Thermodynamics — JEE Mains Physics Formulas and Traps

First Law and Energy Bookkeeping

The first law and its sign convention

First law (work done by the gas positive)

\Delta Q = \Delta U + W \qquad W = \int_{V_1}^{V_2} P\,dV

Energy balances: phase changes, rates and pistons

Energy balance at a steady pressure

\Delta U = Q - P\Delta V \qquad \frac{dU}{dt} = \frac{dQ}{dt} - \frac{dW}{dt}

Common traps

Two sign conventions for work

Physics counts the work done BY the gas as positive and writes ΔQ = ΔU + W. Chemistry counts the work done ON the gas and writes ΔQ = ΔU − W. Both describe the same energy; mixing them flips the sign of W.

Heat added does not mean hotter

If a gas does more work than the heat it receives, its internal energy falls and so does its temperature. Heat in only fixes ΔU = Q − W, not the sign of ΔT.

Subtract the work against the atmosphere

When a liquid boils, ΔU is the latent heat minus PΔV, not the latent heat itself. The PΔV term is often what separates two of the options.

Convert volumes before multiplying

cm³ to m³ is a factor 10⁻⁶ and litres to m³ is 10⁻³. Pressure in Pa times volume in m³ gives joules; so does kPa times litres. Any other pair needs converting first.

Internal Energy and Heat Capacities

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Internal energy from Cv and the degrees of freedom

Internal energy and heat capacities

\Delta U = nC_V\Delta T \qquad C_V = \frac{f}{2}R \qquad C_P - C_V = R \qquad \gamma = 1 + \frac{2}{f}

How heat splits at constant pressure

Isobaric heat split

Q : \Delta U : W = C_P : C_V : R \qquad Q = \frac{\gamma}{\gamma - 1}W \qquad \Delta U = \frac{Q}{\gamma}

Molar heat capacity of any process

Molar heat capacity of a process

C = \frac{Q}{n\Delta T} = C_V + \frac{W}{n\Delta T} \qquad C_{\text{poly}} = C_V + \frac{R}{1 - x}

Common traps

ΔU uses Cv even at constant pressure

The heat at constant pressure is nCpΔT, but the rise in internal energy is still nCvΔT. Using Cp for ΔU is the most common slip on this page.

Each vibrational mode counts twice

A vibration stores both kinetic and potential energy, so each vibrational mode adds 2 to f, not 1.

Calories and joules do not mix

R is about 2 in cal mol⁻¹ K⁻¹ and about 8.3 in J mol⁻¹ K⁻¹. Subtracting R in joules from Cp in calories gives a meaningless Cv.

Only at constant volume does all the heat stay inside

At constant pressure part of the heat, R/Cp of it, leaves as work. ΔU equals Q only when the volume is fixed.

"Rotates but does not oscillate" means f = 5

That phrase describes a rigid diatomic molecule: f = 5, Cv = 5R/2, γ = 7/5. Vibration would add 2 more degrees of freedom.

A heat capacity can be negative

For PV^x = constant with x between 1 and γ, C = Cv + R/(1 − x) is negative: the gas takes in heat yet cools, because it does more work than the heat it gets.

A fraction of Q, not of ΔU

"The gas does work equal to Q/6" means a sixth of the HEAT leaves as work, so ΔU = 5Q/6. Taking the fraction of ΔU instead gives a different heat capacity.

Thermodynamic Processes and Work from P–V Graphs

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Work as the area under a P–V path

Work done by the gas

W = \int_{V_1}^{V_2} P\,dV \qquad W_{\text{iso}} = nRT\ln\frac{V_2}{V_1}

Processes given by an equation such as PV^x = constant

Polytropic process PV^x = constant

W = \frac{P_1V_1 - P_2V_2}{x - 1} = \frac{nR(T_1 - T_2)}{x - 1} \qquad T \propto V^{1 - x}

The standard processes and their W, Q and ΔU

Process	Held fixed	Work by the gas W	Heat into the gas Q	Change ΔU
Isothermal	Temperature	$nRT\ln\dfrac{V_2}{V_1}$	Q = W	0 Heat flows in, yet the temperature does not rise: all of it leaves as work.
Isochoric	Volume	0	$nC_V\Delta T$ , so Q = ΔU	$nC_V\Delta T$
Isobaric	Pressure	$P\Delta V = nR\Delta T$	$nC_P\Delta T$	$nC_V\Delta T$
Adiabatic	No heat exchanged	$\dfrac{nR(T_1 - T_2)}{\gamma - 1}$	0	−W
Cyclic	Start and end state the same	Area enclosed on the P–V diagram	Q = W	0
Free expansion into a vacuum	Insulated, no outside pressure	0	0	0, so an ideal gas keeps its temperature

Physics sign convention: W is work done BY the gas and Q is heat INTO the gas, so Q = ΔU + W in every row.

Common traps

Isothermal does not mean no heat

In an isothermal process the temperature is fixed, but heat still flows: Q = W. The process with no heat is the adiabatic one.

An isochoric line must pass through the origin

P ∝ T describes an isochoric process only when the P–T line passes through absolute zero. A straight line with an intercept, P = a + bT, is not isochoric.

A leg to the left is negative work

Any part of a path that runs towards smaller volume has negative work by the gas, however high the pressure. Add the areas with their signs.

Read each axis on its own scale

For an elliptical or circular arc on a P–V diagram, read the semi-axis along V in m³ and the semi-axis along P in Pa, each from its own axis. A circle on the page is usually an ellipse in physical units.

x = 1 needs the logarithm

The polytropic work formula divides by x − 1, so it fails for an isothermal process. There use W = nRT ln(V₂/V₁).

The index belongs to the process, γ to the gas

x in PV^x = constant is set by the process; γ = Cp/Cv is set by the gas. They are equal only in an adiabatic process, so do not call x by the name γ.

Adiabatic Processes

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The adiabatic relations PV^γ, TV^(γ−1) and P^(1−γ)T^γ

Adiabatic relations

PV^{\gamma} = \text{const} \qquad TV^{\gamma - 1} = \text{const} \qquad P^{1 - \gamma}T^{\gamma} = \text{const}

Work done in an adiabatic process

Adiabatic work (done by the gas)

W = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = -\Delta U

True and false statements about adiabatic processes

Claim about an adiabatic process	True or false	Why
No heat crosses the boundary	True	That is the definition: Q = 0
The internal energy stays constant	False	ΔU = −W, which is not zero unless no work is done Q = 0 does not stop U changing: the work comes out of U.
An adiabatic compression raises the temperature	True	The work done on the gas goes into U, and U rises with T
Its P–V curve is steeper than the isotherm through the same point	True	Its slope is γ times the isothermal slope
The molar heat capacity is zero	True	C = Q/(nΔT) with Q = 0 while ΔT is not zero
The product TV stays constant	False	It is TV^(γ−1) that stays constant
A free expansion into a vacuum follows PV^γ = constant	False	It is irreversible; an ideal gas keeps its temperature

Every row follows from Q = 0 and Q = ΔU + W.

Common traps

A sudden change is adiabatic

There is no time for heat to flow, so use PV^γ = constant. The isothermal answer from PV = constant is always among the options.

Kelvin in TV^(γ−1)

A temperature ratio needs absolute temperatures. A gas at 27 °C is at 300 K; doubling 27 °C is not doubling the temperature.

The sign in a compression

In an adiabatic compression the gas does negative work and warms up. "Work done on the gas" is the positive number; read which one the question asks for.

Divide by γ − 1, not by γ

W = nRΔT/(γ − 1). Dividing by γ gives a much smaller wrong value, and it is often one of the options.

Adiabatic is not isothermal

Q = 0 does not make ΔT = 0. With no heat to make up for the work, the temperature must change in an adiabatic process.

The steeper curve is the adiabat

Through any point the adiabat is steeper than the isotherm by the factor γ. On a P–V diagram showing both, the steeper one is the adiabat.

Cyclic Processes

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Net work of a cycle as the enclosed area

Net work of a cycle

W_{\text{net}} = Q_{\text{net}} = \oint P\,dV \qquad W_{\text{ellipse}} = \pi ab

A cycle worked leg by leg

Net work leg by leg

W_{\text{net}} = \sum_{\text{legs}} W_i = \sum_{\text{legs}} Q_i

Common traps

Half the width, not the width

πab uses semi-axes. Using the full width on one axis doubles the area, and on both axes makes it four times too big.

The direction sets the sign

The area gives only the size of the work. Clockwise on a P–V diagram is positive work by the gas; anticlockwise is negative.

Find the corner pressure before using PΔV

After an isothermal expansion to k times the volume, the pressure is P/k. An isobaric leg that follows runs at that new pressure, not at the starting one.

The volume ratio in the right order

Isothermal work by the gas is nRT ln(V_final/V_initial), which is negative for a compression. Writing the ratio upside down flips the sign.

Heat Engines, Carnot Cycle and Refrigerators

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Carnot efficiency η = 1 − T₂/T₁

Carnot efficiency

\eta = \frac{W}{Q_1} = 1 - \frac{T_2}{T_1}

Heat, work and temperature in engines and refrigerators

Engine and refrigerator

Q_1 = W + Q_2 \qquad \frac{Q_2}{Q_1} = \frac{T_2}{T_1} \qquad \text{COP} = \frac{Q_2}{W} = \frac{T_2}{T_1 - T_2}

Engines in series and entropy

\eta = \eta_1 + \eta_2 - \eta_1\eta_2 = 1 - \frac{T_3}{T_1} \qquad \Delta S = ms\ln\frac{T_2}{T_1}

Common traps

Kelvin, not Celsius

For an engine between 327 °C and 27 °C, η = 1 − 300/600, not 1 − 27/327. Only temperature differences may stay in °C.

Per cent or percentage points?

"Efficiency increases by 30%" from 50% could mean 65% or 80%. Try the relative reading, 1.3 times, first, and check it against the options.

Efficiency divides by the heat taken in

η = W/Q₁, with Q₁ the heat from the hot reservoir. Dividing the work by the rejected heat Q₂ gives a number that is too large.

COP is not an efficiency

A refrigerator's COP = Q₂/W = T₂/(T₁ − T₂) is usually larger than 1. The engine formula (T₁ − T₂)/T₁ is the wrong one for a refrigerator.

Efficiencies in series do not add

Two engines in series have η = η₁ + η₂ − η₁η₂, less than η₁ + η₂. The pair does no better than one Carnot engine across the whole range.

Mass units in ΔS

With s in J kg⁻¹ K⁻¹, the mass must be in kg. A mass given in grams brings a factor of 10⁻³.

Thermodynamics formulas

First Law and Energy Bookkeeping

The first law and its sign convention

Energy balances: phase changes, rates and pistons

Common traps

Internal Energy and Heat Capacities

Internal energy from Cv and the degrees of freedom

How heat splits at constant pressure

Molar heat capacity of any process

Common traps

Thermodynamic Processes and Work from P–V Graphs

Work as the area under a P–V path

Processes given by an equation such as PV^x = constant

The standard processes and their W, Q and ΔU

Common traps

Adiabatic Processes

The adiabatic relations PV^γ, TV^(γ−1) and P^(1−γ)T^γ

Work done in an adiabatic process

True and false statements about adiabatic processes

Common traps

Cyclic Processes

Net work of a cycle as the enclosed area

A cycle worked leg by leg

Common traps

Heat Engines, Carnot Cycle and Refrigerators

Carnot efficiency η = 1 − T₂/T₁

Heat, work and temperature in engines and refrigerators

Common traps

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