf{First product:} \) \[ (4.0\times10^{23})(1.38\times10^{-23})=5.52 \] \( \textbf{Numerator:} \) \[ 5.52\times300=1656\,\text{J} \] \( \textbf{Pressure:} \) \[ P=\frac{1656}{1.0\times10^{-2}}=1.656\times10^5\,\text{Pa} \] \( \textbf{Final answer:} \) The pressure is closest to \(1.66\times10^5\,\text{Pa}\).

Explanation: \( \textbf{Given amount:} \) \(n=1.5\,\text{mol}\). \( \textbf{Given Avogadro constant:} \) \(N_A=6.0\times10^{23}\,\text{mol}^{-1}\). \( \textbf{Required:} \) Molecule count \(N\). \( \textbf{Mole-to-molecule relation:} \) \[ N=nN_A \] \( \textbf{Reason for using it:} \) The molecule-scale ideal-gas equation needs the actual number of molecules. \( \textbf{Substitution:} \) \[ N=(1.5)(6.0\times10^{23}) \] \( \textbf{Calculation:} \) \[ N=9.0\times10^{23} \] \( \textbf{Temperature note:} \) The given \(400\,\text{K}\) is not needed to convert moles into molecules. \( \textbf{Final answer:} \) The sample contains \(9.0\times10^{23}\) molecules.

Explanation: The ideal-gas equation can be written as \(PV=nRT\) or as \(PV=Nk_BT\). The first form counts gas amount in \(\text{mol}\), while the second form counts individual molecules. Since \(N=nN_A\), the two equations must match for the same gas sample. This gives \(R=N_Ak_B\). Thus, \(R\) is a mole-scale constant and \(k_B\) is a molecule-scale constant. The role of \(N_A\) is to connect the two counting scales, not to convert pressure into volume.

Explanation: For \(n\) moles of ideal gas, the equation is \[ PV=nRT \] \( \textbf{Graph form:} \) \[ PV=(nR)T \] \( \textbf{Comparison with } y=mx \textbf{:} \) The vertical quantity \(PV\) plays the role of \(y\), and \(T\) plays the role of \(x\). \( \textbf{Slope:} \) \[ m=nR \] \( \textbf{Condition:} \) This interpretation holds when the amount of gas \(n\) is fixed. \( \textbf{Physical meaning:} \) A larger amount of gas gives a larger slope because more moles produce a larger \(PV\) at the same \(T\). \( \textbf{Final answer:} \) The slope is \(nR\).

Explanation: The ideal-gas equation for a fixed amount of gas is \(PV=nRT\). If \(n\) and \(R\) are constant, then \(PV\propto T\). Therefore, doubling \(PV\) requires doubling the absolute temperature. This reasoning uses \(T\) in \(\text{K}\), not in \(\degree\text{C}\). A doubled Celsius reading does not necessarily mean doubled molecular thermal energy. The proportionality is between \(PV\) and Kelvin temperature.

Explanation: \( \textbf{Initial state:} \) \(P_1=2.0\times10^5\,\text{Pa}\), \(V_1=3.0\times10^{-3}\,\text{m}^3\), and \(T_1=300\,\text{K}\). \( \textbf{Final data:} \) \(P_2=1.5\times10^5\,\text{Pa}\) and \(T_2=450\,\text{K}\). \( \textbf{Condition:} \) The amount of gas is fixed. \( \textbf{Combined ideal-gas relation:} \) \[ \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \] \( \textbf{Rearrange for } V_2 \textbf{:} \) \[ V_2=\frac{P_1V_1T_2}{P_2T_1} \] \( \textbf{Substitution:} \) \[ V_2=\frac{(2.0\times10^5)(3.0\times10^{-3})(450)}{(1.5\times10^5)(300)} \] \( \textbf{Separate ratios:} \) \[ V_2=3.0\times10^{-3}\left(\frac{2.0}{1.5}\right)\left(\frac{450}{300}\right) \] \( \textbf{Ratio values:} \) \[ \frac{2.0}{1.5}=\frac{4}{3}, \qquad \frac{450}{300}=\frac{3}{2} \] \( \textbf{Combined factor:} \) \[ \left(\frac{4}{3}\right)\left(\frac{3}{2}\right)=2 \] \( \textbf{Final answer:} \) \(V_2=6.0\times10^{-3}\,\text{m}^3\).

Explanation: The ideal-gas model treats molecules as very small compared with the volume of the container. It also neglects intermolecular forces except during collisions. These assumptions make the gas laws and kinetic-theory relations simpler. The model still gives molecules mass, motion, and kinetic energy. It does not mean the gas has no molecules or that molecules remain fixed in space. The word “ideal” means a simplified model with conditions under which real gases can be approximated well.

Explanation: Real gases behave more like ideal gases when the molecules are far apart and intermolecular attractions are less important. Low pressure usually means lower density, so the average separation between molecules is larger. High temperature gives molecules greater kinetic energy, making attractive forces less dominant in their motion. At high pressure and low temperature, molecular size and intermolecular forces become more significant. The ideal-gas model works best when real molecular volume and attraction can be neglected.

Explanation: At very high pressure, molecules are crowded closer together, so their own volume is no longer negligible compared with the container volume. At low temperature, molecular kinetic energy is lower, so intermolecular attractions can affect the motion more strongly. These effects are ignored in the simplest ideal-gas model. The gas still has measurable pressure, volume, and temperature. The deviation occurs because real molecules are not perfect point particles with no mutual forces.

Explanation: Row P is correct because low pressure and high temperature make a real gas closer to ideal behaviour. Row Q is correct because high pressure and low temperature often increase real-gas deviations. Row R is correct because high density makes the finite volume of molecules more important. Row S is not correct because near liquefaction, intermolecular attraction becomes important rather than perfectly ignorable. The ideal-gas approximation is condition-dependent, not automatically valid for every gas state.

Explanation: The ideal-gas equation does not explicitly include molecular size because the model assumes molecular volume is negligible. In a real gas at high pressure, the molecules are crowded closer together. Then the actual volume occupied by molecules may become significant compared with the vessel volume. The equation \(PV=nRT\) is useful under ideal-like conditions, but it does not make real molecular size disappear. High pressure is one of the common conditions where real-gas behaviour can differ from the ideal model.

Explanation: A low-density gas far from liquefaction is usually close to ideal behaviour. The molecules are relatively far apart, so intermolecular attractions have a smaller effect on the overall gas behaviour. In the ideal model, forces between molecules are neglected except during collisions. The molecules are not fixed in a lattice and they do not have zero speed at ordinary nonzero temperature. Pressure is explained by molecular impacts on the walls, not only by the weight of a top layer.