201. A gas is heated from \(300\,\text{K}\) to \(1200\,\text{K}\). For the same gas, the rms speed changes by a factor of
ⓐ. \(4\)
ⓑ. \(\frac{1}{2}\)
ⓒ. \(16\)
ⓓ. \(2\)
Correct Answer: \(2\)
Explanation: \( \textbf{Initial temperature:} \) \(T_1=300\,\text{K}\).
\( \textbf{Final temperature:} \) \(T_2=1200\,\text{K}\).
\( \textbf{Same gas condition:} \) Molar mass \(M\) remains constant.
\( \textbf{Rms-speed proportionality:} \)
\[
v_{\text{rms}}\propto\sqrt{T}
\]
\( \textbf{Speed factor:} \)
\[
\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}
\]
\( \textbf{Substitution:} \)
\[
\frac{v_2}{v_1}=\sqrt{\frac{1200}{300}}
\]
\( \textbf{Simplification:} \)
\[
\sqrt{4}=2
\]
\( \textbf{Final answer:} \) The rms speed becomes \(2\) times its initial value, even though the absolute temperature becomes \(4\) times as large.
202. A fixed sample of gas is warmed, and its speed distribution curve becomes broader. This broadening mainly indicates that
ⓐ. molecular speeds are spread over a wider range
ⓑ. every molecule now has the same speed
ⓒ. the number of molecules must become zero
ⓓ. molecular kinetic energy is independent of temperature
Correct Answer: molecular speeds are spread over a wider range
Explanation: A broader speed distribution means that molecular speeds are spread across a wider interval. Heating the gas increases the average translational kinetic energy, so more molecules are found at higher speeds than before. The peak may become lower because the same total number of molecules is distributed over a wider speed range. This does not imply that molecules disappear or that they all take the same speed. The broadening is a statistical sign of increased molecular agitation at higher temperature.
203. For an ideal gas, internal energy is mainly interpreted in kinetic theory as
ⓐ. gravitational potential energy of the container
ⓑ. chemical energy released by burning the gas
ⓒ. total molecular kinetic energy
ⓓ. energy stored only in the container walls
Correct Answer: total molecular kinetic energy
Explanation: In the kinetic theory of an ideal gas, molecules are assumed to have negligible intermolecular potential energy except during collisions. Therefore, the internal energy of an ideal gas is associated with the kinetic energies of its molecules. For a monatomic ideal gas, this means translational kinetic energy only. The gas may have pressure and volume, but its internal energy depends on molecular motion rather than on the container material. This is why temperature becomes the key variable for internal energy in an ideal gas.
204. For \(n\) moles of a monatomic ideal gas at absolute temperature \(T\), the internal energy is
ⓐ. \(U=\frac{1}{2}nRT\)
ⓑ. \(U=\frac{3}{2}nRT\)
ⓒ. \(U=3nRT\)
ⓓ. \(U=\frac{2}{3}nRT\)
Correct Answer: \(U=\frac{3}{2}nRT\)
Explanation: For one molecule of an ideal gas, the average translational kinetic energy is \(\frac{3}{2}k_BT\). For \(N\) molecules, the total translational kinetic energy is \(\frac{3}{2}Nk_BT\). Since \(N=nN_A\) and \(R=N_Ak_B\), this becomes \(\frac{3}{2}nRT\). A monatomic ideal gas has only translational kinetic energy in the simple kinetic-theory treatment. The factor \(\frac{3}{2}\) comes from the three translational directions of motion.
205. Two samples of the same monatomic ideal gas have the same number of moles and the same temperature but different pressures and volumes. Their internal energies are
ⓐ. equal
ⓑ. larger for the sample with larger pressure only
ⓒ. larger for the sample with larger volume only
ⓓ. zero for the sample with smaller volume
Correct Answer: equal
Explanation: For a monatomic ideal gas, \(U=\frac{3}{2}nRT\). This expression contains the amount of gas \(n\) and the absolute temperature \(T\), but not \(P\) or \(V\) separately. If the two samples have the same \(n\) and the same \(T\), their internal energies are equal. Different pressures and volumes can still satisfy the ideal-gas equation \(PV=nRT\) at the same \(T\). The internal energy comparison must be made from temperature and amount of gas, not from pressure or volume alone.
206. A monatomic ideal gas has \(2.0\,\text{mol}\) at \(300\,\text{K}\). Using \(R=8.3\,\text{J mol}^{-1}\text{K}^{-1}\), its internal energy is closest to
ⓐ. \(2.5\times10^3\,\text{J}\)
ⓑ. \(5.0\times10^3\,\text{J}\)
ⓒ. \(7.5\times10^3\,\text{J}\)
ⓓ. \(1.5\times10^4\,\text{J}\)
Correct Answer: \(7.5\times10^3\,\text{J}\)
Explanation: \( \textbf{Given amount:} \) \(n=2.0\,\text{mol}\).
\( \textbf{Temperature:} \) \(T=300\,\text{K}\).
\( \textbf{Gas constant:} \) \(R=8.3\,\text{J mol}^{-1}\text{K}^{-1}\).
\( \textbf{Gas type:} \) Monatomic ideal gas.
\( \textbf{Internal-energy relation:} \)
\[
U=\frac{3}{2}nRT
\]
\( \textbf{Reason for using it:} \) A monatomic ideal gas stores internal energy as translational molecular kinetic energy.
\( \textbf{Substitution:} \)
\[
U=\frac{3}{2}(2.0)(8.3)(300)
\]
\( \textbf{Simplifying the coefficient:} \)
\[
\frac{3}{2}\times2.0=3.0
\]
\( \textbf{Calculation:} \)
\[
U=3.0\times8.3\times300=7470\,\text{J}
\]
\( \textbf{Rounded value:} \)
\[
U\approx7.5\times10^3\,\text{J}
\]
\( \textbf{Final answer:} \) The internal energy is closest to \(7.5\times10^3\,\text{J}\).
207. A monatomic ideal gas changes state through two different paths between the same initial and final temperatures. The change in internal energy is
ⓐ. different for every path even if \(\Delta T\) is the same
ⓑ. determined only by \(\Delta T\) for a fixed amount of gas
ⓒ. always zero for any change of state
ⓓ. determined only by the final volume
Correct Answer: determined only by \(\Delta T\) for a fixed amount of gas
Explanation: For a monatomic ideal gas, \(U=\frac{3}{2}nRT\). If the amount of gas \(n\) is fixed, the change in internal energy is \(\Delta U=\frac{3}{2}nR\Delta T\). This depends only on the change in absolute temperature, not on the path followed between the two states. Different paths may involve different heat and work, but the internal-energy change is fixed by the temperature change. This is a useful connection between kinetic theory and thermodynamics.
208. A \(1.0\,\text{mol}\) monatomic ideal gas is heated from \(300\,\text{K}\) to \(500\,\text{K}\). Taking \(R=8.3\,\text{J mol}^{-1}\text{K}^{-1}\), the change in internal energy is closest to
ⓐ. \(0.83\times10^3\,\text{J}\)
ⓑ. \(1.66\times10^3\,\text{J}\)
ⓒ. \(2.49\times10^3\,\text{J}\)
ⓓ. \(4.15\times10^3\,\text{J}\)
Correct Answer: \(2.49\times10^3\,\text{J}\)
Explanation: \( \textbf{Given amount:} \) \(n=1.0\,\text{mol}\).
\( \textbf{Initial temperature:} \) \(T_1=300\,\text{K}\).
\( \textbf{Final temperature:} \) \(T_2=500\,\text{K}\).
\( \textbf{Temperature change:} \)
\[
\Delta T=T_2-T_1=500-300=200\,\text{K}
\]
\( \textbf{Gas type:} \) Monatomic ideal gas.
\( \textbf{Change in internal energy:} \)
\[
\Delta U=\frac{3}{2}nR\Delta T
\]
\( \textbf{Substitution:} \)
\[
\Delta U=\frac{3}{2}(1.0)(8.3)(200)
\]
\( \textbf{Calculation inside:} \)
\[
8.3\times200=1660\,\text{J}
\]
\( \textbf{Including } \frac{3}{2} \textbf{:} \)
\[
\Delta U=1.5\times1660=2490\,\text{J}
\]
\( \textbf{Final answer:} \) The change in internal energy is \(2.49\times10^3\,\text{J}\).
209. A monatomic ideal gas undergoes an isothermal expansion. Its internal energy change is
ⓐ. positive because volume increases
ⓑ. negative because pressure decreases
ⓒ. zero; temperature is constant
ⓓ. equal to \(nRT\) because the gas expands
Correct Answer: zero; temperature is constant
Explanation: For a monatomic ideal gas, internal energy is \(U=\frac{3}{2}nRT\). In an isothermal process, the temperature \(T\) remains constant. If the amount of gas is fixed, a constant \(T\) means \(U\) does not change. During expansion, pressure and volume may change, but their changes do not separately decide internal energy for an ideal gas. The zero value of \(\Delta U\) comes from \(\Delta T=0\), not from the process being expansion.
210. Study the table for a fixed amount of monatomic ideal gas. Select the row that gives the correct sign of \(\Delta U\).
| Row | Temperature change | Sign of \(\Delta U\) |
|---|
| P | \(T_2\gt T_1\) | Positive |
| Q | \(T_2=T_1\) | Zero |
| R | \(T_2\lt T_1\) | Negative |
| S | \(T_2\gt T_1\) | Always negative |
ⓐ. Rows P, Q, and R
ⓑ. Rows P, Q, and S
ⓒ. Rows P, R, and S
ⓓ. Rows Q, R, and S
Correct Answer: Rows P, Q, and R
Explanation: For a fixed amount of monatomic ideal gas, \(\Delta U=\frac{3}{2}nR\Delta T\). If \(T_2\gt T_1\), then \(\Delta T\) is positive and \(\Delta U\) is positive. If \(T_2=T_1\), then \(\Delta T=0\) and \(\Delta U=0\). If \(T_2\lt T_1\), then \(\Delta T\) is negative and \(\Delta U\) is negative. Row S contradicts the direct proportionality between \(\Delta U\) and \(\Delta T\).
211. A graph of internal energy \(U\) versus absolute temperature \(T\) is drawn for a fixed amount of monatomic ideal gas. The slope of the graph is
ⓐ. \(\frac{2}{3}nR\)
ⓑ. \(\frac{3}{2}RT\)
ⓒ. \(\frac{nR}{T}\)
ⓓ. \(\frac{3}{2}nR\)
Correct Answer: \(\frac{3}{2}nR\)
Explanation: \( \textbf{Internal-energy relation:} \)
\[
U=\frac{3}{2}nRT
\]
\( \textbf{Fixed quantity:} \) The amount of gas \(n\) is fixed.
\( \textbf{Graph variables:} \) Vertical axis is \(U\), and horizontal axis is \(T\).
\( \textbf{Graph form:} \)
\[
U=\left(\frac{3}{2}nR\right)T
\]
\( \textbf{Comparison with } y=mx \textbf{:} \) \(U\) plays the role of \(y\), while \(T\) plays the role of \(x\).
\( \textbf{Slope:} \)
\[
m=\frac{3}{2}nR
\]
\( \textbf{Physical meaning:} \) A larger amount of gas gives a larger rise in internal energy for the same rise in temperature.
\( \textbf{Final answer:} \) The slope is \(\frac{3}{2}nR\).
212. A fixed amount of monatomic ideal gas is compressed so that its temperature rises from \(300\,\text{K}\) to \(450\,\text{K}\). The ratio of final to initial internal energy is
ⓐ. \(2:3\)
ⓑ. \(1:1\)
ⓒ. \(3:2\)
ⓓ. \(9:4\)
Correct Answer: \(3:2\)
Explanation: \( \textbf{Initial temperature:} \) \(T_1=300\,\text{K}\).
\( \textbf{Final temperature:} \) \(T_2=450\,\text{K}\).
\( \textbf{Internal energy for fixed monatomic gas:} \)
\[
U=\frac{3}{2}nRT
\]
\( \textbf{Fixed quantities:} \) \(n\), \(R\), and \(\frac{3}{2}\) remain constant.
\( \textbf{Therefore:} \)
\[
U\propto T
\]
\( \textbf{Energy ratio:} \)
\[
\frac{U_2}{U_1}=\frac{T_2}{T_1}
\]
\( \textbf{Substitution:} \)
\[
\frac{U_2}{U_1}=\frac{450}{300}
\]
\( \textbf{Simplification:} \)
\[
\frac{U_2}{U_1}=\frac{3}{2}
\]
\( \textbf{Final answer:} \) The final-to-initial internal-energy ratio is \(3:2\).
213. Degrees of freedom of a molecule mean the
ⓐ. independent ways to store energy
ⓑ. number of walls in the gas container
ⓒ. number of molecules in one mole
ⓓ. pressure exerted by one molecule only
Correct Answer: independent ways to store energy
Explanation: Degrees of freedom describe the independent modes through which a molecule can store energy. These may include translational, rotational, and vibrational modes depending on the molecular structure and temperature range. A monatomic molecule in three-dimensional space has translational degrees of freedom. More complex molecules can also have rotational and vibrational degrees of freedom under suitable conditions. The idea is about energy storage modes of a molecule, not about the container shape or Avogadro's number.
214. A monatomic molecule moving freely in three-dimensional space has
ⓐ. \(1\) translational degree of freedom
ⓑ. \(2\) translational degrees of freedom
ⓒ. \(3\) translational degrees of freedom
ⓓ. \(6\) translational degrees of freedom
Correct Answer: \(3\) translational degrees of freedom
Explanation: A molecule moving in ordinary three-dimensional space can translate independently along the \(x\), \(y\), and \(z\) directions. These three independent translational motions correspond to three translational degrees of freedom. A monatomic molecule is treated as a point-like particle for this purpose, so it does not have rotational energy in the simple ideal-gas treatment. Its translational energy is therefore associated with three independent directions. This is the same three-direction idea that gives the factor \(\frac{3}{2}\) in monatomic ideal-gas energy.
215. In the expression \(U=\frac{f}{2}nRT\), the symbol \(f\) represents
ⓐ. pressure force on one wall
ⓑ. active degrees of freedom
ⓒ. number of moles only
ⓓ. frequency of wall collision only
Correct Answer: active degrees of freedom
Explanation: The symbol \(f\) represents the number of active degrees of freedom of the molecule. Each active quadratic degree of freedom contributes energy according to the equipartition idea used later in kinetic theory. For \(n\) moles, the internal energy is written as \(U=\frac{f}{2}nRT\). For a monatomic ideal gas at ordinary conditions, \(f=3\), which gives \(U=\frac{3}{2}nRT\). The symbol \(f\) must not be confused with force, even though force is also often denoted by a similar letter in other contexts.
216. Consider the following statements about degrees of freedom.
I. They represent independent ways of storing molecular energy.
II. Translational degrees of freedom exist for molecules moving in space.
III. They always equal Avogadro's number for any gas.
ⓐ. II and III only
ⓑ. I and III only
ⓒ. I, II, and III
ⓓ. I and II only
Correct Answer: I and II only
Explanation: Statement I is correct because degrees of freedom describe independent energy modes. Statement II is also correct because translational motion along independent coordinate directions gives translational degrees of freedom. Statement III is not correct because \(N_A\) is the number of particles in one mole, not a count of energy modes for one molecule. The number of degrees of freedom depends on molecular structure and active modes. A monatomic gas and a diatomic gas can have different \(f\) values even though one mole of each contains \(N_A\) molecules.
217. Match the molecular motion type with the best description.
| Motion type | Description |
|---|
| P. Translational | 1. Motion of the molecule's centre of mass through space |
| Q. Rotational | 2. Turning motion about an axis |
| R. Vibrational | 3. Periodic relative motion of atoms within a molecule |
| S. Mole count | 4. Number of molecules in a sample |
ⓐ. P-2, Q-1, R-3, S-4
ⓑ. P-1, Q-3, R-2, S-4
ⓒ. P-4, Q-2, R-3, S-1
ⓓ. P-1, Q-2, R-3, S-4
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: Translational motion is the motion of the molecule's centre of mass through space. Rotational motion is turning motion about an axis. Vibrational motion involves atoms in a molecule oscillating relative to one another. Mole count is not a type of mechanical motion, but it describes how many molecules are present in the sample. Separating these ideas helps avoid mixing molecular energy modes with particle-count quantities.
218. A diatomic molecule at ordinary temperature is often treated as having translational and rotational degrees of freedom, while vibrational modes are neglected. This treatment means that
ⓐ. all possible molecular motions are always active equally
ⓑ. only active energy modes are counted
ⓒ. molecules have no translational motion
ⓓ. the gas must have zero internal energy
Correct Answer: only active energy modes are counted
Explanation: The number \(f\) counts active degrees of freedom, not every motion that can be imagined mathematically. At ordinary temperatures, a diatomic molecule is commonly treated as having three translational and two rotational degrees of freedom active. Vibrational modes are often neglected treatment at such temperatures. This does not mean the molecule has no motion or no internal energy. It means the energy calculation includes only the modes that are active under the stated conditions.
219. According to the law of equipartition of energy, each active quadratic degree of freedom contributes an average energy of
ⓐ. \(\frac{1}{2}k_BT\) per molecule
ⓑ. \(\frac{3}{2}k_BT\) per molecule
ⓒ. \(k_BT^2\) per molecule
ⓓ. \(\frac{1}{2}RT\) per molecule
Correct Answer: \(\frac{1}{2}k_BT\) per molecule
Explanation: The law of equipartition of energy assigns equal average energy to each active quadratic degree of freedom. For one molecule, each such degree of freedom contributes \(\frac{1}{2}k_BT\). If a molecule has \(f\) active degrees of freedom, its average energy is \(\frac{f}{2}k_BT\). For \(n\) moles, the corresponding internal energy is \(U=\frac{f}{2}nRT\). The expression \(\frac{3}{2}k_BT\) belongs to a molecule with three active translational degrees of freedom, not to one single degree of freedom.
220. A monatomic ideal gas has \(f=3\) active degrees of freedom. The average energy per molecule is therefore
ⓐ. \(\frac{1}{2}k_BT\)
ⓑ. \(k_BT\)
ⓒ. \(\frac{3}{2}k_BT\)
ⓓ. \(3k_BT\)
Correct Answer: \(\frac{3}{2}k_BT\)
Explanation: A monatomic molecule in a gas can translate independently along three coordinate directions. Thus, it has \(f=3\) active degrees of freedom in the simple kinetic-theory model. By equipartition, each active degree of freedom contributes \(\frac{1}{2}k_BT\) per molecule. The total average energy per molecule is therefore \(3\times\frac{1}{2}k_BT=\frac{3}{2}k_BT\). This is the same result obtained earlier from the pressure relation and the ideal-gas equation.
