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Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 3 50 b Solution by
Codex 0 2026-10-20
Radical polymerization, a type of chain-growth polymerization. Peroxide bond cleavage initiates reactive chains; propagation adds monomer to a radical end while regenerating a radical end on the longer chain. Termination and chain transfer can then stop or redirect growth. In the propagation step there is no obligatory small-molecule condensate, so this is not a step-growth condensation.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 3 50 a Solution by
Codex 0 2026-10-20
Addition consumes the carbon–carbon double bond of vinyl chloride and retains chlorine as a side substituent. The interior repeat of A is thereforenamely polyvinyl chloride. Benzoyl peroxide supplies initiating free radicals; head-to-tail addition gives the usual backbone connectivity. End groups and tacticity are not uniquely specified by this repeat formula.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 3 48 f Solution by
Codex 0 2026-10-20
The crystalline expression diverges at , where the cage free area vanishes. This is physically different from the spurious pole produced by extrapolating the low-density fluid approximation. A real compression path cannot be inferred by following that fluid pole and then abruptly switching curves. At low density it follows a fluid equation of state, while at sufficiently high density an ordered arrangement can leave more accessible rattling configurations than a disordered crowded arrangement. This is entropy-driven hard-sphere crystallization, here in two dimensions: allowed hard-core configurations have the same interaction energy, so configurational entropy drives the phase competition. An equilibrium first-order coexistence interval would have common pressure and chemical potential, giving a constant-pressure segment as mean density changes. The two asymptotic expressions alone neither locate that interval nor determine the detailed two-dimensional melting sequence. They require a consistent free-energy comparison and trustworthy approximations in the transition region.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 3 48 e Solution by
Codex 0 2026-10-20
Insert the stipulated near-close-packing relation for into the hard-disc cage approximation:Writing , we have and therefore . From the supplied crystalline canonical partition function,Its close-packing scale is set by the triangular lattice packing of discs, with in the ideal geometry.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 3 48 d Solution by
Codex 0 2026-10-20
In the circular hard-disc cage approximation, moving the central disc towards a neighbour brings contact after a centre displacement . ThusAt close packing, and . The radius concerns the accessible positions of the moving centre; subtracting only one disc radius would incorrectly leave a finite rattling area at contact. The real cage is bounded by curved exclusion arcs, so the circle is an inscribed approximation.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 3 48 c Solution by
Codex 0 2026-10-20
The surface pressure follows by differentiating the logarithm of the canonical partition function with respect to container area:Its formal divergence is therefore . This lies outside the low-density regime used to derive the expression. It is a model pole, not an actual geometrical close-packing fraction.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 3 48 b Solution by
Codex 0 2026-10-20
At sufficiently low packing fraction, neglect overlaps of the excluded regions around different disc centres. Sequential insertion gives the configurational integralTaking logarithms and expanding to first order in density,For large , replace by . The expression has the same leading-density expansion when . Thus the hard-disc excluded-area approximation yieldsThe factor accounts for indistinguishable particles and for two-dimensional momentum integration. The factor one-half between pair-excluded area and avoids counting every pair twice. Equality of leading expansions does not make this free-volume resummation exact at high density.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 3 48 a Solution by
Codex 0 2026-10-20
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 3 47 d Solution by
Codex 0 2026-10-20
The volume integral is a gamma function integral,For the prescribed bridge relation, , so . Since , this gives the macroscopic ideal-gas resultHeating at fixed pressure requires both an increase in internal energy and expansion work; at fixed volume the latter contribution is absent.
The finite-size normalization of the isothermal-isobaric ensemble matters here. Applying part (a) literally to the unprefactored instead gives and . The two answers differ because . The normalized continuous-volume distribution has ; its variance agrees with the unprefactored result in part (b). The bridge convention subtracts an order-one contribution and recovers the usual extensive thermodynamic answer. Both agree to leading order in the thermodynamic limit, but their exact finite- derivatives should not be mixed.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 3 47 c Solution by
Codex 0 2026-10-20
Away from singular regimes, assume enthalpy and heat capacity at constant pressure are extensive: and , with nonzero intensive . The enthalpy fluctuation formula then givesAbsolute fluctuations grow as , while the mean grows as . In the thermodynamic limit, enthalpy per particle therefore concentrates around its mean; it is relative, rather than absolute, fluctuations that vanish. This is the usual relative fluctuations of an extensive observable scaling. It assumes ordinary extensive heat capacity and a nonzero mean with a stated energy reference; a critical divergence or an artificially zero mean requires separate analysis.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 3 47 b Solution by
Codex 0 2026-10-20
Differentiate the normalized average, including the derivative of its denominator. The enthalpy fluctuation formula follows fromSince ,Thus a larger equilibrium spread of enthalpy corresponds to a larger response to heating at fixed pressure. The variance is nonnegative, so this finite-system equilibrium description has . The relation connects a response coefficient to fluctuations of the appropriate exchanged quantity; at fixed volume the corresponding quantity is internal energy.
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