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Question 1Physical Chemistry

One mole of an ideal gas expands reversibly and isothermally from 10 L to 20 L at 300 K. Which quantity of the system is exactly zero for this process?

  1. Change in entropy, ΔS
  2. Change in internal energy, ΔU Correct answer.
  3. Work done on the system, w
  4. Heat absorbed by the system, q

Explanation

For an ideal gas the internal energy is a function of temperature alone, so an isothermal process gives ΔU = 0 irrespective of the volume change. The expansion is spontaneous, giving ΔS = nR ln(V₂/V₁) = +5.76 J K⁻¹, and q = −w = nRT ln 2 = +1.73 kJ; neither of these vanishes.

Question 2Physical Chemistry

The number of radial (spherical) nodes in a 3p atomic orbital of a hydrogen-like atom is

  1. 1 Correct answer.
  2. 3
  3. 2
  4. 0

Explanation

The number of radial nodes is given by n − l − 1; for a 3p orbital n = 3 and l = 1, which gives exactly one radial node. The single angular node of a p orbital is a nodal plane passing through the nucleus and is not counted as a radial node.

Question 3Physical Chemistry

For a reaction whose rate constant has the units mol L⁻¹ s⁻¹, the overall order of the reaction is

  1. Second order
  2. Third order
  3. First order
  4. Zero order Correct answer.

Explanation

For an n-th order rate law, rate = k[A]ⁿ, the units of k are (mol L⁻¹)^(1−n) s⁻¹. Setting 1 − n = 1 gives n = 0, so a rate constant expressed in concentration per unit time belongs to a zero-order reaction, whose rate is independent of concentration.

Question 4Physical Chemistry

A galvanic cell is written as Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s), with E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V. The standard cell emf and the sign of ΔG° for the cell reaction are, respectively

  1. −1.10 V and positive
  2. +0.42 V and negative
  3. −0.42 V and positive
  4. +1.10 V and negative Correct answer.

Explanation

E°cell = E°cathode − E°anode = 0.34 − (−0.76) = +1.10 V, with copper acting as the cathode because it has the higher reduction potential. A positive cell emf means the cell reaction is spontaneous as written, so ΔG° = −nFE°cell is negative.

Question 5Physical Chemistry

Two moles of ideal gas A and three moles of ideal gas B are mixed at constant temperature and pressure. The entropy of mixing is closest to

  1. 28 J K⁻¹ Correct answer.
  2. 11.2 J K⁻¹
  3. 56 J K⁻¹
  4. 5.6 J K⁻¹

Explanation

ΔS_mix = −R Σ nᵢ ln xᵢ, with x_A = 0.4 and x_B = 0.6. Substituting gives −8.314 × (2 ln 0.4 + 3 ln 0.6) = 8.314 × 3.365 ≈ 28 J K⁻¹, positive as expected for the spontaneous mixing of ideal gases at constant T and p.

Question 6Physical Chemistry

The pure rotational microwave spectrum of a rigid diatomic molecule consists of equally spaced lines. If the rotational constant is B = 1.93 cm⁻¹, the spacing between adjacent lines is

  1. 3.86 cm⁻¹ Correct answer.
  2. 7.72 cm⁻¹
  3. 5.79 cm⁻¹
  4. 1.93 cm⁻¹

Explanation

For a rigid rotor the energies are BJ(J+1) and the selection rule is ΔJ = ±1, so the lines appear at 2B, 4B, 6B and so on. The constant spacing is therefore 2B = 3.86 cm⁻¹, and measuring it gives the rotational constant directly.

Question 7Physical Chemistry

A reaction follows first-order kinetics with a half-life of 20 min. The time required for the reaction to reach 75% completion is

  1. 40 min Correct answer.
  2. 20 min
  3. 30 min
  4. 60 min

Explanation

For a first-order reaction the fraction remaining after time t is (1/2)^(t/t½). Reaching 75% completion leaves 25% of the reactant, which is two successive half-lives, so t = 2 × 20 = 40 min.

Question 8Physical Chemistry

A metal crystallises in a face-centred cubic lattice with unit-cell edge length a = 400 pm. The atomic radius derived from the touching condition is

  1. 100 pm
  2. 200 pm
  3. 173 pm
  4. 141 pm Correct answer.

Explanation

In an fcc cell the atoms touch along the face diagonal, so 4r = a√2 and hence r = a/(2√2). Substituting a = 400 pm gives r = 400/2.828 ≈ 141 pm, which is the correct fcc radius for this cell edge.

Question 9Physical Chemistry

The total number of microstates arising from a d³ electron configuration is

  1. 252
  2. 45
  3. 120 Correct answer.
  4. 90

Explanation

Five d orbitals provide ten spin-orbitals, and the number of ways of placing three electrons among them is C(10,3) = 10!/(3!7!) = 120. This counting automatically includes all the terms and microstates of the d³ configuration without imposing any coupling scheme.

Question 10Physical Chemistry

A reaction has an activation energy of 100 kJ mol⁻¹. On raising the temperature from 300 K to 310 K, the rate constant increases by a factor closest to (R = 8.314 J K⁻¹ mol⁻¹)

  1. 2.7
  2. 2.0
  3. 3.6 Correct answer.
  4. 1.3

Explanation

Using ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) = (100000/8.314)(1/300 − 1/310) gives ln(k₂/k₁) ≈ 1.29, so k₂/k₁ ≈ e^1.29 ≈ 3.6. This strong temperature sensitivity of about a factor of three per 10 K is why reaction rates are so sensitive to temperature control.

Question 11Physical Chemistry

For the adiabatic free expansion of an ideal gas into a vacuum, which combination of thermodynamic quantities is correct for the system?

  1. q = 0, w < 0, ΔU < 0, and ΔS = 0
  2. q = 0, w > 0, ΔU > 0, and ΔS = 0
  3. q = 0, w = 0, ΔU = 0, and ΔS > 0 Correct answer.
  4. q > 0, w = 0, ΔU > 0, and ΔS > 0

Explanation

In an adiabatic free expansion the vessel is insulated, so q = 0, and the gas expands against zero external pressure, so w = 0. For an ideal gas U depends only on T, hence ΔU = 0 and ΔT = 0. The process is irreversible, so the entropy of the gas increases even though no heat enters.

Question 12Physical Chemistry

Which quantum number determines the magnitude of the orbital angular momentum of an electron in a hydrogen-like atom?

  1. Principal quantum number n
  2. Magnetic quantum number ml
  3. Azimuthal quantum number l Correct answer.
  4. Spin quantum number ms

Explanation

The magnitude of orbital angular momentum is given by L = [l(l + 1)]^(1/2)ℏ, so it depends on the azimuthal quantum number l. The principal quantum number fixes the energy in a hydrogen-like atom, while ml and ms describe orientation and spin.

Question 13Physical Chemistry

At constant temperature and pressure, a closed system can undergo a spontaneous process only when which condition is satisfied?

  1. ΔG = 0
  2. ΔS_total < 0
  3. ΔG < 0 Correct answer.
  4. ΔG > 0

Explanation

At constant T and P, the Gibbs energy change is the criterion of spontaneity for a closed system. A negative ΔG corresponds to a spontaneous process, while ΔG = 0 describes equilibrium. The second law also requires the total entropy change to be positive for an irreversible spontaneous process.

Question 14Physical Chemistry

What is the osmotic pressure of an ideal 0.10 mol L⁻¹ glucose solution at 300 K? Use R = 0.0821 L atm mol⁻¹ K⁻¹.

  1. 2.46 atm Correct answer.
  2. 4.92 atm
  3. 1.23 atm
  4. 0.25 atm

Explanation

For dilute ideal solutions, osmotic pressure is π = iCRT. Glucose is a nonelectrolyte, so i = 1, giving π = (1)(0.10)(0.0821)(300) = 2.46 atm.

Question 15Physical Chemistry

For a second-order reaction 2A → products with rate law rate = k[A]² and initial concentration [A]₀, what is the half-life?

  1. 1/(k[A]₀) Correct answer.
  2. 2/(k[A]₀)
  3. [A]₀/(2k)
  4. ln 2/k

Explanation

The integrated rate law for a second-order reaction with rate = k[A]² is 1/[A] = 1/[A]₀ + kt. Setting [A] = [A]₀/2 gives t₁/₂ = 1/(k[A]₀).

Question 16Physical Chemistry

For the concentration cell Ag(s) | Ag⁺(0.010 M) || Ag⁺(0.10 M) | Ag(s) at 298 K, what is the cell potential closest to?

  1. 0.118 V
  2. 0.059 V Correct answer.
  3. 0.000 V
  4. −0.059 V

Explanation

Both electrodes are Ag/Ag⁺, so E° = 0 and the cell potential is set by the concentration ratio. Using E = (0.0591/1) log([Ag⁺]cathode/[Ag⁺]anode) gives 0.0591 log(0.10/0.010) = 0.059 V. The higher concentration is the cathode, so the potential is positive.

Question 17Physical Chemistry

Which diatomic molecule gives a pure rotational microwave absorption spectrum?

  1. N₂
  2. O₂
  3. H₂
  4. CO Correct answer.

Explanation

A pure rotational microwave transition requires a permanent electric dipole moment. CO is heteronuclear and polar, whereas H₂, N₂, and O₂ are homonuclear and have no permanent dipole moment.

Question 18Physical Chemistry

In a body-centred cubic unit cell, the number of atoms per unit cell and the coordination number are respectively

  1. 2 and 8 Correct answer.
  2. 1 and 6
  3. 4 and 12
  4. 2 and 6

Explanation

A BCC cell has one atom at each of eight corners, each shared by eight cells, plus one atom at the body centre. This gives 8(1/8) + 1 = 2 atoms per unit cell, and each atom has eight nearest neighbours along the body diagonals.

Question 19Physical Chemistry

For a hydrogenic 1s orbital with nuclear charge Z, what is the expectation value ⟨r⟩?

  1. a₀/(2Z)
  2. 2a₀/Z
  3. 3a₀/(2Z) Correct answer.
  4. a₀/Z

Explanation

The radial probability density for a hydrogenic 1s orbital is proportional to r²e^(−2Zr/a₀). Evaluating the radial integral gives ⟨r⟩ = 3a₀/(2Z), so the mean radius decreases as Z increases.

Question 20Physical Chemistry

Consider Pt(s) | H₂(g, 1 bar) | H⁺(aq, pH = 3.00) || Cu²⁺(aq, 0.10 M) | Cu(s) at 298 K. Given E°(Cu²⁺/Cu) = +0.34 V, what is Ecell closest to?

  1. 0.49 V Correct answer.
  2. 0.16 V
  3. 0.52 V
  4. 0.31 V

Explanation

The hydrogen electrode potential is E = −0.0591 pH = −0.177 V at pH 3. The copper electrode potential is 0.34 − (0.0591/2) log(1/0.10) = 0.310 V. Hence Ecell = 0.310 − (−0.177) = 0.49 V.

Question 21Physical Chemistry

For one mole of an ideal gas, which relation between the molar heat capacities at constant pressure and constant volume is correct?

  1. Cp,m − Cv,m = 0
  2. Cp,m − Cv,m = R Correct answer.
  3. Cp,m − Cv,m = RT
  4. Cp,m − Cv,m = R ln T

Explanation

For an ideal gas, H = U + PV, and differentiating with respect to temperature at constant pressure gives Cp = Cv + nR. For one mole, Cp,m − Cv,m = R.

Question 22Physical Chemistry

The number of angular nodes present in a 4d atomic orbital is

  1. 3
  2. 1
  3. 2 Correct answer.
  4. 0

Explanation

The number of angular nodes in an atomic orbital is equal to the azimuthal quantum number l. For a d orbital, l = 2, so a 4d orbital has two angular nodes.

Question 23Physical Chemistry

For a reaction obeying first-order kinetics, a plot of ln[A] against time t gives

  1. a straight line with intercept −k
  2. a straight line with slope +k
  3. a parabola with slope −k
  4. a straight line with slope −k Correct answer.

Explanation

The integrated first-order rate law is ln[A] = ln[A]0 − kt. Therefore a plot of ln[A] versus t is linear with slope −k and intercept ln[A]0.

Question 24Physical Chemistry

For a cell reaction with n = 2 and standard cell potential E° = +0.50 V at 298 K, the standard Gibbs energy change ΔG° is closest to (F = 96500 C mol⁻¹)

  1. −96.5 kJ mol⁻¹ Correct answer.
  2. +96.5 kJ mol⁻¹
  3. +48.3 kJ mol⁻¹
  4. −48.3 kJ mol⁻¹

Explanation

The relation is ΔG° = −nFE°. Substituting n = 2, F = 96500 C mol⁻¹ and E° = 0.50 V gives −96500 J mol⁻¹, or −96.5 kJ mol⁻¹.

Question 25Physical Chemistry

An aqueous solution is 0.10 mol kg⁻¹ in NaCl, and NaCl is assumed to dissociate completely. If Kb for water is 0.512 K kg mol⁻¹, the boiling-point elevation is closest to

  1. 0.205 K
  2. 0.512 K
  3. 0.102 K Correct answer.
  4. 0.051 K

Explanation

For complete dissociation, NaCl gives i = 2 particles per formula unit. Thus ΔTb = i Kb m = 2 × 0.512 × 0.10 = 0.102 K.

Question 26Physical Chemistry

X-rays of wavelength 154 pm are diffracted by a set of crystal planes at a glancing angle θ = 30.0° in first order. The interplanar spacing d is closest to

  1. 178 pm
  2. 308 pm
  3. 154 pm Correct answer.
  4. 77 pm

Explanation

Bragg's law is nλ = 2d sinθ. With n = 1, λ = 154 pm and sin 30.0° = 0.500, d = 154/(2 × 0.500) = 154 pm.

Question 27Physical Chemistry

A vibrational mode of a molecule is infrared active only when it produces

  1. a change in dipole moment Correct answer.
  2. a change in polarizability
  3. a change in bond order
  4. a change in electron count

Explanation

Infrared absorption requires an oscillating dipole moment during the vibration. Therefore the mode must produce a change in the molecular dipole moment; polarizability changes govern Raman activity.

Question 28Physical Chemistry

For the reaction N2(g) + 3H2(g) ⇌ 2NH3(g), the relation between Kp and Kc at temperature T is

  1. Kp = Kc
  2. Kp = Kc(RT)^2
  3. Kp = Kc(RT)^−2 Correct answer.
  4. Kp = Kc(RT)^−1

Explanation

In general Kp = Kc(RT)^Δn, where Δn is the change in moles of gas. Here Δn = 2 − (1 + 3) = −2, so Kp = Kc(RT)^−2.

Question 29Physical Chemistry

The ground-state term symbol for a free d² ion, according to Hund's rules, is

  1. ¹G
  2. ¹D
  3. ³P
  4. ³F Correct answer.

Explanation

For d², Hund's first rule favours maximum spin, giving a triplet with S = 1. Among the triplet terms, Hund's second rule selects the largest L, which is L = 3, so the ground term is ³F.

Question 30Physical Chemistry

From the fundamental equation dG = −S dT + V dp, which Maxwell relation is obtained?

  1. (∂T/∂p)_S = (∂V/∂S)_p
  2. (∂S/∂V)_T = (∂p/∂T)_V
  3. (∂S/∂p)_T = (∂V/∂T)_p
  4. (∂S/∂p)_T = −(∂V/∂T)_p Correct answer.

Explanation

For dG = −S dT + V dp, equality of the mixed second derivatives gives (∂S/∂p)_T = −(∂V/∂T)_p. The minus sign comes from the coefficient of dT being −S.

Question 31Physical Chemistry

For a closed system undergoing a reversible process at constant pressure with only pressure–volume work, the heat exchanged is equal to which thermodynamic quantity?

  1. ΔH Correct answer.
  2. ΔA
  3. ΔG
  4. ΔU

Explanation

At constant pressure, q_p = ΔH because H = U + pV and, with only pV work, ΔH = ΔU + pΔV = q_p. The other thermodynamic potentials correspond to different constraints, such as constant volume or constant temperature.

Question 32Physical Chemistry

How many radial nodes does a 3p atomic orbital possess?

  1. 2
  2. 3
  3. 0
  4. 1 Correct answer.

Explanation

The number of radial nodes is given by n − l − 1. For a 3p orbital n = 3 and l = 1, so the orbital has exactly one radial node.

Question 33Physical Chemistry

For a reaction that is zero order in each reactant, what is the unit of the rate constant when concentration is expressed in mol L⁻¹ and time in seconds?

  1. L² mol⁻² s⁻¹
  2. s⁻¹
  3. L mol⁻¹ s⁻¹
  4. mol L⁻¹ s⁻¹ Correct answer.

Explanation

For a zero-order reaction the rate law is rate = k, so the rate constant has the units of reaction rate. With concentration in mol L⁻¹ and time in seconds, those units are mol L⁻¹ s⁻¹.

Question 34Physical Chemistry

Two moles of an ideal gas expand isothermally and reversibly from 10.0 L to 20.0 L at 300 K. What is ΔS for the gas? Use R = 8.314 J mol⁻¹ K⁻¹.

  1. 5.76 J K⁻¹
  2. 11.5 J K⁻¹ Correct answer.
  3. 23.1 J K⁻¹
  4. 0.693 J K⁻¹

Explanation

For an isothermal reversible expansion of an ideal gas, ΔS = nR ln(V₂/V₁). Substituting n = 2 mol gives 2 × 8.314 × ln 2 = 11.5 J K⁻¹.

Question 35Physical Chemistry

The molar conductivity of a 0.010 mol L⁻¹ acetic acid solution is 16.0 S cm² mol⁻¹, and its limiting molar conductivity is 390.0 S cm² mol⁻¹. What is the degree of dissociation of acetic acid in this solution?

  1. 0.410
  2. 0.00410
  3. 0.980
  4. 0.0410 Correct answer.

Explanation

The degree of dissociation is α = Λm/Λm°. Substituting 16.0/390.0 gives α = 0.0410, so about 4.1% of the acid is dissociated.

Question 36Physical Chemistry

In a cubic close-packed (ccp) arrangement of identical spheres, how many octahedral voids are present per unit cell?

  1. 8
  2. 4 Correct answer.
  3. 2
  4. 12

Explanation

In ccp the unit cell contains four atoms, and the number of octahedral voids equals the number of atoms. Therefore a ccp unit cell contains four octahedral voids.

Question 37Physical Chemistry

In the rigid-rotor approximation, what is the spacing in wavenumber between successive lines of a pure rotational spectrum of a diatomic molecule with rotational constant B?

  1. 2B Correct answer.
  2. B/2
  3. 4B
  4. B

Explanation

Rigid-rotor levels have energies F(J) = BJ(J+1). The selection rule ΔJ = ±1 gives transition wavenumbers 2B(J+1), so adjacent lines are separated by 2B.

Question 38Physical Chemistry

For a reaction at 298 K with standard Gibbs energy change ΔG° = +10.0 kJ mol⁻¹, what is the equilibrium constant K closest to? Use R = 8.314 J mol⁻¹ K⁻¹.

  1. 0.018 Correct answer.
  2. 0.98
  3. 56
  4. 1.0

Explanation

Using ΔG° = −RT ln K, ln K = −10000/(8.314 × 298) = −4.04. Therefore K = e^(−4.04) ≈ 0.018, well below unity for a positive ΔG°.

Question 39Physical Chemistry

Two non-degenerate energy levels of a molecule are separated by 1.0 kJ mol⁻¹. At 300 K, what is the ratio of the population of the upper level to that of the lower level? Use R = 8.314 J mol⁻¹ K⁻¹.

  1. 0.67 Correct answer.
  2. 1.5
  3. 0.99
  4. 0.40

Explanation

The Boltzmann distribution gives N_upper/N_lower = exp(−ΔE/RT). With ΔE = 1000 J mol⁻¹, RT = 8.314 × 300 = 2494 J mol⁻¹, and exp(−0.401) ≈ 0.67.

Question 40Physical Chemistry

An electron is confined in a one-dimensional box of length 1.0 nm. What is the energy of the first excited state (n = 2) closest to? Use h = 6.626×10⁻³⁴ J s, m_e = 9.109×10⁻³¹ kg, and 1 eV = 1.602×10⁻¹⁹ J.

  1. 0.094 eV
  2. 6.02 eV
  3. 1.50 eV Correct answer.
  4. 0.376 eV

Explanation

The particle-in-a-box energy is E_n = n²h²/(8mL²). The n = 1 level lies at about 0.376 eV, so the n = 2 first excited state is four times higher, about 1.50 eV.

Question 41Physical Chemistry

The total number of degenerate orbitals in the hydrogenic shell with principal quantum number n = 3, ignoring spin, is

  1. 3
  2. 9 Correct answer.
  3. 18
  4. 6

Explanation

For a hydrogenic atom the energy depends only on n, so all orbitals with the same n are degenerate. A shell with principal quantum number n contains n^2 orbitals, giving 3^2 = 9 orbitals when spin is ignored.

Question 42Physical Chemistry

For a fixed amount of an ideal gas undergoing a reversible adiabatic expansion, which relation is valid, with γ = Cp/Cv?

  1. T V^γ = constant
  2. p V^(1−γ) = constant
  3. T V^(γ−1) = constant Correct answer.
  4. T p^γ = constant

Explanation

For a reversible adiabatic process in an ideal gas, q = 0 and the entropy is constant. Combining dU = −p dV with the ideal-gas law gives T V^(γ−1) = constant, where γ = Cp/Cv.

Question 43Physical Chemistry

In an operating galvanic cell, the electrode at which oxidation takes place is called the

  1. anode Correct answer.
  2. salt bridge
  3. cathode
  4. reference electrode

Explanation

Oxidation is loss of electrons and occurs at the anode in both galvanic and electrolytic cells. In a galvanic cell the anode is negative, while reduction occurs at the cathode.

Question 44Physical Chemistry

For a reaction A → products that is second order in A, the half-life t1/2 is given by

  1. 2 / (k[A]0)
  2. ln 2 / k
  3. 1 / (k[A]0) Correct answer.
  4. 1 / (k[A]0^2)

Explanation

For a second-order reaction with rate law −d[A]/dt = k[A]^2, integration gives 1/[A] = 1/[A]0 + kt. Setting [A] = [A]0/2 yields t1/2 = 1/(k[A]0).

Question 45Physical Chemistry

For a cell reaction with n = 2 at 298 K, E° = +0.40 V and reaction quotient Q = 10, the cell potential E is closest to

  1. 0.43 V
  2. 0.40 V
  3. 0.34 V
  4. 0.37 V Correct answer.

Explanation

The Nernst equation at 298 K is E = E° − (0.0592 V/n) log Q. With n = 2 and Q = 10, the correction is 0.0296 V, so E = 0.40 − 0.0296 ≈ 0.37 V.

Question 46Physical Chemistry

The number of normal modes of vibration for a nonlinear molecule containing five atoms is

  1. 15
  2. 6
  3. 12
  4. 9 Correct answer.

Explanation

A nonlinear molecule with N atoms has 3N − 6 vibrational normal modes because three translations and three rotations are removed. For N = 5 this gives 15 − 6 = 9 modes.

Question 47Physical Chemistry

A crystal plane intercepts the crystallographic axes at a, 2a and 3a. Its Miller indices (hkl) are

  1. (1 2 3)
  2. (2 3 6)
  3. (3 2 1)
  4. (6 3 2) Correct answer.

Explanation

Miller indices are obtained by taking reciprocals of the fractional intercepts and clearing fractions. Intercepts a, 2a, 3a give reciprocals 1, 1/2, 1/3, which multiply by 6 to yield (6 3 2).

Question 48Physical Chemistry

The zero-point energy of a quantum harmonic oscillator of force constant k and reduced mass μ is

  1. kT
  2. hν
  3. ½hν Correct answer.
  4. zero

Explanation

A quantum harmonic oscillator has allowed energies E_v = (v + ½)hν, with ν = (1/2π)√(k/μ). Even at v = 0 the energy is ½hν, the zero-point energy.

Question 49Physical Chemistry

In the pure rotational Raman spectrum of a rigid linear rotor with rotational constant B (in cm⁻¹), the separation between successive Stokes lines is

  1. 2B
  2. 6B
  3. 4B Correct answer.
  4. 8B

Explanation

Rotational Raman selection rules give ΔJ = ±2, so the Stokes shift is B[(J+2)(J+3) − J(J+1)] = B(4J+6). Successive Stokes lines therefore differ by 4B.

Question 50Physical Chemistry

The Boyle temperature of a van der Waals gas, expressed in terms of its molar constants a and b and the gas constant R, is

  1. 2a/(Rb)
  2. a/(Rb) Correct answer.
  3. b/(Ra)
  4. a/(2Rb)

Explanation

For a van der Waals gas the second virial coefficient is B2 = b − a/(RT). The Boyle temperature is where B2 = 0, giving T_B = a/(Rb).