Thermodynamics NEET PYQReal NEET & AIPMT 2005–2019 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Thermodynamics question from the NEET & AIPMT papers on this site (2005–2019), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Thermodynamics previous year questions
24 questions. Tap an option to check your answer.
Q1NEET 2019 · Q40
In which of the following processes, heat is neither absorbed nor released by a system?
Not quite — the answer is B.
An adiabatic process is by definition one with no heat exchange (Q = 0). In an isothermal process heat does flow, exactly balancing the work done, so that temperature remains constant.
Q2NEET 2018 · Q4
The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is
Not quite — the answer is C.
Carnot efficiency η = 1 − T₂/T₁ = 1 − 273/373 = 100/373 ≈ 26.8%. Temperatures must be in kelvin; using Celsius would wrongly give 100%.
Q3NEET 2018 · Q45
A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 10⁵ N m⁻²) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is
Not quite — the answer is B.
Q = 54 cal ≈ 54 × 4.18 ≈ 225.7 J. Work by the vapour W = PΔV = 1.013 × 10⁵ × (167.1 − 0.1) × 10⁻⁶ ≈ 16.9 J. ΔU = Q − W ≈ 208.8 J ≈ 208.7 J.
Q4NEET 2017 · Q148
A carnot engine having an efficiency of 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is:
Not quite — the answer is B.
For a Carnot cycle the coefficient of performance as a refrigerator is β = (1 − η)/η = (9/10)/(1/10) = 9. Heat drawn from the cold reservoir Q₂ = βW = 9 × 10 = 90 J. 100 J is the heat rejected to the hot reservoir (Q₂ + W).
Q5NEET 2017 (Gujarati paper) · Q163
One mole of a gas obeying the equation of state P(V − b) = RT is made to expand from a state with coordinates (P₁, V₁) to a state with (P₂, V₂) along a process that is depicted by a straight line on a P–V diagram. Then, the work done is given by:
Not quite — the answer is A.
Work done is the area under the P–V path, whatever the equation of state. For a straight line from (V₁, P₁) to (V₂, P₂) the area is a trapezium: W = ½(P₁ + P₂)(V₂ − V₁). The constant b does not enter, because the work uses the actual volume change.
Q6NEET 2016 Phase 2 · Q17
One mole of an ideal monatomic gas undergoes a process described by the equation PV³ = constant. The heat capacity of the gas during this process is
Not quite — the answer is D.
For a polytropic process PVⁿ = constant, C = C_v + R/(1 − n). With C_v = 3R/2 and n = 3, C = 3R/2 − R/2 = R. 3R/2 is just C_v, which applies only at constant volume.
Q7NEET 2016 Phase 2 · Q18
The temperature inside a refrigerator is t₂ °C and the room temperature is t₁ °C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be
Not quite — the answer is B.
For an ideal (Carnot) refrigerator, heat delivered to the hot reservoir per unit work is Q₁/W = T₁/(T₁ − T₂), with T₁ = t₁ + 273 K and T₁ − T₂ = t₁ − t₂. The expression with t₂ + 273 in the numerator is the heat extracted from inside (coefficient of performance), not the heat delivered to the room.
Q8NEET 2016 Phase 1 · Q11
A refrigerator works between 4°C and 30°C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is: (Take 1 cal = 4.2 Joules)
Not quite — the answer is C.
Coefficient of performance = T₂/(T₁ − T₂) = 277/26 ≈ 10.65. Heat removed per second = 600 × 4.2 = 2520 W, so power = 2520/10.65 ≈ 236.5 W. Forgetting to convert calories to joules gives about 56 W, and slips in the powers of ten give the other values.
Q9NEET 2016 Phase 1 · Q30
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then:
Not quite — the answer is B.
On a P–V diagram the adiabatic curve is steeper, so during adiabatic compression the pressure rises faster and the area under the curve (work done on the gas) is larger for the same volume change. This holds for any γ > 1, so it does not depend on atomicity.
Q10AIPMT 2015
A Carnot engine, having an efficiency of η = 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is:
Not quite — the answer is D.
For the Carnot cycle η = W/Q_H = 1/10, so with W = 10 J, Q_H = 100 J. The heat taken from the cold reservoir is Q_L = Q_H − W = 90 J. 100 J is the heat rejected to the hot reservoir, not that absorbed from the cold one.
Q11AIPMT 2014
A monoatomic gas at a pressure P, having a volume V expands isothermally to a volume 2V and then adiabatically to a volume 16V. The final pressure of the gas is: (take γ = 5/3)
Not quite — the answer is B.
Isothermal doubling halves the pressure to P/2. Adiabatic from 2V to 16V: P_f = (P/2)(2/16)^(5/3) = (P/2)(1/32) = P/64. Pressure must fall on expansion, so the large multiples are impossible.
Q12NEET 2013
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of Cp/Cv for the gas is
Not quite — the answer is D.
For an adiabatic process PV^γ = constant combined with PV = nRT gives P ∝ T^(γ/(γ − 1)). Setting γ/(γ − 1) = 3 gives γ = 3γ − 3, so γ = 3/2. 5/3 is the monatomic value and does not satisfy the given relation.
Q13AIPMT 2011 Mains
A mass of diatomic gas (γ = 1.4) at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from 27°C to 927°C. The pressure of the gas in the final state is
Not quite — the answer is A.
For an adiabatic process P^(1−γ)T^γ is constant, so P ∝ T^(γ/(γ−1)) = T^3.5. T rises from 300 K to 1200 K, a factor of 4, so P₂ = 2 × 4^3.5 = 2 × 128 = 256 atm. Using Celsius temperatures or P ∝ T (isochoric) gives the wrong values.
Q14AIPMT 2011 Prelims
When 1 kg of ice at 0°C melts to water at 0°C, the resulting change in its entropy, taking latent heat of ice to be 80 cal/g, is
Not quite — the answer is A.
Heat absorbed Q = mL = 1000 g × 80 cal/g = 8 × 10⁴ cal at constant T = 273 K. ΔS = Q/T = 80000/273 ≈ 293 cal/K. Giving 8 × 10⁴ cal/K forgets to divide by the temperature.
Q15AIPMT 2010 Mains · Q80
If cp and cv denote the specific heats (per unit mass) of an ideal gas of molecular weight M, then (where R is the molar gas constant)
Not quite — the answer is C.
For one mole, Cp − Cv = R (Mayer's relation). Specific heats per unit mass are molar heats divided by M, so cp − cv = R/M. 'cp − cv = R' applies to molar specific heats, not per unit mass.
Q16AIPMT 2010 Mains · Q90
A monoatomic gas at pressure P₁ and volume V₁ is compressed adiabatically to 1/8th of its original volume. What is the final pressure of the gas?
Not quite — the answer is D.
For an adiabatic process PV^γ is constant with γ = 5/3 for a monoatomic gas. P₂ = P₁(V₁/V₂)^γ = P₁ × 8^(5/3) = P₁ × 2⁵ = 32P₁. 64P₁ would correspond to γ = 2, and 16P₁ to γ = 4/3.
Q17AIPMT 2010 Prelims · Q17
If ΔU and ΔW represent the increase in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true?
Not quite — the answer is B.
First law: Q = ΔU + ΔW with ΔW the work done by the system. In an adiabatic process Q = 0, so ΔU = −ΔW. In an isothermal process of an ideal gas ΔU = 0, so neither isothermal relation holds.
Q18AIPMT 2009 Prelims · Q3
In thermodynamic processes which of the following statements is not true?
Not quite — the answer is A.
In an isochoric process the volume is held constant; constant pressure describes an isobaric process. The other statements (constant temperature in isothermal, PV^γ = constant and thermal insulation in adiabatic) are all correct.
Q19AIPMT 2009 Prelims · Q37
The internal energy change in a system that has absorbed 2 kcal of heat and done 500 J of work is:
Not quite — the answer is C.
First law: ΔU = Q − W = 2 × 4200 − 500 = 7900 J (1 cal = 4.2 J). Adding the work instead of subtracting gives 8900 J.
Q20AIPMT 2008 Prelims · Q14
If Q, E and W denote respectively the heat added, change in internal energy and the work done in a closed cycle process, then
Not quite — the answer is D.
Internal energy is a state function, so after a complete cycle its change E is zero. From Q = E + W this gives Q = W, but neither heat nor work need be zero: a heat engine running in a cycle does net work.
Q21AIPMT 2007 Prelims · Q42
An engine has an efficiency of 1/6. When the temperature of sink is reduced by 62°C, its efficiency is doubled. Temperature of the source is
Not quite — the answer is C.
1/6 = 1 − T₂/T₁ and 1/3 = 1 − (T₂ − 62)/T₁. Subtracting, 62/T₁ = 1/6, so T₁ = 372 K = 99°C. 62°C confuses the temperature drop with the source temperature.
Q22AIPMT 2006 Prelims · Q4
A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency ?
Not quite — the answer is C.
With the sink at 300 K, 0.4 = 1 − 300/T₁ gives T₁ = 500 K. The new efficiency is 0.4 + 0.5 × 0.4 = 0.6, so 300/T₁' = 0.4 and T₁' = 750 K. The source must be raised by 750 − 500 = 250 K; reading '50% more' as an efficiency of 90% gives an impossible value.
Q23AIPMT 2005 Prelims · Q20
Which of the following processes is reversible ?
Not quite — the answer is D.
A quasi-static isothermal compression can be retraced through the same equilibrium states, so it is reversible. Heat flow by conduction or radiation across a finite temperature difference and Joule heating are irreversible.
Q24AIPMT 2005 Prelims · Q28
An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C. It absorbs 6 × 10⁴ cal of heat at higher temperature. Amount of heat converted to work is :
Not quite — the answer is C.
η = 1 − T₂/T₁ = 1 − 400/500 = 0.2 (temperatures in kelvin). W = 0.2 × 6 × 10⁴ = 1.2 × 10⁴ cal. Using Celsius temperatures or the rejected heat (4.8 × 10⁴ cal) gives the wrong choices.
How often Thermodynamics came up
Questions from this chapter in each paper on this site.
- NEET 2019: 1
- NEET 2018: 2
- NEET 2017: 1
- NEET 2017 (Gujarati paper): 1
- NEET 2016 Phase 2: 2
- NEET 2016 Phase 1: 2
- AIPMT 2015: 1
- AIPMT 2014: 1
- NEET 2013: 1
- AIPMT 2011 Mains: 1
- AIPMT 2011 Prelims: 1
- AIPMT 2010 Mains: 2
- AIPMT 2010 Prelims: 1
- AIPMT 2009 Prelims: 2
- AIPMT 2008 Prelims: 1
- AIPMT 2007 Prelims: 1
- AIPMT 2006 Prelims: 1
- AIPMT 2005 Prelims: 2
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