JEE Mains Physics · Thermodynamics

Internal Energy and Heat Capacities

The internal energy of an ideal gas changes by nCvΔT in every process; the heat needed depends on the process, and the heat per mole per kelvin is that process's molar heat capacity.

All Thermodynamics notes JEE Mains Physics strategy

Why this matters

Twenty PYQs, nine of them asking for a number, and five from 2026. Six find a change in internal energy from Cv, which is often first worked out from Cp or from the degrees of freedom. Ten split the heat at constant pressure into internal energy and work. Four find a molar heat capacity from the first law: for a process that turns a set fraction of its heat into work, for a polytropic process, or for a gas that is not ideal.

Concept 1 of 3: Internal energy from Cv and the degrees of freedom

The internal energy of an ideal gas depends only on its temperature. So ΔU = nCvΔT holds in every process, not only at constant volume: the heat needed changes from path to path, but ΔU does not. Cv itself comes from the degrees of freedom, each of which holds ½RT of energy per mole.

Definition

$U = \dfrac{f}{2}nRT$ , so $\Delta U = nC_V\Delta T$ with $C_V = \dfrac{f}{2}R$ , in any process.
f = 3 for a monoatomic gas; f = 5 for a diatomic gas that rotates but does not vibrate ("rigid"); f = 6 for a rigid non-linear molecule. Each vibrational mode adds 2.
$C_P - C_V = R$ (Mayer's relation) and $\gamma = \dfrac{C_P}{C_V} = 1 + \dfrac{2}{f}$ .
$R \approx 8.3\ \text{J mol}^{-1}\text{K}^{-1} \approx 2\ \text{cal mol}^{-1}\text{K}^{-1}$ . Keep Cp, Cv and R in the same unit.
A temperature change has the same size in °C and in K.
With a specific heat per kilogram at constant volume, $\Delta U = mc_V\Delta T$ .

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}

Worked example

3 mol of a gas is heated at constant pressure from 20 °C to 50 °C. Its

C_P = 5\ \text{cal mol}^{-1}\,^{\circ}\text{C}^{-1}

, and take

R = 2\ \text{cal mol}^{-1}\,^{\circ}\text{C}^{-1}

. Find ΔU.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q24Moderate

Example 1 · Thermodynamics · Internal Energy and Heat Capacities

10 mole of oxygen is heated at constant volume from

30^{\circ}C

40^{\circ}C

. The change in the internal energy of the gas is

\_\_\_\_

cal. (The molecular specific heat of oxygen at constant pressure,

C_{p} = 7cal./{mol}^{\circ}C

and

R = 2cal./{mol}^{\circ}C

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.

Drill 5 more on internal energy from cv and the degrees of freedom

Concept 2 of 3: How heat splits at constant pressure

At constant pressure the heat is Q = nCpΔT, the internal energy rises by nCvΔT and the work is PΔV = nRΔT. All three share the factor nΔT, so they always stand in the ratio Cp : Cv : R. Knowing any one of them fixes the other two.

Definition

$Q : \Delta U : W = C_P : C_V : R = \left(\tfrac{f}{2} + 1\right) : \tfrac{f}{2} : 1$ .
Monoatomic gas (f = 3): 5 : 3 : 2.
From the work: $Q = \dfrac{\gamma}{\gamma - 1}W$ and $\Delta U = \dfrac{W}{\gamma - 1}$ .
From the heat: $\Delta U = \dfrac{Q}{\gamma}$ and $W = Q\left(1 - \dfrac{1}{\gamma}\right)$ .
$W = P\Delta V = nR\Delta T$ 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}

Worked example

A monoatomic ideal gas expands at constant pressure and does

60\ \text{J}

of work. Find the heat supplied and the rise in internal energy.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q93Moderate

Example 2 · Thermodynamics · Internal Energy and Heat Capacities

Heat energy of

735\text{ }J

is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but do not oscillate. The increase in the internal energy of the gas will be:

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.

Drill 9 more on how heat splits at constant pressure

Concept 3 of 3: Molar heat capacity of any process

A molar heat capacity is heat per mole per kelvin, and its value depends on the path. At constant volume it is Cv and at constant pressure it is Cp. For any other process, use the first law: find what share of the heat stays as internal energy, then divide the heat by nΔT.

Definition

$C = \dfrac{Q}{n\Delta T}$ . Since $\Delta U = nC_V\Delta T$ always, $C = C_V + \dfrac{W}{n\Delta T}$ .
If the gas turns a fraction 1/k of the heat into work, $\Delta U = \left(1 - \tfrac{1}{k}\right)Q$ , so $C = \dfrac{C_V}{1 - 1/k}$ .
Polytropic process $PV^{x} = \text{const}$ : $C = C_V + \dfrac{R}{1 - x}$ . x = 0 gives Cp; x = γ (adiabatic) gives 0; x = 1 (isothermal) gives an infinite C.
Two processes between the same two temperatures: the one doing more work needs more heat, so it has the larger C.
For a gas that is not ideal (U still depending on T alone), $C_P - C_V = P\left(\dfrac{\partial V}{\partial T}\right)_{P}$ . Find $(\partial V/\partial T)_P$ from its equation of state; for PV = RT it gives R.

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}

Worked example

A monoatomic ideal gas does work equal to one-third of the heat it receives. Find its molar heat capacity in this process.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 26 · Q22Moderate

Example 3 · Thermodynamics · Internal Energy and Heat Capacities

1 mole of rigid diatomic gas performs a work of

\frac{Q}{5}

when heat

Q

is supplied to it. The molar heat capacity of the gas during this transformation is

\frac{xR}{8}

. The value of

x

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.

Drill 3 more on molar heat capacity of any process

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

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

Watch out for (7)

ΔU uses Cv even at constant pressure
→ Internal energy from Cv and the degrees of freedom
Each vibrational mode counts twice
→ Internal energy from Cv and the degrees of freedom
Calories and joules do not mix
→ Internal energy from Cv and the degrees of freedom
Only at constant volume does all the heat stay inside
→ How heat splits at constant pressure
"Rotates but does not oscillate" means f = 5
→ How heat splits at constant pressure
A heat capacity can be negative
→ Molar heat capacity of any process
A fraction of Q, not of ΔU
→ Molar heat capacity of any process

Test yourself on Thermodynamics

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