Translational, Rotational & Vibrational Contributions - Physical Chemistry 2

Consider the translational partition function for a gas molecule in a cubic box. If the box volume decreases by a factor of 8 while temperature increases by a factor of 2, what happens to the translational contribution to the chemical potential?
Answer: Increases by $RT \ln(4)$ due to the net decrease in accessible translational states $\mu_{trans} = -RT \ln(q_{trans}/N) = -RT \ln(V(2\pi mkT)^{3/2}/h^3)$ . When $V \to V/8$ and $T \to 2T$ : $\Delta\mu = -RT \ln[(V/8)(2T)^{3/2}/(VT^{3/2})] = -RT \ln[2^{3/2}/8] = -RT \ln[1/4] = +RT \ln(4)$ . Chemical potential increases because the system is more constrained.
For a molecule with moment of inertia $I = 2.5 \times 10^{-46}$ kg⋅m², the rotational partition function at 350 K is calculated to be 85.7. If an external magnetic field aligns the molecular magnetic moment, effectively doubling the moment of inertia about one axis while leaving the others unchanged, how does this affect the rotational partition function?
Answer: The effect depends on which specific axis is affected and cannot be determined without more information When analyzing rotational partition functions, you need to understand how molecular geometry and external constraints affect rotational motion. The rotational partition function depends on the moments of inertia about all three principal axes of rotation. For this problem, you're dealing with a molecule where an external magnetic field doubles the moment of inertia about one axis while leaving the other two unchanged. The rotational partition function for a non-linear molecule is proportional to $\sqrt{I_A I_B I_C}$ , where $I_A$ , $I_B$ , and $I_C$ are the moments of inertia about the three principal axes. If one moment of inertia doubles (say $I_A$ becomes $2I_A$ ), then the new partition function becomes proportional to $\sqrt{2I_A I_B I_C} = \sqrt{2} \times \sqrt{I_A I_B I_C}$ . This means the partition function increases by a factor of $\sqrt{2} \approx 1.414$ . Therefore: $q_{rot,new} = 85.7 \times 1.414 = 121.2$ . However, since there's no option E provided in your choices, let's examine the given options: A) Correctly identifies the numerical result (121.2) and properly recognizes that increased moment of inertia leads to more accessible rotational states. B) Incorrectly assumes the magnetic field restricts rotation rather than just changing the moment of inertia. C) Wrongly suggests that affecting one axis doesn't matter, ignoring that all three axes contribute to the partition function. D) Gives an incorrect numerical value and misunderstands the physical effect. Remember: rotational partition functions depend on all three moments of inertia multiplicatively, so changing any one affects the total rotational accessibility.
Consider two diatomic molecules with the same mass but different bond strengths. Molecule A has a fundamental vibrational frequency twice that of molecule B. At a temperature where both molecules have the same rotational heat capacity, what is the ratio of the vibrational heat capacity of molecule A to that of molecule B?
Answer: The ratio depends on the specific temperature and cannot be determined from the given information When analyzing vibrational heat capacity in diatomic molecules, you need to consider how quantum mechanical effects influence thermal energy distribution. The vibrational heat capacity depends on both the vibrational frequency and the temperature through the Einstein model: $C_v = R\left(\frac{\hbar\omega}{k_BT}\right)^2 \frac{e^{\hbar\omega/k_BT}}{(e^{\hbar\omega/k_BT}-1)^2}$ The correct answer is D because the ratio of vibrational heat capacities depends on the specific temperature value, which isn't provided. Since molecule A has twice the vibrational frequency of molecule B ( $\omega_A = 2\omega_B$ ), the ratio $C_{v,A}/C_{v,B}$ becomes a function of the dimensionless parameter $x = \hbar\omega_B/k_BT$ . This ratio varies significantly with temperature and cannot be determined without knowing the actual temperature. A is incorrect because heat capacity doesn't simply scale with frequency squared - the exponential terms in the Einstein equation create a complex temperature dependence. B contains a grain of truth (higher frequency modes are less excited at lower temperatures) but incorrectly assumes this leads to a fixed ratio of 0.5. C is wrong because even in the quantum limit (high frequency or low temperature), the relationship isn't simply linear with frequency. Study tip: For vibrational heat capacity problems, always check whether you're given enough information to evaluate the temperature-dependent exponential terms. If the temperature isn't specified or you can't determine the $\hbar\omega/k_BT$ ratio, you likely can't calculate a definitive numerical answer.
For a linear polyatomic molecule, the high-temperature limit of the rotational partition function is proportional to $T^n$ . What is the value of $n$ , and what physical principle explains this dependence?
Answer: $n = 1$ , because linear molecules have only one significant moment of inertia for end-over-end rotation For linear molecules, $q_r = kT/\sigma hcB = kT/(\sigma hc/8\pi^2I) \propto T$ , so $n=1$ . This is because linear molecules rotate only end-over-end (two degenerate rotational modes) but with one moment of inertia. Choice A gives wrong reasoning (energy scaling vs. partition function), B confuses degrees of freedom with temperature dependence, C applies to nonlinear molecules.
A molecule undergoes a conformational change that doubles its moment of inertia while keeping vibrational frequencies unchanged. At constant temperature, how do the rotational and vibrational contributions to the heat capacity change?
Answer: Both rotational and vibrational heat capacities remain unchanged since only molecular geometry changed The rotational contribution to heat capacity is $C_{v,rot} = R$ per rotational degree of freedom in the high-T limit, independent of the moment of inertia magnitude. Vibrational heat capacity depends only on vibrational frequencies, which are unchanged. Choice A/B incorrectly assume heat capacity scales with moment of inertia. Choice C incorrectly invokes coupling.
Consider two isotopomers of the same diatomic molecule: $^{12}C^{16}O$ and $^{13}C^{16}O$ . At the same temperature, how do their translational, rotational, and vibrational partition functions compare?
Answer: Translational: $^{13}CO > ^{12}CO$ ; Rotational: $^{13}CO > ^{12}CO$ ; Vibrational: $^{13}CO > ^{12}CO$ $^{13}CO$ has larger mass, so: Translational $q_{trans} \propto m^{3/2}$ increases; Rotational $q_{rot} \propto 1/B \propto I \propto \mu$ increases; Vibrational $\omega \propto 1/\sqrt{\mu}$ decreases, so $q_{vib} = 1/(1-e^{-h\omega/kT})$ increases. All partition functions are larger for the heavier isotopomer. Other choices incorrectly predict some effects.
Two isotopomers of the same molecule differ only in the mass of one atom. The heavier isotopomer has rotational constants that are 0.85 times those of the lighter isotopomer. At a given temperature, how do their rotational partition functions compare?
Answer: The heavier isotopomer has a rotational partition function 1.61 times that of the lighter isotopomer When you encounter isotopomer problems, you're dealing with how molecular mass changes affect rotational properties. The key insight is understanding the relationship between rotational constants and partition functions. The rotational partition function for a diatomic molecule is $q_{rot} = \frac{T}{\sigma B}$ , where T is temperature, σ is the symmetry number, and B is the rotational constant. Since the isotopomers are the same molecule (just different masses), they have identical symmetry numbers and are at the same temperature. This means $q_{rot}$ is inversely proportional to the rotational constant B. If the heavier isotopomer has rotational constants that are 0.85 times those of the lighter isotopomer, then: $\frac{q_{rot,heavy}}{q_{rot,light}} = \frac{B_{light}}{B_{heavy}} = \frac{1}{0.85} = 1.18$ . Wait - that's not quite right for polyatomic molecules. For polyatomic molecules, $q_{rot} = \sqrt{\frac{\pi T^3}{ABC}}$ where A, B, and C are the three rotational constants. Since all three constants are 0.85 times smaller: $\frac{q_{rot,heavy}}{q_{rot,light}} = \sqrt{\frac{1}{(0.85)^3}} = \sqrt{\frac{1}{0.614}} = 1.61$ Choice A incorrectly assumes direct proportionality. Choice B uses the diatomic formula or considers only one rotational constant. Choice D appears to square the 0.85 factor incorrectly. Choice C correctly accounts for all three rotational constants in the polyatomic partition function. Remember: rotational partition functions are inversely related to rotational constants, and for polyatomic molecules, you must consider all three principal moments of inertia simultaneously.
A nonlinear molecule undergoes a temperature increase from 298 K to 598 K. The translational partition function increases by a factor of 7.3, and the rotational partition function increases by a factor of 4.0. If the total molecular partition function increases by a factor of 45.2, what can be concluded about the vibrational modes?
Answer: The molecule has intermediate-frequency modes (800-1500 cm⁻¹) that become significantly excited over this temperature range When analyzing partition function temperature dependence, you need to consider how each molecular motion responds differently to thermal energy. The total partition function is the product of translational, rotational, and vibrational contributions: $q_{total} = q_{trans} \times q_{rot} \times q_{vib}$ . Given the data, you can calculate the vibrational contribution: $q_{vib}$ increases by a factor of $45.2 ÷ (7.3 × 4.0) = 1.55$ . This modest but significant increase tells you about the vibrational mode frequencies. Answer B is correct because a 55% increase in the vibrational partition function indicates modes that are partially excited at 298 K but become substantially more populated at 598 K. This behavior is characteristic of intermediate-frequency vibrations (800-1500 cm⁻¹), where the energy gaps are comparable to thermal energy at these temperatures. Answer A is wrong because high-frequency modes (>2000 cm⁻¹) would show negligible change in their partition functions over this temperature range, since $k_BT$ remains much smaller than the vibrational energy spacing. Answer C is incorrect because low-frequency modes (<600 cm⁻¹) would already be highly excited at 298 K and wouldn't show such a significant proportional increase. Answer D is wrong because if low-frequency modes dominated, you'd expect either a much larger increase (if going from quantum to classical) or a smaller increase (if already classical at both temperatures). Remember: vibrational partition function changes reveal mode frequencies—dramatic changes suggest low frequencies, modest changes indicate intermediate frequencies, and minimal changes point to high frequencies.
For an ideal gas at constant volume, the internal energy increases by 5.2 kJ when heated from 298 K to 398 K. If the translational contribution to this energy change is 2.5 kJ and the rotational contribution is 1.7 kJ, what can be concluded about the molecular structure?
Answer: The molecule is linear polyatomic because it has both rotational and vibrational contributions of significant magnitude When analyzing molecular structure from energy changes, you need to examine how internal energy is distributed among translational, rotational, and vibrational modes. Each type of motion contributes differently based on molecular geometry. Let's calculate the vibrational contribution: Total energy change = 5.2 kJ, with 2.5 kJ translational and 1.7 kJ rotational, leaving 5.2 - 2.5 - 1.7 = 1.0 kJ vibrational. This significant vibrational contribution (about 19% of total energy change) indicates multiple atoms are present, ruling out monatomic molecules. The substantial rotational contribution (1.7 kJ) confirms the molecule can rotate in space, which also eliminates monatomic possibilities. The presence of both significant rotational and vibrational contributions points to a polyatomic molecule. To determine if it's linear or nonlinear, consider that linear molecules have fewer rotational degrees of freedom than nonlinear ones, but the key evidence here is the vibrational contribution magnitude, which suggests multiple vibrational modes typical of linear polyatomic molecules like CO₂ or acetylene. Choice A incorrectly focuses on comparing vibrational to translational energy rather than recognizing that any significant vibrational contribution indicates polyatomic structure. Choice B applies an incorrect ratio analysis - the 2/3 relationship isn't a diagnostic criterion for diatomic molecules. Choice D misinterprets the vibrational contribution as evidence for nonlinear structure, when 1.0 kJ actually aligns well with linear polyatomic molecules having multiple vibrational modes. Remember: significant contributions from all three energy modes (translation, rotation, vibration) typically indicate linear polyatomic molecules, especially when vibrational energy represents a substantial fraction of the total change.
A linear triatomic molecule (such as CO₂) has a vibrational frequency of 2349 cm⁻¹ for one normal mode and 667 cm⁻¹ for a doubly degenerate bending mode. At 298 K, which contribution dominates the total vibrational heat capacity?
Answer: The 667 cm⁻¹ modes contribute more because lower frequency modes are more easily excited at 298 K When analyzing vibrational heat capacity contributions, you need to consider both the degeneracy (number of modes) and how "activated" each mode is at the given temperature. The key is calculating the vibrational heat capacity using $C_v = R \left(\frac{\Theta_v}{T}\right)^2 \frac{e^{\Theta_v/T}}{(e^{\Theta_v/T}-1)^2}$ , where $\Theta_v = \frac{h\nu}{k}$ is the vibrational temperature. For the 2349 cm⁻¹ mode: $\Theta_v = \frac{(6.626 \times 10^{-34})(2349 \times 3 \times 10^{12})}{1.381 \times 10^{-23}} = 3380$ K. At 298 K, $\Theta_v/T = 11.3$ , giving a negligible heat capacity contribution since this mode is barely excited. For the 667 cm⁻¹ modes: $\Theta_v = 960$ K, so $\Theta_v/T = 3.22$ . This gives a substantial contribution per mode, and since there are two degenerate bending modes, the total contribution is doubled. The 667 cm⁻¹ modes dominate because they're much more thermally accessible at room temperature, and this effect is amplified by the degeneracy factor of 2. Choice A incorrectly assumes higher frequencies always give larger heat capacities—the opposite is true at moderate temperatures. Choice C is backwards; lower frequency modes are closer to classical activation. Choice D incorrectly suggests the effects cancel when the lower frequency modes clearly dominate. Remember: vibrational heat capacity depends on thermal accessibility, not just frequency. Lower frequency modes contribute more at moderate temperatures, especially when degeneracy multiplies their effect.
A linear molecule has three vibrational modes: a symmetric stretch at 1388 cm⁻¹, an antisymmetric stretch at 2349 cm⁻¹, and a doubly degenerate bend at 667 cm⁻¹. At 400 K, which statement correctly describes the relative contributions to the vibrational internal energy?
Answer: The bending modes contribute most because lower frequency modes have larger populations of excited states When analyzing vibrational contributions to internal energy, you need to consider both the energy of each mode and how much it's actually populated at the given temperature. The vibrational contribution depends on the quantum mechanical expression involving $\frac{hv}{kT}$ , where higher frequency modes require more thermal energy to be significantly excited. At 400 K, calculate the thermal parameter $\frac{hv}{kT}$ for each mode. For the antisymmetric stretch (2349 cm⁻¹), this gives the largest value, meaning this high-frequency mode remains largely in its ground state with minimal thermal excitation. The bending modes (667 cm⁻¹) have the smallest $\frac{hv}{kT}$ value, making them most thermally accessible, and their degeneracy factor of 2 doubles their contribution. The bending modes contribute most to the vibrational internal energy because they combine favorable thermal accessibility (low frequency means easier excitation at 400 K) with a degeneracy advantage (two equivalent modes instead of one). Option A incorrectly assumes higher frequency automatically means higher contribution—while each quantum has more energy, far fewer high-frequency quanta are populated at this temperature. Option C misidentifies which mode achieves the optimal balance of frequency and accessibility. Option D ignores that different modes contribute unequally due to their different frequencies and degeneracies, even though the total degrees of freedom determines the classical limit. Remember: vibrational contributions to thermodynamic properties depend on both the intrinsic energy spacing and thermal population of states. Lower frequency modes with degeneracy often dominate at moderate temperatures.
Consider a gas mixture containing equal molar amounts of two isotopomers of the same molecule. The lighter isotopomer has a rotational constant $B_1 = 1.2$ cm⁻¹ and the heavier has $B_2 = 1.0$ cm⁻¹. At 300 K, both have identical translational and vibrational partition functions. What is the ratio of the partial pressures in the mixture?
Answer: $P_1/P_2 = 1.00$ because the mixture contains equal molar amounts regardless of partition functions When you encounter problems about gas mixtures and partition functions, remember that partial pressure depends on the number of moles present, not on molecular properties like rotational constants. The key insight is understanding what determines partial pressure in an ideal gas mixture. According to Dalton's law, the partial pressure of each component depends only on its mole fraction: $P_i = x_i \cdot P_{total}$ , where $x_i$ is the mole fraction. Since the problem states the mixture contains "equal molar amounts" of both isotopomers, each has a mole fraction of 0.5, making their partial pressures equal regardless of their molecular properties. Answer A is correct because partial pressures depend solely on the amounts present, not on partition functions or rotational constants. The partition functions affect thermodynamic properties like internal energy and entropy, but don't change the fact that equal molar amounts produce equal partial pressures. Answer B incorrectly assumes that rotational partition functions affect partial pressure. While the lighter isotopomer does have a larger rotational partition function (since $q_{rot} \propto 1/B$ ), this doesn't influence pressure ratios. Answer C mistakenly equates the pressure ratio to the rotational constant ratio. Rotational constants affect molecular energy levels, not gas law behavior. Answer D incorrectly applies partition function ratios to pressure calculations. Partition functions relate to thermodynamic properties, not to the fundamental gas law relationships that govern partial pressures. Remember: partial pressures in gas mixtures depend only on composition (mole fractions), not on molecular properties. Partition functions affect thermodynamic properties but don't override basic gas law relationships.
The rotational partition function for a symmetric top molecule depends on two moments of inertia $I_\parallel$ and $I_\perp$ . If $I_\parallel < I_\perp$ , which statement correctly describes the temperature dependence of $q_{rot}$ ?
Answer: $q_{rot} \propto T^{3/2}$ with coefficient proportional to $\sqrt{I_\parallel I_\perp^2}$ from the full rotational state density For symmetric tops, $q_{rot} = \frac{\sqrt{\pi}}{\sigma}\left(\frac{8\pi^2 kT}{h^2}\right)^{3/2}\sqrt{I_\parallel I_\perp^2}$ , giving $T^{3/2}$ dependence with coefficient involving $\sqrt{I_\parallel I_\perp^2}$ . Choice A/B incorrectly emphasize single moments, D has wrong temperature power and moment dependence.
A diatomic molecule has a vibrational frequency of $1500 \text{ cm}^{-1}$ and a rotational constant $B = 2.0 \text{ cm}^{-1}$ . At $T = 300 \text{ K}$ , which statement best describes the relative contributions to the total partition function?
Answer: Translational contribution dominates, with vibrational being negligible and rotational being intermediate At 300 K: Translational partition function is typically ~ $10^{30}$ (very large), rotational $q_r = kT/hcB \approx 7$ (moderate), and vibrational $q_v = 1/(1-e^{-hc\nu/kT}) \approx 1.001$ since $hc\nu/kT \approx 7.2 >> 1$ (nearly negligible). The translational contribution vastly dominates.
At 800 K, the population ratio $N(J=2)/N(J=0)$ for the rotational states of a diatomic molecule is 2.45. If the same molecule is studied at 1600 K, what will be the new population ratio $N(J=2)/N(J=0)$ ?
Answer: 3.50 because the exponential factor depends on the square root of the temperature ratio When you encounter rotational population ratios at different temperatures, you're dealing with the Boltzmann distribution for rotational energy levels. The key insight is understanding how rotational energy depends on the quantum number J and how temperature affects the distribution. For a diatomic molecule, rotational energy is $E_J = BJ(J+1)$ where B is the rotational constant. The population ratio follows: $\frac{N(J=2)}{N(J=0)} = \frac{g_2}{g_0} \exp\left(-\frac{E_2 - E_0}{kT}\right)$ Since $g_J = 2J+1$ , we have $g_2/g_0 = 5/1 = 5$ . The energy difference is $E_2 - E_0 = 6B$ , so: $\frac{N(J=2)}{N(J=0)} = 5\exp\left(-\frac{6B}{kT}\right)$ At 800 K, this ratio equals 2.45. When temperature doubles to 1600 K, the exponential argument is halved: $\exp(-6B/k \cdot 1600) = \sqrt{\exp(-6B/k \cdot 800)}$ . From the original data: $2.45 = 5\exp(-6B/k \cdot 800)$ , so $\exp(-6B/k \cdot 800) = 0.49$ . At 1600 K: $\frac{N(J=2)}{N(J=0)} = 5\sqrt{0.49} = 5 \times 0.7 = 3.50$ Answer C is correct because the exponential factor depends on the square root of the original exponential when temperature doubles. Answer A incorrectly suggests populations decrease with temperature. Answer B wrongly assumes a simple doubling relationship. Answer D misapplies the square root relationship to the entire Boltzmann factor rather than recognizing it emerges from the temperature change in the exponential. Remember: when temperature changes affect Boltzmann distributions, focus on how the exponential argument transforms—often involving square root relationships when temperature doubles.
A diatomic molecule has two vibrational modes with frequencies $\nu_1 = 2000 \text{ cm}^{-1}$ and $\nu_2 = 500 \text{ cm}^{-1}$ . At $T = 600 \text{ K}$ , which approximation is most appropriate for calculating the vibrational partition function?
Answer: Mode 1 requires quantum treatment, mode 2 can use classical limit, requiring mixed approach At 600 K: For $\nu_1$ : $h\nu_1/kT = hc(2000)/kT \approx 4.8 > 1$ (quantum), For $\nu_2$ : $h\nu_2/kT \approx 1.2 \sim 1$ (borderline, but classical approximation reasonable). Mode 1 needs quantum treatment, mode 2 can use classical. Choice A assumes both classical (wrong for mode 1), C is overly conservative, D reverses the assignments.
A molecule at 500 K has a vibrational mode with frequency 1000 cm⁻¹. The population of the v=1 state relative to the v=0 state is 0.135. If a second vibrational mode of the same frequency is coupled to the first such that the total vibrational energy is $E = h\nu(v_1 + v_2 + 1)$ instead of treating them as independent modes, how does this affect the partition function?
Answer: The coupled partition function equals the product of individual partition functions because the Hamiltonian is still separable When you encounter vibrational coupling problems, the key insight is understanding what "coupling" actually means mathematically and how it affects the separability of the Hamiltonian. The given energy expression $E = h\nu(v_1 + v_2 + 1)$ reveals that despite being called "coupled," the Hamiltonian remains separable. You can rewrite this as $E = h\nu v_1 + h\nu v_2 + h\nu$ , which is simply the sum of two independent harmonic oscillators plus a constant zero-point energy term. The partition function becomes $q = \sum_{v_1,v_2} e^{-h\nu(v_1+v_2+1)/k_BT} = e^{-h\nu/k_BT} \sum_{v_1} e^{-h\nu v_1/k_BT} \sum_{v_2} e^{-h\nu v_2/k_BT}$ , which factors into the product of individual oscillator partition functions. Answer A is incorrect because it claims coupling doesn't change energy levels, but the reasoning about separability is wrong—the Hamiltonian's separability, not energy level changes, determines factorization. Answer B incorrectly suggests coupling creates degeneracies; this energy expression doesn't introduce any new degeneracies beyond what independent oscillators would have. Answer D is wrong because no energy levels are removed—all combinations of $v_1$ and $v_2$ are still allowed. Remember: partition function factorization depends on Hamiltonian separability, not whether modes are labeled as "coupled." Always check if you can separate variables mathematically, regardless of the terminology used in the problem.
A symmetric top molecule has a vibrational mode that transforms as the E representation of its point group, making it doubly degenerate. At 500 K, each component of this degenerate mode has a vibrational partition function of 1.8. What is the contribution of this vibrational mode to the molecular heat capacity?
Answer: $C_v = R \times 1.44$ treating the degeneracy as two independent oscillators When you encounter degenerate vibrational modes in statistical thermodynamics, the key principle is that each degenerate component contributes independently to the thermodynamic properties. A doubly degenerate E mode behaves as two separate oscillators that happen to have the same frequency. For vibrational heat capacity, you use the standard formula: $C_v = R \left(\frac{hν/kT}{e^{hν/kT}-1}\right)^2 \frac{e^{hν/kT}}{(e^{hν/kT}-1)^2}$ . However, when given the vibrational partition function $q_{vib} = 1.8$ , you can use the relationship: $C_v = R \left(\frac{\partial \ln q_{vib}}{\partial T}\right)^2 + R \frac{\partial^2 \ln q_{vib}}{\partial T^2}$ . For typical vibrational modes, this simplifies to give a contribution factor that, with $q_{vib} = 1.8$ , yields approximately 0.72 per oscillator. Since you have two degenerate components, the total contribution is $2 \times 0.72 = 1.44$ , making the heat capacity $C_v = R \times 1.44$ . Answer A correctly treats this as two independent oscillators. Answer B gives only the single-oscillator contribution, incorrectly ignoring the degeneracy. Answer C squares the contribution factor, which has no physical basis—degeneracy adds contributions, it doesn't square them. Answer D incorrectly assumes an inverse relationship with the partition function, which contradicts the actual mathematical relationship between partition functions and heat capacity. Remember: degeneracy always means independent addition of contributions. Each degenerate component contributes fully and separately to all thermodynamic properties.
For a quantum harmonic oscillator with frequency $\omega$ , the vibrational contribution to the internal energy is $U_{vib} = \frac{\hbar\omega}{2} + \frac{\hbar\omega}{e^{\hbar\omega/kT}-1}$ . Which term represents the quantum mechanical zero-point energy, and why is it temperature-independent?
Answer: The first term, because it represents the irreducible quantum mechanical energy of the ground vibrational state When analyzing quantum mechanical systems, you need to distinguish between temperature-dependent thermal contributions and fundamental quantum mechanical effects that exist even at absolute zero. The vibrational energy expression $U_{vib} = \frac{\hbar\omega}{2} + \frac{\hbar\omega}{e^{\hbar\omega/kT}-1}$ contains two distinct terms. The first term, $\frac{\hbar\omega}{2}$ , is completely independent of temperature T and represents the zero-point energy - the minimum energy that a quantum harmonic oscillator must possess even in its ground state (v=0). This arises because the uncertainty principle prevents the oscillator from being completely at rest, requiring some residual motion even at T=0K. The second term, $\frac{\hbar\omega}{e^{\hbar\omega/kT}-1}$ , is the average thermal energy above the ground state and clearly depends on temperature - as T increases, this term grows, representing thermal excitation to higher vibrational levels. Option A incorrectly identifies the temperature-dependent thermal term as zero-point energy. The second term represents thermal excitation, not the minimum quantum energy. Option B is wrong because zero-point energy specifically refers to the ground state energy, not the total expression. Option D mischaracterizes the thermal term - while uncertainty principle considerations do lead to zero-point energy, the second term represents thermal population of excited states, not uncertainty effects. Remember: Zero-point energy is always temperature-independent because it's a fundamental quantum mechanical property of the ground state, while thermal contributions always involve temperature-dependent Boltzmann factors like $e^{-E/kT}$ .
At what temperature does the vibrational heat capacity of a harmonic oscillator with frequency 2000 cm⁻¹ reach exactly half of its classical value of R?
Answer: T = 2070 K, where the quantum parameter equals 1.39 When you encounter vibrational heat capacity problems, you're dealing with quantum mechanics applied to molecular motion. The key is understanding how quantum effects diminish at higher temperatures through the quantum parameter $\Theta_v/T$ , where $\Theta_v = hc\tilde{\nu}/k_B$ is the vibrational temperature. First, calculate the vibrational temperature: $\Theta_v = \frac{(6.626 \times 10^{-34})(3.0 \times 10^{10})(2000)}{1.381 \times 10^{-23}} = 2870 \text{ K}$ The quantum vibrational heat capacity is $C_v = R\left(\frac{\Theta_v}{T}\right)^2 \frac{e^{\Theta_v/T}}{(e^{\Theta_v/T}-1)^2}$ , while the classical value is simply $R$ . You need to find where the quantum expression equals $R/2$ , so: $\left(\frac{\Theta_v}{T}\right)^2 \frac{e^{\Theta_v/T}}{(e^{\Theta_v/T}-1)^2} = 0.5$ Let $x = \Theta_v/T$ . This equation becomes $x^2 \frac{e^x}{(e^x-1)^2} = 0.5$ . Solving numerically gives $x \approx 1.39$ , so $T = \Theta_v/1.39 = 2870/1.39 = 2065 \text{ K}$ . Answer B gives T = 2070 K with quantum parameter 1.39, matching our calculation perfectly. Answer A uses the wrong quantum parameter (2.0 instead of 1.39). Answer C incorrectly assumes the answer occurs when $\Theta_v/T = 1$ , but this gives $C_v = 0.92R$ , not $0.5R$ . Answer D uses 0.69, which would give a much higher heat capacity than half the classical value. Remember: vibrational heat capacity problems always involve finding the right quantum parameter $\Theta_v/T$ that satisfies the given condition—calculate $\Theta_v$ first, then solve for the temperature.

Start this quiz