States of Matter Questions and Answers

Answer: A

The order is HF \(\gt\) HI \(\gt\) HBr \(\gt\) HCl.

HF boils far higher than the others (about 20 °C) because its molecules are linked by strong hydrogen bonds, F being small and highly electronegative. Among HCl, HBr and HI hydrogen bonding is negligible, and boiling point rises with molar mass and polarisability, because London dispersion forces grow with the size of the electron cloud: HI (about -35 °C) \(\gt\) HBr (about -67 °C) \(\gt\) HCl (about -85 °C).

HF \(\gt\) HCl \(\gt\) HBr \(\gt\) HI reverses the dispersion trend. HI \(\gt\) HBr \(\gt\) HCl \(\gt\) HF ignores hydrogen bonding in HF. HI \(\gt\) HF puts HI above HF, which understates the effect of hydrogen bonding.

Answer: C

Cl₂ is most nearly ideal at 200 °C and 0.50 atm.

A real gas approaches ideal behaviour when its molecules are far apart (low pressure) and move fast enough for intermolecular attractions to matter little (high temperature). Both conditions make the molecular volume and the attractive forces negligible.

Of the options, 200 °C with 0.50 atm combines the highest temperature with the lowest pressure. 0 °C and 0.50 atm has the same low pressure but a lower temperature, where chlorine's fairly strong attractions show more. The 10.0 atm options compress the gas, and −100 °C at 10.0 atm is close to conditions where chlorine liquefies.

Answer: B

The potential is lowest at \(r = 2^{1/6}\sigma\).

Set the derivative of the potential to zero:

\[ \begin{aligned} \frac{dV_{LJ}}{dr} &= 4\epsilon\left[-\frac{12\sigma^{12}}{r^{13}} + \frac{6\sigma^{6}}{r^{7}}\right] = 0 \\ \frac{12\sigma^{12}}{r^{13}} &= \frac{6\sigma^{6}}{r^{7}} \\ r^{6} &= 2\sigma^{6} \\ r &= 2^{1/6}\sigma \approx 1.12\,\sigma \end{aligned} \]

The option \(\sigma\) is where the potential crosses zero, not where it is lowest: at \(r = \sigma\) the two bracketed terms are equal. The other two expressions cannot be a distance at all: \(\sigma^{3/2}\) has units of length to the power 3/2, and \(\epsilon^{1/6}\) is an energy term with no \(\sigma\) in it.

Answer: D

The value of Y is 4.

Two hard spheres of radius \(r\) cannot approach closer than \(2r\) centre to centre, so each pair of molecules excludes a sphere of radius \(2r\) from the other's centre:

\[ \begin{aligned} V_{pair} &= \frac{4}{3}\pi(2r)^{3} \\ &= 8 \times \frac{4}{3}\pi r^{3} = 8\,V_{molecule} \end{aligned} \]

That excluded volume belongs to a pair, so each molecule is charged half of it, \(4\,V_{molecule}\). For a mole of gas, \(b = 4 \times V_{molecule} \times N_{A}\).

The answer 8 is the excluded volume of the pair without halving it. 2 and 16 come from halving or doubling at the wrong step; the correct factor is four times the actual volume of the molecules.

Answer: E

The highest enthalpy of fusion is that of \(NaCl\).

Melting NaCl means breaking up an ionic lattice held by strong electrostatic forces, so its enthalpy of fusion is large, about 28.8 kJ mol^-1.

The other substances are molecular solids held by weaker intermolecular forces: water (hydrogen bonds) about 6.0, benzene about 10.6, carbon tetrachloride about 2.5 and carbon monoxide about 0.8 kJ mol^-1.

The general rule: the stronger the forces holding the particles in the solid, the more energy is needed to melt one mole of it.

Answer: D

The increasing order is CCl₄ \(\lt\) CH₃COCH₃ \(\lt\) C₆H₆.

The standard enthalpies of fusion are about 2.5 kJ mol^-1 for carbon tetrachloride, 5.7 for acetone and close to 10 for benzene. Benzene's flat, symmetrical molecules pack efficiently in the solid, so more energy is needed to break up the crystal, while the near-spherical CCl₄ molecules form a solid that melts easily.

Each other option puts one of these values in the wrong place.

Answer: E

The reason is that f is appreciable and V is not negligible.

The ideal gas model assumes there are no intermolecular forces and that the molecules' own volume is negligible. At high pressure and low temperature the molecules are crowded together and move slowly, so attractive forces become significant and the molecules' own volume is a sizeable part of the total.

Both assumptions fail, which is why van der Waals added the corrections a and b. The options saying f or V is negligible describe the conditions in which a gas behaves ideally.

Answer: D

Cooling the gas leads to a decrease in the average velocity of the gas.

The mean speed of gas molecules depends only on temperature and molar mass, \(\bar{c} = \sqrt{8RT/\pi M}\). Cooling from 50°C (323 K) to 25°C (298 K) lowers T, so the molecules move more slowly on average.

At constant volume the number of molecules per unit volume does not change, so the mean free path stays the same, which rules out option B. Slower molecules collide less often, so the number of collisions per unit time falls, which rules out option A. An increase in average velocity would require heating, not cooling.

Answer: B

The correct equation is dP = −RT/(V−b)² dV + R/(V−b) dT.

For a gas obeying P = RT/(V − b) (the van der Waals equation with only the excluded-volume correction), P is a function of V and T, and its total differential is

\[ \begin{aligned} dP &= \left(\frac{\partial P}{\partial V}\right)_{T} dV \\ &+ \left(\frac{\partial P}{\partial T}\right)_{V} dT \end{aligned} \]

\[ \begin{aligned} \left(\frac{\partial P}{\partial V}\right)_{T} &= -\frac{RT}{(V-b)^{2}} \\ \left(\frac{\partial P}{\partial T}\right)_{V} &= \frac{R}{V-b} \end{aligned} \]

Pressure falls as volume rises, so the dV coefficient must be negative, which rules out the options with +2RT. The dT coefficient has (V − b) to the first power, not squared, which rules out the last two options.