Is Thermodynamics part of the EAPCET Chemistry syllabus?

Yes. Thermodynamics is taught in Intermediate first year Chemistry, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.

How many Chemistry questions are there in EAPCET?

40 of the 160 questions in the Engineering paper are Chemistry, spread across 24 chapters.

What mistakes do students make in Thermodynamics?

Including liquids and solids in Δn(g) — only gaseous moles count when converting between ΔH and ΔU. Mixing joules and kilojoules — R = 8.314 J K⁻¹ mol⁻¹ must be divided by 1000 when ΔH is in kJ. Assuming every exothermic reaction is spontaneous — a reaction with ΔH < 0 and ΔS < 0 is spontaneous only below T = ΔH/ΔS. Taking ΔH°f of O₃, diamond or Br₂(g) as zero — only the most stable form of an element in its standard state (O₂, graphite, Br₂ liquid) has zero enthalpy of formation.

EAPCET Chemistry · Intermediate first year

Thermodynamics

Q: How should I prepare Thermodynamics for EAPCET?

Learn the four sign combinations of ΔH and ΔS and what each means for spontaneity at high and low temperature.

Thermodynamics in Chemistry covers internal energy and enthalpy, Hess’s law, entropy, Gibbs energy and the criteria for spontaneity.

Subject: Chemistry
Taught in: Intermediate first year
Chemistry in the paper: 40 of 160 questions

Key formulas and results

$\Delta H = \Delta U + \Delta n_{(g)}RT$ , where $\Delta n_{(g)}$ counts gaseous moles only
Hess’s law: enthalpy change is independent of the path
$\Delta G = \Delta H - T\Delta S$ ; a process is spontaneous when $\Delta G < 0$
$\Delta G^\circ = -RT \ln K$
Standard enthalpy of formation of an element in its standard state is zero
First law: $\Delta U = q + w$ , with work done on the system taken as positive
Irreversible work against a constant external pressure: $w = -P_{\text{ext}}\,\Delta V$
Reversible isothermal work: $w = -2.303\,nRT \log\frac{V_2}{V_1} = -2.303\,nRT \log\frac{P_1}{P_2}$
Reversible adiabatic process: $PV^{\gamma} = \text{constant}$ , $TV^{\gamma - 1} = \text{constant}$ , and $w = \frac{P_2V_2 - P_1V_1}{\gamma - 1} = \frac{nR\Delta T}{\gamma - 1}$
Molar heat capacities: $C_p - C_v = R$ and $\gamma = \frac{C_p}{C_v}$ ; $\Delta U = nC_v\Delta T$ and $\Delta H = nC_p\Delta T$
Entropy change of an ideal gas: $\Delta S = nC_v \ln\frac{T_2}{T_1} + nR \ln\frac{V_2}{V_1}$
Entropy of a phase change: $\Delta S = \frac{\Delta H_{\text{transition}}}{T}$
Free energy away from equilibrium: $\Delta G = \Delta G^\circ + RT \ln Q$

What this chapter covers

System, surroundings and state functions
First law, internal energy and enthalpy
Hess’s law
Enthalpies of reaction
Entropy
Gibbs energy and spontaneity

How it is usually asked

ΔH from Hess’s law
Converting between ΔH and ΔU
Conditions for spontaneity
Relating ΔG° and K

Understanding Thermodynamics

Thermodynamics tracks energy as a reaction happens. The first law says energy is conserved: ΔU = q + w, where heat absorbed by the system and work done on it are positive. Internal energy U, enthalpy H, entropy S and Gibbs energy G are state functions — their changes depend only on the start and end states — whereas heat and work depend on the path. At constant pressure, the heat exchanged equals ΔH.

ΔH and ΔU differ only because of gas-phase expansion work: ΔH = ΔU + Δn(g)RT. Hess’s law lets you add or subtract equations, with their ΔH values, to reach a target reaction. Enthalpies of formation of elements in their standard states (O₂ gas, graphite, and so on) are zero, so ΔH°(reaction) = ΣΔH°f(products) − ΣΔH°f(reactants). Bond enthalpies give an estimate the other way round: bonds broken minus bonds formed.

Entropy measures the spread of energy and matter; it rises from solid to liquid to gas and when gas moles increase. Spontaneity at constant T and P is decided by ΔG = ΔH − TΔS: negative ΔG means spontaneous. If ΔH < 0 and ΔS > 0, the reaction is always spontaneous; if both are positive, it becomes spontaneous above T = ΔH/ΔS. ΔG° = −RT ln K connects thermodynamics to equilibrium — ΔG° < 0 means K > 1.

Worked example

EAPCET-style question

For $\ce{N2(g) + 3H2(g) -> 2NH3(g)}$ , $\Delta H = -92.4\ \text{kJ}$ at 298 K. What is $\Delta U$ for the reaction? ( $R = 8.314\ \text{J K}^{-1}\,\text{mol}^{-1}$ )

(1)−87.4 kJ
(2)−97.4 kJ
(3)−92.4 kJ
(4)−89.9 kJ

Solution

$\Delta n_{(g)}$ = moles of gaseous products − moles of gaseous reactants $= 2 - 4 = -2$ .
$\Delta U = \Delta H - \Delta n_{(g)}RT$ .
$\Delta n_{(g)}RT = (-2)(8.314 \times 10^{-3}\ \text{kJ K}^{-1}\,\text{mol}^{-1})(298\ \text{K}) = -4.96\ \text{kJ}$ .
$\Delta U = -92.4 - (-4.96) = -87.4\ \text{kJ}$ .
−97.4 kJ comes from a sign error, and −89.9 kJ from taking $\Delta n = -1$ .

Answer: −87.4 kJ

Shortcuts and results to remember

$C_v$ , $C_p$ and $\gamma$ : monoatomic $\frac{3R}{2}$ , $\frac{5R}{2}$ , $\frac{5}{3}$ ; diatomic (and linear triatomic, ignoring vibration) $\frac{5R}{2}$ , $\frac{7R}{2}$ , $\frac{7}{5}$ ; non-linear triatomic $3R$ , $4R$ , $\frac{4}{3}$ .
For the same expansion, the magnitude of work done is isobaric > isothermal > adiabatic, and reversible work is always greater than irreversible work.
Isothermal process of an ideal gas: $\Delta U = 0$ so $q = -w$ ; adiabatic process: $q = 0$ so $\Delta U = w$ .
Free expansion of an ideal gas into vacuum: $w = 0$ , $q = 0$ and $\Delta U = 0$ .
In a cyclic process $\Delta U = 0$ so $q = -w$ , and the area enclosed on the P–V diagram equals the net work.

Common mistakes to avoid

Including liquids and solids in Δn(g) — only gaseous moles count when converting between ΔH and ΔU.
Mixing joules and kilojoules — R = 8.314 J K⁻¹ mol⁻¹ must be divided by 1000 when ΔH is in kJ.
Assuming every exothermic reaction is spontaneous — a reaction with ΔH < 0 and ΔS < 0 is spontaneous only below T = ΔH/ΔS.
Taking ΔH°f of O₃, diamond or Br₂(g) as zero — only the most stable form of an element in its standard state (O₂, graphite, Br₂ liquid) has zero enthalpy of formation.

How to prepare it

Learn the four sign combinations of ΔH and ΔS and what each means for spontaneity at high and low temperature.

Once the formulas feel familiar, test them the way the exam does: practise Thermodynamics questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Chemistry chapters

See every chapter in the EAPCET Chemistry syllabus.

Practise Thermodynamics chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Thermodynamics part of the EAPCET Chemistry syllabus?: Yes. Thermodynamics is taught in Intermediate first year Chemistry, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Thermodynamics for EAPCET?: Learn the four sign combinations of ΔH and ΔS and what each means for spontaneity at high and low temperature.
How many Chemistry questions are there in EAPCET?: 40 of the 160 questions in the Engineering paper are Chemistry, spread across 24 chapters.
What mistakes do students make in Thermodynamics?: Including liquids and solids in Δn(g) — only gaseous moles count when converting between ΔH and ΔU. Mixing joules and kilojoules — R = 8.314 J K⁻¹ mol⁻¹ must be divided by 1000 when ΔH is in kJ. Assuming every exothermic reaction is spontaneous — a reaction with ΔH < 0 and ΔS < 0 is spontaneous only below T = ΔH/ΔS. Taking ΔH°f of O₃, diamond or Br₂(g) as zero — only the most stable form of an element in its standard state (O₂, graphite, Br₂ liquid) has zero enthalpy of formation.