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Equations of State Questions and Answers

Answer: A

Subtracting b from the molar volume reduces the space available for movement, which is the excluded volume correction. The parameter a is the one that carries the attractive contribution. Together they turn the ideal law into a two-parameter one.

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Answer: B

A molecule about to strike the wall is pulled back by its neighbours, so it arrives with less momentum than an ideal calculation predicts. The term is subtracted from the pressure and scales with the square of density.

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3.

Assertion (A): Van der Waals parameters can be obtained from critical constants alone.
Reason (R): The critical isotherm has both a zero first derivative and a zero second derivative at the critical point.

Answer: C

Both are true and the reason gives the method: two conditions on the same isotherm give two equations, which is exactly enough to fix the two parameters without any further data. Critical temperature and pressure are tabulated for most fluids.

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4.

Van der Waals predicts a critical compressibility factor of three eighths. What decimal value is that?

Answer: D

Three divided by eight is three hundred and seventy-five thousandths. Real fluids cluster nearer two hundred and seventy thousandths, which is the clearest single measure of how crude the equation is at the critical point.

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5.

State whether the following is true or false.
The van der Waals equation predicts the critical compressibility factor accurately for most real fluids.

Answer: A

It gives the same value for every substance while real fluids lie considerably lower and vary among themselves. That failure is the reason the later cubic equations were developed. Its value for water is under a quarter.

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6.

Match each cubic equation of state with its distinguishing feature.
List I
A. Van der Waals
B. Redlich-Kwong
C. Soave-Redlich-Kwong
D. Peng-Robinson
List II
1. an alpha function fitted to vapour pressure data
2. the first cubic, with a temperature-independent attraction term
3. a modified volume term giving better liquid densities
4. an attraction term divided by the square root of temperature

Answer: B

The family is a sequence of repairs to one idea, each addressing a specific failing of its predecessor. Knowing which repair each made is what guides the choice of equation for a particular duty. The lineage runs from eighteen seventy-three to the modern packages.

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Answer: C

Correct answer: a temperature-dependent alpha function containing the acentric factor

Soave replaced Redlich-Kwong’s simple \( 1/\sqrt{T} \) dependence in the attraction term with a fitted function \( \alpha(T,\omega) \) built around the acentric factor. The gain was decisive: the equation then reproduces vapour pressures accurately, which is what makes it usable for vapour–liquid equilibrium rather than for gas density alone. The equation stays cubic and gains no new parameter beyond the acentric factor, which is already tabulated.

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Answer: D

Its modified volume dependence was chosen specifically to improve the liquid root, where Soave is noticeably poor. For vapour pressures alone the two perform very much alike. Simulators usually offer both and let the user choose.

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9.

Consider the following statements about cubic equations of state.
1. They can be solved analytically or by iteration for volume.
2. They can represent both liquid and vapour phases with one equation.
3. They are more accurate than the virial equation at all conditions.

Answer: A

The first two are correct and are the reasons cubics dominate process simulation. The third overstates the case, because a truncated virial equation with good coefficients is more accurate in the moderate pressure gas region.

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Answer: B

The middle root lies on a branch where volume increases with pressure, which is mechanically unstable and cannot be realised. The two outer roots are the saturated phases at that pressure. Their difference vanishes as the critical point is approached.

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