MDCAT Physics chapter 7: Physics · Thermodynamics MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
18 exam-style questions on Physics · Thermodynamics, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
Ten Physics · Thermodynamics questions with answers
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. In a certain process an ideal gas gives out 300 J of heat, and its internal energy falls by exactly 300 J. Which kind of process is this?
A
isothermal
B
adiabatic
C
isobaric
D
isochoric
Show answer
Answer: D. By the first law Q = ΔU + W: −300 = −300 + W, so W = 0 and the volume does not change. An isothermal change would have ΔU = 0, an adiabatic one Q = 0, and cooling at constant pressure involves work as the volume shrinks.
2. The absolute temperature of an ideal gas in a sealed rigid container is made four times as large. The pressure of the gas and the rms speed of its molecules change by the factors
A
pressure × 4, rms speed × 2
B
pressure × 2, rms speed × 4
C
pressure × 4, rms speed × 4
D
pressure × 2, rms speed × 2
Show answer
Answer: A. At constant volume P ∝ T, so the pressure becomes 4 times as large. vrms ∝ √T, so the rms speed becomes √4 = 2 times as large. Forgetting the square root gives a factor of 4 for the speed.
3. When water at 0 °C freezes into ice at 0 °C, the entropy of the water
A
decreases, as its molecules become more ordered
B
increases, as it gives out latent heat to the surroundings
C
stays the same, as its temperature does not change
D
becomes zero, as the ice is a perfect crystal
Show answer
Answer: A. ΔS = ΔQ/T, and ΔQ is negative because latent heat leaves the water, so its entropy falls: the solid is more ordered. The entropy of the colder surroundings rises by more than this, so the total entropy still increases.
4. Two moles of an ideal gas are heated at constant pressure, and the temperature rises by 50 K. Taking R = 8.31 J mol−1 K−1, the work done by the gas is
A
415.5 J
B
1246.5 J
C
831 J
D
zero
Show answer
Answer: C. At constant pressure W = PΔV = nRΔT = 2 × 8.31 × 50 = 831 J. 415.5 J is the work for one mole only, and 1246.5 J is (3/2)nRΔT, the rise in internal energy of a monatomic gas, not the work done.
5. Which of these would violate the second law of thermodynamics?
A
an engine that works in a cycle and turns all the heat it takes in into work
B
a refrigerator that uses electrical work to move heat from its cold inside to a warm room
C
an ideal gas that expands isothermally, doing work equal to all the heat it absorbs
D
an engine working between 600 K and 300 K that takes in 1000 J and gives out 700 J per cycle
Show answer
Answer: A. The Kelvin statement forbids an engine working in a cycle that converts heat completely into work. A single isothermal expansion can do this because it is not a cycle (the gas ends in a different state); a refrigerator is allowed because work is supplied to it; and 30% is below the Carnot limit of 1 − 300/600 = 50%.
6. An ideal gas is at 127 °C. To what temperature must it be heated so that the average translational kinetic energy of its molecules is doubled?
A
254 °C
B
800 °C
C
527 °C
D
1327 °C
Show answer
Answer: C. The average kinetic energy is proportional to the absolute temperature, so T must double from 400 K to 800 K, which is 527 °C. Doubling the Celsius value gives 254 °C, and 800 °C forgets to convert back from kelvin.
7. A real heat engine, with friction and heat losses, works between the same two temperatures as a Carnot engine. Its efficiency is
A
greater than the Carnot efficiency
B
equal to the Carnot efficiency
C
greater or less, depending on its working substance
D
less than the Carnot efficiency
Show answer
Answer: D. No engine working between two given temperatures can be more efficient than a reversible (Carnot) engine. A real engine has irreversible losses, so its efficiency is lower. The Carnot efficiency does not depend on the working substance.
8. In each cycle a heat engine takes in 800 J of heat from its source and gives out 600 J of heat to its sink. Its efficiency is
A
75%
B
33%
C
25%
D
14%
Show answer
Answer: C. Work done = 800 − 600 = 200 J, and efficiency = work ÷ heat taken in = 200/800 = 25%. 75% is the fraction of the heat that is rejected, and 33% divides the work by the heat rejected.
9. An ideal gas is compressed slowly at constant temperature, and 250 J of work is done on it. During the compression
A
250 J of heat flows into the gas
B
250 J of heat flows out of the gas
C
no heat flows, as the temperature is constant
D
500 J of heat flows out of the gas
Show answer
Answer: B. In an isothermal change ΔU = 0 for an ideal gas, so Q = ΔU + W = 0 + (−250 J) = −250 J: the gas gives out as much heat as the work done on it. A constant temperature does not mean that no heat flows.
10. A rigid container holds an ideal gas at 300 K and pressure P. At constant temperature, half of the gas leaks out; the gas left behind is then heated to 600 K. Its pressure is now
A
2P
B
P
C
P/2
D
4P
Show answer
Answer: B. P = ⅔N0⟨½mv2⟩ with ⟨½mv2⟩ ∝ T, so at fixed volume P ∝ N0T. Halving the number of molecules halves P and doubling T doubles it, so the pressure is back to P. 2P allows only for the heating.
18, each with the answer and a short explanation, across 2 topics: Kinetic theory and the first law, Heat engines and entropy. Every chapter test draws ten of them, so you can take it many times.
Is the MDCAT Physics chapter 7 test free?
Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.
How is the chapter test marked?
The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
