Solids, Liquids and Gases Worked ExamplesIGCSE Chemistry 0620

(a)(i) In a solid the particles touch and are in a regular pattern — box R. (ii) In a gas the particles are far apart and randomly arranged — box Q. (Box P is the liquid: particles touch but are irregular.)

(b) For a liquid: separation — particles are close together / touching; arrangement — irregular (random); motion — particles move around (slide) past each other.

(c) A gas has no fixed volume — it spreads out to fill its container — and it can be compressed easily. A liquid has a fixed volume and is almost incompressible. (Both flow and both take the shape of the container, so those are not differences.)

(a) 1: solid → liquid is melting. 2: liquid → solid is freezing. 4: gas → liquid is condensing (condensation).

(b) Liquid → gas is either boiling or evaporating.

Difference: boiling takes place throughout the liquid and only at the boiling point; evaporating takes place only at the surface of the liquid and happens at any temperature below the boiling point.

The flat sections are changes of state. Q–R, at $80^{\circ}\text{C}$, is melting; S–T, at $218^{\circ}\text{C}$, is boiling.

Energy is supplied at a constant rate, so the energy absorbed is proportional to the time. S–T lasts $33 - 17 = 16$ min; Q–R lasts $11 - 5 = 6$ min. More energy is absorbed during boiling. Answer: C.

A is wrong: during Q–R the substance is melting, so solid and liquid are both present. B is wrong: energy is still absorbed — it is used to overcome the forces of attraction between particles instead of raising the temperature. D is wrong: at U the substance is a gas, so the particles are much further apart than in the liquid at R.

(a) The flat section is at $54^{\circ}\text{C}$. A pure substance freezes and melts at the same temperature, so the melting point is $54^{\circ}\text{C}$.

(b)(i) A–B: liquid (the temperature falls). (ii) B–C: liquid and solid — Z is freezing. (iii) D: solid.

(c) As Z freezes, the particles are pulled into fixed positions as forces of attraction form between them. This releases energy, which replaces the energy lost to the surroundings, so the temperature (the average kinetic energy of the particles) does not fall until all the liquid has frozen.

(d) At A (hot liquid) the particles move around, sliding past each other, and move fast. At D (solid) the particles only vibrate about fixed positions, and more slowly because they have less kinetic energy.

The substance is cooled, so the temperature must fall overall. That rules out A, which is a heating curve.

Two changes of state happen: condensing (gas → liquid) and then freezing (liquid → solid). Each gives a flat section, so there must be two plateaus. That rules out C, which shows only freezing.

A pure substance condenses at the same temperature as it boils ($218^{\circ}\text{C}$) and freezes at the same temperature as it melts ($80^{\circ}\text{C}$). B shows condensing at a lower temperature than the boiling point, so it is wrong. Answer: D.

(a) The average kinetic energy decreases — the molecules move more slowly as the gas cools.

(b) Slower molecules hit the walls of the cylinder less often (lower collision frequency) and with less force.

So the pressure decreases. The volume cannot change because the cylinder is rigid.

(c) No change — the cylinder is sealed, so no molecules can enter or leave. Only their speed changes.

(a) The volume increases as the temperature increases — by $4\text{ cm}^3$ for every $20^{\circ}\text{C}$ rise.

(b) At a higher temperature the particles have more kinetic energy and move faster. They hit the plunger more often and with more force, so the plunger is pushed out. The gas expands until its pressure again equals atmospheric pressure — the particles end up further apart.

(c) The volume rises by the same amount for each equal rise in temperature, so $50^{\circ}\text{C}$, halfway between $40^{\circ}\text{C}$ and $60^{\circ}\text{C}$, gives a volume halfway between $64\text{ cm}^3$ and $68\text{ cm}^3$.

(d) So that the air in the syringe has time to reach the same temperature as the water bath before the volume is read.

(e) Friction stops the plunger moving out fully, so the volume readings would be too low (especially at higher temperatures). The student could twist or gently tap the plunger before each reading (or use a lubricated / glass syringe) so it moves freely.

Particles closer together means a smaller volume. Decreasing the temperature makes the particles move more slowly, so a gas at constant pressure contracts. Increasing the pressure at constant temperature squashes the particles closer together.

Only D has both changes acting to reduce the volume, so it always works. Answer: D.

In A and B the two changes act in opposite directions, so the outcome depends on how big each change is — it is not "always". C makes the volume increase: both changes spread the particles out.

(a) At a constant rate of heating, energy supplied $\propto$ time. Find the time for each change of state.

Divide to compare them.

(b) On melting, the particles only need enough energy to break out of their fixed positions and slide past each other — they stay close together, so only some of the attractive forces are overcome.

On boiling, the particles must be separated completely and move far apart, so (almost) all of the forces of attraction between the particles must be overcome. This needs much more energy.

(c) During boiling, all the energy supplied is used to overcome the forces of attraction between the particles, not to increase their kinetic energy. The temperature stays at the boiling point until all the liquid has turned into gas.

A substance is a solid below its melting point, a liquid between its melting and boiling points, and a gas above its boiling point.

(a) Only J has a boiling point below $25^{\circ}\text{C}$ ($-183^{\circ}\text{C}$), so J is a gas.

(b) Liquids have melting point $< 25^{\circ}\text{C} <$ boiling point: K ($-7$ to $59$) and M ($-114$ to $78$). N melts at $44^{\circ}\text{C}$, so it is still solid at $25^{\circ}\text{C}$.

(c) $-10^{\circ}\text{C}$ is below $-7^{\circ}\text{C}$ (it is colder), so K is a solid. $70^{\circ}\text{C}$ is above $59^{\circ}\text{C}$, so K is a gas.

(d) Liquid range = boiling point − melting point.

So L has the largest liquid range, $1413 - 801 = 612^{\circ}\text{C}$. (e) M melts at $-114^{\circ}\text{C}$ and boils at $78^{\circ}\text{C}$, so it is liquid at both $0^{\circ}\text{C}$ and $70^{\circ}\text{C}$: M. (K has boiled by $70^{\circ}\text{C}$.)

(a) Draw three sections: a smooth curve through the falling points from $0$ to about $4$ min, a horizontal straight line at the freezing point, and a smooth curve through the falling points after about $9$ min. Ignore the anomalous point; the line does not need to pass through every point.

(b) At 6 min the reading of $72.0^{\circ}\text{C}$ lies well above the points on either side ($69.0^{\circ}\text{C}$ at $5$ min and at $7$ min). During the flat section the temperature should have been $69^{\circ}\text{C}$.

(c) The temperature stays constant at $69^{\circ}\text{C}$ from about $4$ min to $9$ min while energy is still being lost. A constant temperature while cooling means a change of state — the liquid is freezing — so the freezing point is $69^{\circ}\text{C}$.

(d) At $2$ min ($75^{\circ}\text{C}$, above the freezing point) W is liquid. At $12$ min ($56.5^{\circ}\text{C}$, after freezing has finished) W is solid.

(e) The flat section would be at the same temperature ($69^{\circ}\text{C}$), because the freezing point does not depend on mass, but it would last longer, because more particles must be attracted into fixed positions and more energy must be lost before all of W has frozen.

(a) In warmer air the particles have more kinetic energy and move faster. They hit the inside walls of the tyre more often and with more force. The volume cannot increase much, so the pressure increases.

(b)(i) Cooling makes the helium atoms move more slowly, so they hit the inside of the balloon less often and with less force. The balloon would contract until the pressure inside again balances the pressure outside: the volume decreases (atoms closer together).

(ii) With less pressure from the air particles outside, the helium atoms hitting the inside of the balloon push the balloon outwards. The balloon expands until the pressures balance: the volume increases (atoms further apart).

(iii) The two effects act in opposite directions. The balloon actually gets larger, so the effect of the decrease in pressure must be the larger effect.