Thermodynamic Property Relations Questions and Answers

Answer: A

Writing the change in internal energy as temperature times the entropy change minus pressure times the volume change makes entropy and volume its natural variables. Each of the four energies has its own pair, and using the wrong pair loses information.

Answer: B

The Legendre transformation that defines each potential swaps one variable for its conjugate, which is why the four pairs form a neat pattern. Choosing the potential whose variables you control is what makes a problem tractable.

Answer: C

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the cause. Each thermodynamic potential is natural to a particular pair of variables, and the Gibbs energy is natural to temperature and pressure - which are precisely what a plant controls and measures. The Helmholtz energy would be preferred if volume rather than pressure were fixed, and internal energy if entropy and volume were, but neither situation is common in practice.

Answer: D

Because each potential is an exact differential, its cross derivatives are equal in either order, and each equality is a Maxwell relation. They hold for any substance whatever its equation of state. Only exactness of the differential is assumed anywhere.

Answer: A

They follow purely from the mathematics of exact differentials, so they hold for real gases, liquids and solids alike. That generality is what makes them useful for converting unmeasurable derivatives into measurable ones.

Answer: B

Entropy cannot be measured directly, but pressure, volume and temperature can, so a relation converting one into the other is exactly what a property calculation needs. Heat capacity data are still required for the temperature dependence.

Answer: C

Correct answer: 1 and 2 only

Statements 1 and 2 are the criterion of spontaneity and equilibrium at constant temperature and pressure: the Gibbs energy falls as the process proceeds and is stationary at a minimum when it stops. Statement 3 confuses two different quantities - \( G = H - TS \), so they coincide only if the entropy term vanishes, which it does not at any finite temperature.

Answer: D

Correct answer: the Helmholtz energy is a minimum

Fixing temperature and volume rather than temperature and pressure changes which potential is minimised: it is the Helmholtz energy \( A = U - TS \). The pattern is general - each potential is minimised at equilibrium when its own natural variables are the ones held constant - and applying the Gibbs criterion to a constant-volume system is a common and consequential error.

Answer: A

Subtracting the temperature times the entropy change, that is thirty kilojoules, from forty leaves ten kilojoules per mole. Failing to convert joules to kilojoules is what produces answers wrong by a factor of a thousand.

Answer: B

Both are true and the reason gives the derivation: differentiating that equality along the line produces the relation directly, with the volume change of the transition appearing in the denominator. It holds for melting and sublimation lines equally well.