Past exam of the physics B course of the University of Cambridge / 2018 / Paper 2 / Section C / C10

Treat the film at fixed material content, with area $A$ the coordinate conjugate to the surface tension in the stated work term. The Helmholtz free energy is a Legendre transform of the internal energy:Its natural independent variables are therefore $T,A$, soEquality of mixed second partial derivatives gives the Maxwell relation for the thermodynamics of a liquid film:The area convention matters: the supplied fundamental relation already specifies the coefficient of $dA$. There is no additional factor of two unless one changes from total interfacial area to a projected film area and changes the work term consistently.

Write ${\gamma }^{\prime }=d\gamma /dT<0$. The Maxwell relation givesThus isothermal stretching increases the entropy. For an infinitesimal increase, $dS=-{\gamma }^{\prime }dA$; if the derivative is independent of area, this integrates to $\Delta S=-{\gamma }^{\prime }\Delta A$ at fixed temperature. Mechanical energy used to create extra surface is not, in general, identical to the increase of internal energy at fixed temperature, because the stretching can also absorb heat.

The definition of heat capacity at constant area and the Maxwell relation give the entropy differentialFor a reversible adiabatic process, $dS=0$. Hencefor positive heat capacity $C_A$. Reversible adiabatic stretching cools the film. Reversibility, or negligible entropy production, is essential to this inference. Adiabatic isolation alone only sets the heat transfer to zero; an irreversible stretch can generate entropy, so its temperature change is not fixed by this isentropic derivative.

Substitute the entropy differential into the fundamental relation:ThereforeThe coefficient of $dA$ is the change of internal energy per added area at fixed temperature, whereas $\gamma$ is the change of Helmholtz free energy per added area. Their difference $-T{\gamma }^{\prime }$ is the heat absorbed per added area in a reversible isothermal stretch. Both $\gamma$ and $T{\gamma }^{\prime }$ have units $\mathrm{J}{\mathrm{m}}^{-2}=\mathrm{N}{\mathrm{m}}^{-1}$.

Reading the original PDF graph gives approximatelyFor example, the points near $330\mathrm{K}$ and $450\mathrm{K}$ are about $22$ and $6.5\mathrm{mN}{\mathrm{m}}^{-1}$, respectively, giving this slope. The graph supports approximate values rather than exact fitted constants. The area increment is $\Delta A=10^{-4}{\mathrm{m}}^2$.

At fixed temperature, the surface tension is constant during this stretch. The work done on the film is positive:For a reversible isothermal process, the heat supplied is $Q=T\Delta S=-T{\gamma }^{\prime }\Delta A$. ThusThe film absorbs heat while its temperature stays constant. The first law gives the useful consistency check $\Delta U=Q+W_{\mathrm{on}}\simeq 6.5\mu \mathrm{J}$, agreeing with $(\gamma -T{\gamma }^{\prime })\Delta A$.

Over the plotted range, approximate the surface tension by $\gamma (T)={\gamma }_0-\beta T$ with constant positive $\beta =-{\gamma }^{\prime }$, and take $C_A=70\mathrm{mJ}{\mathrm{K}}^{-1}=0.070\mathrm{J}{\mathrm{K}}^{-1}$. A reversible adiabat hasIntegrating givesThe arbitrary fixed reference temperature $T_{\ast }$ makes the logarithm dimensionless; changing it only shifts the integration constant. This is the dimensionally precise version of writing $C_A\ln T+\beta A$ as constant. Each term has units of entropy. Between two states,If the graph had substantial curvature, the entropy differential would still hold, but this particular constant-$\beta$ exponential would no longer follow.

For the specified increment,Using the adiabat, or its first-order Taylor expansion,This is a very small cooling, about $61\mu \mathrm{K}$. There is no heat exchange, and the film is stretched against positive surface tension, so the internal energy increases, withThe cooling does not contradict this increase. In the linear model, $\gamma -T{\gamma }^{\prime }={\gamma }_0\simeq 0.065\mathrm{J}{\mathrm{m}}^{-2}$, soCreation of surface raises the internal energy more than the cooling lowers its temperature-dependent part. The change in surface tension during this tiny temperature fall changes the work only at a negligible higher order.