Thermodynamics: Unit Diagnostic (Physics 2 Unit 9)

What this diagnostic is for

Use this short check to find which topics in Unit 9 (Thermodynamics) you should revisit. It has eight multiple-choice and three short-answer questions. These are original Marlbridge practice questions, not past exam questions. They are not calibrated against real exam results, and your total does not predict an exam score. What matters is which questions you miss.

How to take it:

Work without notes, in about 30 minutes. Write a reason for each answer.
A scientific calculator is allowed. Data: k_B = 1.38 × 10⁻²³ J/K; R = 8.31 J/(mol·K); c for water = 4180 J/(kg·K). Every gas is ideal and monatomic, so U = (3/2)nRT = (3/2)PV. W is the work done on the gas, so ΔU = Q + W. Add 273 to convert °C to K.
Open each answer only after you have committed to your own.

Question 1 (multiple choice · 9.1)

A sealed, rigid container of gas is heated. Which statement best explains, in terms of atoms, why the pressure of the gas increases?

(A) The atoms move faster on average, so each collision with a wall gives a larger impulse and collisions happen more often.
(B) The atoms expand when heated, so they take up more of the container and hit the walls more often.
(C) Heating adds more atoms to the gas, so more atoms hit each square metre of wall.
(D) Each collision gives a larger impulse, but the number of collisions per second stays the same.

Answer and explanation

Answer: (A). A higher temperature means a higher average kinetic energy, so the atoms move faster. A faster atom changes its momentum by more in each bounce, and it reaches the walls more often. Both effects raise the force per unit area.

(B) In the ideal gas model the atoms’ own volume is negligible and does not change.
(C) cannot happen: the container is sealed, so N is fixed.
(D) has half the reasoning. Faster atoms also reach the walls more often.

If you missed this: read “Pressure from collisions” in the Topic 9.1 study guide.

Question 2 (multiple choice · 9.2)

A 5.0 L container holds 1.2 × 10²³ gas molecules at 7 °C. What is the pressure of the gas?

(A) 9.27 × 10⁴ Pa
(B) 2.32 × 10³ Pa
(C) 92.7 Pa
(D) 5.58 × 10²⁸ Pa

Answer and explanation

Answer: (A). Convert first: T = 7 + 273 = 280 K and V = 5.0 × 10⁻³ m³. You are given a number of molecules, so use PV = Nk_BT: P = (1.2 × 10²³)(1.38 × 10⁻²³ J/K)(280 K) ÷ (5.0 × 10⁻³ m³) = 9.27 × 10⁴ Pa.

(B) uses 7 °C instead of 280 K.
(C) leaves the volume in litres.
(D) uses R with a number of molecules. R goes with moles; k_B goes with molecules.

If you missed this: read “The ideal gas law” in the Topic 9.2 study guide.

Question 3 (multiple choice · 9.3)

Object A is placed in thermal contact with object C and left until there is no net energy transfer between them. Separately, object B is placed in thermal contact with C and left in the same way. A and B are then placed in contact with each other. Which statement is correct?

(A) There is no net energy transfer between A and B, because they are at the same temperature.
(B) There is no net energy transfer between A and B only if they have the same internal energy.
(C) Net energy flows from the larger object to the smaller one until their internal energies are equal.
(D) A and B stop exchanging any energy at all as soon as they touch.

Answer and explanation

Answer: (A). Each object reached thermal equilibrium with C, so each is at C’s temperature. Systems at the same temperature have no net energy transfer. This is how a thermometer (C) works.

(B) and (C) use internal energy to decide equilibrium. Temperature decides it.
(D) confuses “no net transfer” with “no transfer”. Atoms at the boundary still collide and pass energy both ways; the flows just balance.

If you missed this: read “Thermal equilibrium” in the Topic 9.3 study guide.

Question 4 (multiple choice · 9.4)

A gas in a cylinder is compressed slowly by a piston while the cylinder sits in a large water bath, so the temperature of the gas stays constant. Which row is correct for the gas?

(A) W > 0, ΔU = 0, Q < 0
(B) W > 0, ΔU > 0, Q = 0
(C) W < 0, ΔU = 0, Q > 0
(D) W = 0, ΔU = 0, Q = 0

Answer and explanation

Answer: (A). The piston pushes in, so work is done on the gas: W > 0. T is constant, so U = (3/2)nRT is constant: ΔU = 0. Then Q = ΔU − W = −W < 0. Energy leaves the gas into the bath.

(B) describes an adiabatic compression. The bath keeps T constant, so energy can leave by heating.
(C) uses the “work done by the gas” sign. It describes an isothermal expansion.
(D) ignores the work done by the piston.

If you missed this: read “Four special processes” in the Topic 9.4 study guide.

Question 5 (multiple choice · 9.4)

A gas is in a rigid container of volume 2.5 × 10⁻³ m³. It is cooled, and its pressure falls from 4.0 × 10⁵ Pa to 1.6 × 10⁵ Pa. What is the energy transferred to the gas by heating?

(A) −900 J
(B) −600 J
(C) +900 J
(D) −1500 J

Answer and explanation

Answer: (A). Rigid means ΔV = 0, so W = 0 and Q = ΔU. With U = (3/2)PV: ΔU = 1.5 × (1.6 × 10⁵ − 4.0 × 10⁵) Pa × 2.5 × 10⁻³ m³ = −900 J. Energy leaves the gas.

(B) is Δ(PV) without the factor 3/2.
(C) has the right size but the wrong sign. The gas cools, so Q < 0.
(D) uses (5/2)Δ(PV), the constant-pressure result.

If you missed this: read “Energy inside a gas” and the special-process table in the Topic 9.4 study guide.

Question 6 (multiple choice · 9.5)

Blocks X and Y have the same mass and are made of different materials. Each receives the same energy by heating. The temperature of X rises three times as much as the temperature of Y. How do their specific heats compare?

(A) c_X = c_Y ÷ 3
(B) c_X = 3c_Y
(C) c_X = c_Y ÷ 9
(D) c_X = c_Y, because the masses and energies are equal

Answer and explanation

Answer: (A). From Q = mcΔT, with Q and m the same, c ∝ 1/ΔT. X has three times the temperature rise, so it has one-third of the specific heat.

(B) treats a large temperature rise as a sign of a large specific heat. A high c means a small temperature change.
(C) squares the factor. The relationship is inverse, not inverse square.
(D) forgets that c belongs to the material.

If you missed this: read “Energy to change temperature” in the Topic 9.5 study guide.

Question 7 (multiple choice · 9.5)

The foam wall of a cool box is 3.0 cm thick. The temperature difference across it is 15 K. On a hotter day the difference rises to 25 K. What foam thickness, with the same area and material, would keep the rate of energy conduction the same as before?

(A) 5.0 cm
(B) 1.8 cm
(C) 3.0 cm
(D) 8.3 cm

Answer and explanation

Answer: (A). Q/Δt = kAΔT/L. With k and A fixed, the rate stays the same if ΔT/L stays the same. So L must rise by the same factor as ΔT: 3.0 cm × 25/15 = 5.0 cm.

(B) multiplies by 15/25, as if a thicker wall conducted faster.
(C) assumes the thickness does not matter.
(D) squares the factor. The rate is proportional to 1/L, not 1/L².

If you missed this: read “Conduction through a slab” in the Topic 9.5 study guide.

Question 8 (multiple choice · 9.6)

In which process does the entropy of the named system decrease?

(A) A sealed jar of hot jam (the system) cools on a shelf in a room.
(B) An ice cube (the system) melts in a warm room.
(C) A gas (the system) expands freely into an empty chamber inside an insulated box.
(D) A sliding book (the system) comes to rest on a table because of friction and warms slightly.

Answer and explanation

Answer: (A). The jam is a closed system. Energy leaves it by cooling, so its entropy can fall. The room gains more entropy than the jam loses, so the total still increases.

(B) Energy enters the ice by heating, so its entropy rises.
(C) No energy enters, but the gas spreads into more space, so its entropy rises.
(D) The book’s kinetic energy spreads into random motion of its atoms, so its entropy rises.

If you missed this: read “Isolated systems and closed systems” in the Topic 9.6 study guide.

Question 9 (short answer · 9.1)

A gas is made of molecules of mass 5.0 × 10⁻²⁶ kg at 320 K.

(a) Calculate the average kinetic energy of a molecule. (b) Calculate the rms speed of the molecules. (c) At what temperature would the rms speed be twice as large?

Worked answer

(a) K_avg = (3/2)k_BT = 1.5 × (1.38 × 10⁻²³ J/K)(320 K) = 6.62 × 10⁻²¹ J.
(b) ½m v_rms² = K_avg, so v_rms = √(2K_avg/m) = √(2 × 6.624 × 10⁻²¹ ÷ 5.0 × 10⁻²⁶) = 515 m/s.
(c) v_rms ∝ √T, so doubling v_rms needs 4 times the kelvin temperature: 1280 K.

The most common slip is 640 K in (c).

If you missed this: read “Temperature and the motion of atoms” in the Topic 9.1 study guide.

Question 10 (short answer · 9.4)

0.050 mol of gas at 450 K is in a cylinder with insulated walls. It expands and does 120 J of work on a piston.

(a) State Q, W and ΔU for the gas. (b) Calculate the final temperature of the gas.

Worked answer

(a) Insulated walls, so Q = 0. The gas does 120 J of work, so the work done on it is W = −120 J. Then ΔU = Q + W = −120 J.
(b) ΔU = (3/2)nRΔT, so ΔT = −120 J ÷ [1.5 × 0.050 mol × 8.31 J/(mol·K)] = −193 K. Final temperature = 450 K − 193 K = 257 K.

An adiabatic expansion cools the gas.

If you missed this: read “Two ways to change internal energy” in the Topic 9.4 study guide.

Question 11 (short answer · 9.6)

A gas in a cylinder is compressed slowly while it sits in cold water, so its temperature stays the same. A student says: “The entropy of the gas goes down, so this process breaks the second law.” Evaluate the claim.

Worked answer

The first part is right: energy leaves the gas by heating (Q < 0), so its entropy decreases. The conclusion is wrong. The gas is a closed system, not an isolated one. The second law says only that the total entropy of an isolated system never decreases. The water gains energy, and its entropy rises by at least as much as the gas’s falls.

If you missed this: read “The second law” and “Isolated systems and closed systems” in the Topic 9.6 study guide.

Your next step

Mark each question. Count a short answer as wrong if any part is missing, and any answer as wrong if it was a guess.

Topic	Question(s)	If you missed it, read
9.1 Kinetic theory of temperature and pressure	1, 9	Topic 9.1 study guide
9.2 The ideal gas law	2	Topic 9.2 study guide
9.3 Thermal energy transfer and equilibrium	3	Topic 9.3 study guide
9.4 The first law of thermodynamics	4, 5, 10	Topic 9.4 study guide
9.5 Specific heat and thermal conductivity	6, 7	Topic 9.5 study guide
9.6 Entropy and the second law	8, 11	Topic 9.6 study guide

How to use your result

Start with the topic where you missed the most. Read its study guide, then do that topic’s practice questions.
Look at why you missed it. The explanation names the mistake behind each distractor. A unit or sign slip (Celsius, litres, the sign of W) needs a habit, not a full re-read.
Missed nothing? Go straight to the Unit 9 mixed review. Its questions combine topics, which is harder.
Come back to this check after a week of revision, and answer again before opening the explanations.