JEE Mains Chemistry · Chemical Thermodynamics

Heat Capacity, Calorimetry and Enthalpy vs Internal Energy

How much heat a temperature rise takes (q = nCΔT), why a bomb calorimeter measures ΔU, and how ΔH follows from ΔU through the change in moles of gas.

All Chemical Thermodynamics notes JEE Mains Chemistry strategy

Why this matters

Nineteen PYQs, sixteen of them numerical — the most calculation-heavy page in the chapter — and three from 2026. Five use heat capacities, seven read a bomb calorimeter, and seven convert between ΔH and ΔU for a reaction or a vaporisation.

Concept 1 of 3: Heat capacity: Cp, Cv and q = nCΔT

Heat capacity says how much heat raises the temperature by one kelvin. At constant volume all the heat goes into internal energy. At constant pressure some of it pushes the surroundings back, so more heat is needed for the same rise: Cp is larger than Cv by R per mole.

Definition

$q = C\Delta T = nC_m\Delta T = mc\Delta T$ .
Constant volume: $q_V = nC_v\Delta T = \Delta U$ . Constant pressure: $q_p = nC_p\Delta T = \Delta H$ .
Ideal gas: $C_p - C_v = R$ per mole, so $C_p > C_v$ always.
Monatomic ideal gas: $C_v = \tfrac{3}{2}R$ , $C_p = \tfrac{5}{2}R$ . For an ideal gas $\Delta U = nC_v\Delta T$ in ANY process.
Molar heat capacities near 298 K (J K⁻¹ mol⁻¹): He(g) about 21 ( $C_p$ ), Cu(s) about 25 (roughly $3R$ ), $\mathrm{Br_2}(l)$ about 76. A liquid of molecules has the most ways to store energy.
Electrical heating: $q = Pt$ (watts × seconds).

Heat and heat capacity

q = nC_m\Delta T,\qquad C_p - C_v = R

Worked example

2 mol of helium is heated at constant pressure from 320 K to 370 K. Find

q

\Delta U

and

w

. (

R = 8.314

J K⁻¹ mol⁻¹)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q38Moderate

Example 1 · Chemical Thermodynamics · Heat Capacity, Calorimetry and Enthalpy vs Internal Energy

500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm . The final temperature and the change in internal energy respectively are : Given :

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

Cv for heat added at constant pressure

Heat supplied at constant pressure is

nC_p\Delta T

. Dividing it by

nC_v

overstates the temperature rise. And

\Delta U

is less than the heat supplied, because part of the heat did expansion work.

Drill 4 more on heat capacity: cp, cv and q = ncδt

Concept 2 of 3: Bomb calorimeter: heat at constant volume

A bomb calorimeter is a sealed steel vessel, so the volume cannot change and no expansion work is done. The heat it soaks up is the ΔU of the reaction with its sign reversed. An open vessel works at constant pressure and measures ΔH instead.

Definition

Heat taken up by the calorimeter: $q = C_{\mathrm{cal}}\Delta T$ .
Per mole of fuel: $\Delta_c U = -\frac{C_{\mathrm{cal}}\Delta T}{n}$ , with $n = \frac{m}{M}$ .
Then $\Delta_c H = \Delta_c U + \Delta n_g RT$ .
An open vessel (constant pressure) measures $\Delta H$ directly.
Heat released $= n \times |\Delta_c H|$ , so a known heat gives the mass burnt.

Bomb calorimeter

\Delta_c U = -\frac{C_{\mathrm{cal}}\,\Delta T}{n}

Worked example

1.28 g of naphthalene,

\mathrm{C_{10}H_8}

(M = 128), burns in a bomb calorimeter of heat capacity 10.0 kJ K⁻¹, and the temperature rises by 5.15 K. Find

\Delta_c U

and

\Delta_c H

at 298 K. (

R = 8.314

J K⁻¹ mol⁻¹)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 June 2022 · Q51Moderate

Example 2 · Chemical Thermodynamics · Heat Capacity, Calorimetry and Enthalpy vs Internal Energy

For complete combustion of methanol

CH_{3}OH(l) +\frac{3}{2}O_{2}(\text{ }g) \rightarrow CO_{2}(\text{ }g) + 2H_{2}O(l)

the amount of heat produced as measured by bomb calorimeter is

726\text{ }kJ{\text{ }mol}^{- 1}

27^{\circ}C

. The enthalpy of combustion for the reaction is

- xkJmol^{- 1}

, where

x

\_\_\_\_

. (Nearest integer) (Given:

R = 8.3JK^{- 1}{\text{ }mol}^{- 1}

)

Bomb heat is ΔU, not ΔH

When a question quotes heat measured in a bomb calorimeter and asks for an enthalpy, it needs the

\Delta n_g RT

step. Skipping it is a planted option, a few kJ away from the answer.

Positive heat, negative ΔU

The calorimeter warms because the reaction gives out heat.

C\Delta T

is positive, but

\Delta_c U

of the reaction is negative.

Drill 6 more on bomb calorimeter: heat at constant volume

Concept 3 of 3: ΔH and ΔU through the change in gas moles

ΔH and ΔU differ by the work of making room for gas. One mole of gas at constant pressure needs RT of work to push the surroundings back, so each extra mole of gas made adds RT. Solids and liquids take up almost no volume, so they do not count.

Definition

$\Delta H = \Delta U + \Delta n_g RT$ , with $\Delta n_g$ = moles of gaseous products − moles of gaseous reactants.
Liquid water among the products does not count; water vapour does.
$\Delta n_g = 0 \Rightarrow \Delta H = \Delta U$ , as for $\mathrm{C(s) + O_2(g) \to CO_2(g)}$ .
With $\Delta H$ in kJ, use $R = 8.314\times10^{-3}$ kJ K⁻¹ mol⁻¹.
Vaporising 1 mol of a liquid: $\Delta n_g = +1$ , so $\Delta U = \Delta_{\mathrm{vap}}H - RT$ .
For a gas heated or cooled at constant pressure: $\Delta U = \Delta H - p\Delta V$ .

Enthalpy and internal energy

\Delta H = \Delta U + \Delta n_g RT

Worked example

For

\mathrm{N_2(g) + 3H_2(g) \to 2NH_3(g)}

\Delta H = -92.4

kJ at 298 K. Find

\Delta U

. (

R = 8.314

J K⁻¹ mol⁻¹)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 8 April 2024 · Q141Moderate

Example 3 · Chemical Thermodynamics · Heat Capacity, Calorimetry and Enthalpy vs Internal Energy

\Delta_{\text{vap~}}H^{\ominus}

for water is

+ 40.49\text{ }kJ{\text{ }mol}^{- 1}

at 1 bar and

100^{\circ}C

. Change in internal energy for this vapourisation under same condition is _______

J

mol^{- 1}

. (Integer answer) (Given

R = 8.3JK^{- 1}{\text{ }mol}^{- 1}

)

Counting liquid water as a gas

In a combustion that makes

\mathrm{H_2O(l)}

, water is not a gas. For

\mathrm{C_2H_6(g) + \tfrac{7}{2}O_2(g) \to 2CO_2(g) + 3H_2O(l)}

\Delta n_g = 2 - 4.5 = -2.5

, not

+0.5

R in joules beside ΔH in kilojoules

8.314 \times 300 = 2494

J must be written as 2.494 kJ before it is added to a

\Delta H

in kJ. Otherwise the correction is a thousand times too big.

Drill 6 more on δh and δu through the change in gas moles

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)

Heat capacity: Cp, Cv and q = nCΔT
Heat and heat capacity
$q = nC_m\Delta T,\qquad C_p - C_v = R$
Bomb calorimeter: heat at constant volume
Bomb calorimeter
$\Delta_c U = -\frac{C_{\mathrm{cal}}\,\Delta T}{n}$
ΔH and ΔU through the change in gas moles
Enthalpy and internal energy
$\Delta H = \Delta U + \Delta n_g RT$

Watch out for (5)

Cv for heat added at constant pressure
→ Heat capacity: Cp, Cv and q = nCΔT
Bomb heat is ΔU, not ΔH
→ Bomb calorimeter: heat at constant volume
Positive heat, negative ΔU
→ Bomb calorimeter: heat at constant volume
Counting liquid water as a gas
→ ΔH and ΔU through the change in gas moles
R in joules beside ΔH in kilojoules
→ ΔH and ΔU through the change in gas moles

Test yourself on Chemical Thermodynamics

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Related notes

The first law — why q at constant volume is ΔU →