NEET PhysicsUpdated 19 Oct 2026

Thermodynamics: NEET PYQs and weightage

Thermodynamics carries 0.9 of 100 Physics marks in NEET by past-paper share: 7 of the 758 tagged NEET Physics PYQs in the ViJEEta bank (NEET 2013–2026). It ranks 29 of 30 Physics chapters.

Take the 25-question Thermodynamics test Formula sheet ↓All NEET Physics chapters

0.9/100 Weight in Physics

Exam importance (1/5, by past-paper share)

7 NEET PYQs

107 Practice questions

100 With measured difficulty

How the weight is computed: this chapter's tagged NEET PYQs divided by all tagged NEET Physics PYQs in the bank (7 of 758), times 100, with repeated questions counted once. The bank does not store the year of each NEET question, so none is printed. Computed from the bank at build time. Updated 19 Oct 2026.

Thermodynamics: formula sheet for NEET

14 formulas, each with when to use it. Copy it onto one page of your notebook: that page is this chapter's handbook.

First law
$Δ Q = Δ U + Δ W (Δ Q = heat given to gas, Δ W = work done BY gas)$
Use it when: any process asks for heat, work or internal energy and you know two of the three
Work by gas
$W = \int P d V = area under the P - V curve$
Use it when: a $P$ - $V$ graph is given; expansion gives $W > 0$ , compression gives $W < 0$
Change in internal energy (ideal gas)
$Δ U = n C_{V} Δ T$
Use it when: finding $Δ U$ in ANY process of an ideal gas, because $U$ depends only on $T$
Ideal gas equation
$P V = n R T$
Use it when: you need a missing $P$ , $V$ or $T$ at a state; apply it state by state ( $P Δ V = n R Δ T$ only if $P$ is constant)
Isothermal work
$W = n R T ln \frac{V _{2}}{V _{1}}$
$Δ U = 0$
$Q = W$
Use it when: $T$ is constant ( $P V = constant$ ); $ln$ is natural log
Adiabatic relations
$P V^{γ} = constant$
$T V^{γ - 1} = constant$
Use it when: an ideal gas goes through a quasi-static (reversible) adiabatic process and you need the final $P$ , $V$ or $T$ ; not for free expansion, which has $Q = 0$ but obeys neither
Adiabatic work
$W = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} = \frac{n R ( T _{1} - T _{2} )}{γ - 1}$
Use it when: work in a $Q = 0$ process; the gas cools when $W > 0$ and heats up when $W < 0$
Isobaric work
$W = P (V_{2} - V_{1}) = n R (T_{2} - T_{1})$
Use it when: $P$ is constant; here $Q = Δ U + W = n C_{P} Δ T$
Isochoric
$W = 0$
$Q = Δ U = n C_{V} Δ T$
Use it when: $V$ is constant (rigid container), so all the heat goes into raising $U$
Mayer's relation
$C_{P} - C_{V} = R (molar heat capacities, ideal gas only)$
$γ = \frac{C _{P}}{C _{V}}$
Use it when: for an ideal gas, you know one molar heat capacity (or $γ$ ) and need the other
Heat capacities
$s = \frac{Δ Q}{m Δ T} (in J k g^{- 1} K^{- 1})$
$C = \frac{Δ Q}{n Δ T} (in J mo l^{- 1} K^{- 1})$
Use it when: heat to warm a mass or a number of moles; take $m$ in kg, and $1 cal = 4.186 J$
Cyclic process
$Δ U = 0$
$Q_{net} = W_{net} = area enclosed by the loop$
Use it when: the gas returns to its starting state; a clockwise loop on a $P$ - $V$ graph means $W > 0$
Engine efficiency
$η = \frac{W}{Q _{1}} = 1 - \frac{Q _{2}}{Q _{1}} (every engine)$
$η = 1 - \frac{T _{2}}{T _{1}} (Carnot only)$
Use it when: $W / Q_{1}$ or $1 - Q_{2} / Q_{1}$ for any engine; $1 - T_{2} / T_{1}$ only for a Carnot (reversible) engine between source $T_{1}$ and sink $T_{2}$ , and it is the maximum any engine there can reach; $T$ in kelvin, never $^{\circ} C$
Carnot heat ratio
$\frac{Q _{2}}{Q _{1}} = \frac{T _{2}}{T _{1}}$
Use it when: you know the heat absorbed from the source and need the heat rejected to the sink, or the reverse

Trap

Students mix up the sign of W. In ΔQ = ΔU + ΔW, W is work done BY the gas. In a compression, or whenever work is done ON the gas, W is negative, and heat given out by the gas is negative. Getting this sign wrong flips the whole answer.

How does NEET ask Thermodynamics?

5 real previous-year questions from this chapter, as printed in the paper, each with its worked solution. Answer the first one here with no sign-in.

Q1PhysicsNEET UG PYQThermodynamic Processes

Try right here · No sign-in

In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas (

γ = 5/3

) decreases from 60 K to 50 K . The work done by the gas in the process is: (Take the universal gas constant as

R = 8.3 J mol^{- 1} K^{- 1}

)

Answer key (B) & worked solution

Correct answer: Option B

Work done in adiabatic process

W = \frac{n R Δ T}{1 - γ}

Δ T = - 10 K

W = \frac{1 \times 8.3 \times ( - 10 )}{1 - \frac{5}{3}}

W = \frac{83 \times 3}{2}

W = 124.5 J

Q2PhysicsNEET UG PYQFirst Law of Thermodynamics

Two gases A and B are filed at the same pressure in separate cylinders with movable pistons of radius

r_{A}

and

r_{B},

respectively. On supplying an equal amount of heat to both the systems reversibly under constant pressure, the pistons of gas A and B are displaced by 16 cm and 9 cm, respectively. If the change in their internal energy is the same, then the ratio

\frac{r _{A}}{r _{B}}

is equal to

\frac{2}{3}

\frac{3}{2}

\frac{4}{3}

\frac{3}{4}

Answer key (D) & worked solution

Correct answer: Option D

Δ U

are same

Δ θ

are same.

∴ Δ ω

should be same

P π r_{A}^{2} L_{A} = P π r_{B}^{2} L_{B}

\frac{r _{A}}{r _{B}} = \frac{L _{B}}{L _{A}} = \frac{9}{16} = \frac{3}{4}

Q3PhysicsNEET UG PYQFirst Law of Thermodynamics

An electric heater supplies heat to a system at a rate of 100 W . If the system performs work at a rate of

75 J / s

, then the rate at which internal energy increases will be:

A125 W

B100 W

C25 W

D75 W

Answer key (C) & worked solution

Correct answer: Option C

Using

1^{st}

law of thermodynamics for electric heater,

Q = Δ U + W \Rightarrow \frac{d Q}{d t} = \frac{d ( Δ U )}{d t} + \frac{d W}{d t} \Rightarrow 100 W = \frac{d ( Δ U )}{d t} + 75 W ∴ \frac{d ( Δ U )}{d t} = 25 W

Q4PhysicsNEET UG PYQSpecific Heat Capacity

The molar specific heats of an ideal gas at constant pressure and volume are denoted by C

_{P}

and C

_{V}

respectively. If

γ = \frac{C _{P}}{C _{V}}

and R is the universal gas constant, then

C_{V}

is equal to

\frac{1 + γ}{1 - γ}

\frac{R}{( γ - 1 )}

\frac{( 1 - γ )}{R}

γ R

Answer key (B) & worked solution

Correct answer: Option B

Difference between two molar specific heats is C

_{P}

- C

_{V}

= R Dividing both the sides by C

_{V}

, we get

\frac{C _{P}}{C _{V}} - \frac{C _{V}}{C _{V}} = \frac{R}{C _{V}}

\frac{C _{P}}{C _{V}} = γ

∴ γ - 1 = \frac{R}{C _{V}}

∴ C_{V} = \frac{R}{γ - 1}

Q5PhysicsNEET UG PYQSecond Law of Thermodynamics and Reversible and Irreversible Processes

The temperature inside a refrigerator is

t_{2}^{o} C

and the room temperature is

t_{1}^{o} C

. The amount of heat delivered to the room from each joule of electrical energy consumed ideally will be

\frac{t _{1} + 273}{t _{1} - t _{2}}

\frac{t _{2} + 273}{t _{1} - t _{2}}

\frac{t _{1} + t _{2}}{t _{1} + 273}

\frac{t _{1}}{t _{1} - t _{2}}

Answer key (A) & worked solution

Correct answer: Option A

Let Heat delivered = Q

_{1}

The coefficient of performance (

β

) is defined as the ratio of the heat removed per cycle from the content of the refrigerator to the energy per cycle to remove this heat

C O P (β) = \frac{t _{2} + 273}{t _{1} - t _{2}} = \frac{Q _{2}}{W} = \frac{Q _{1} - W}{W} = \frac{Q _{1}}{W} - 1

\Rightarrow \frac{Q _{1}}{W} = 1 + \frac{t _{2} + 273}{t _{1} - t _{2}} = \frac{t _{1} + 273}{t _{1} - t _{2}}

Which topics in Thermodynamics carry the most questions?

All 8 topics, by tagged PYQs, then practice questions. Counted from the bank. Updated 19 Oct 2026.

Topic	PYQs	Practice questions
First Law of Thermodynamics	3	19
Thermodynamic Processes	1	53
Thermal Equilibrium	1	9
Second Law of Thermodynamics and Reversible and Irreversible Processes	1	4
Specific Heat Capacity	1	2
Carnot Engine*	0	10
Heat, Internal Energy, and Work	0	9
Thermodynamic State Variables and Equation of State	0	1

Why does Thermodynamics show "–" until you answer 3 questions?

Unknown is not zero. Until you answer at least 3 graded questions in Thermodynamics, ViJEEta shows – instead of a zero or a 50% default. Book exercises can raise chapter mastery up to 0.60, adaptive practice up to 0.75, and only a timed chapter test or mock can certify 0.80 and above.

How ViJEEta scores you

What happens when you test Thermodynamics?

Every chapter test is 25 questions. It starts at moderate difficulty and adapts after each saved answer. If you open a solution before answering, that attempt earns no credit, so your mastery stays honest.

Start the Thermodynamics test

Chapters either side of Thermodynamics by weight

Magnetism and Matter Weight 1.7/100 Mechanical Properties of Solids Weight 1.6/100 Practical Physics Weight 0.8/100

Thermodynamics: NEET PYQs and weightage

Thermodynamics: formula sheet for NEET

First law

Work by gas

Change in internal energy (ideal gas)

Ideal gas equation

Isothermal work

Adiabatic relations

Adiabatic work

Isobaric work

Isochoric

Mayer's relation

Heat capacities

Cyclic process

Engine efficiency

Carnot heat ratio

How does NEET ask Thermodynamics?

Which topics in Thermodynamics carry the most questions?

Why does Thermodynamics show "–" until you answer 3 questions?

What happens when you test Thermodynamics?

Chapters either side of Thermodynamics by weight