JEE Main & Advanced

Thermodynamics

Thermodynamics for JEE Main & Advanced

1
Module 1

Laws of Thermodynamics

Zeroth, First Law and Thermodynamic ProcessesTopic 1

Zeroth Law: If two systems are each in thermal equilibrium with a third, they are in thermal equilibrium with each other (defines temperature).

First Law of Thermodynamics: Energy conservation including heat. $$\Delta Q = \Delta U + \Delta W$$ where $\Delta Q$ = heat added to system, $\Delta U$ = change in internal energy, $\Delta W$ = work done by the system. Sign convention: $Q > 0$ if heat added; $W > 0$ if system does work on surroundings.

For an ideal gas: $\Delta U = nC_V\Delta T$ (depends only on temperature).

Work done in expansion: $W = \int P\,dV$ (area under P-V curve).

Thermodynamic Processes (Quasi-static):

ProcessCondition$W$$\Delta U$$Q$
Isothermal$T$ constant$nRT \ln(V_2/V_1)$$0$$W$
Isobaric$P$ constant$P\Delta V = nR\Delta T$$nC_V\Delta T$$nC_P\Delta T$
Isochoric$V$ constant$0$$nC_V\Delta T$$\Delta U$
Adiabatic$Q = 0$$-\Delta U = \dfrac{P_1V_1 - P_2V_2}{\gamma-1}$$nC_V\Delta T$$0$
CyclicReturns to initial state$\oint P\,dV$ = area$0$$W$

Adiabatic relation: $PV^\gamma = \text{const}$; also $TV^{\gamma-1} = \text{const}$ and $TP^{(1-\gamma)/\gamma} = \text{const}$, where $\gamma = C_P/C_V$.

Polytropic process: $PV^n = \text{const}$, generalizing all of the above.

Worked Examples
1

A gas absorbs $200$ J of heat and does $80$ J of work. Find the change in internal energy.

Show solution

$$\Delta U = Q - W = 200 - 80 = 120 \text{ J}$$

Final Answer: $\Delta U = 120$ J.

2

$1$ mole of an ideal monatomic gas expands isothermally at $T = 300$ K from volume $V$ to $2V$. Find work done. ($R = 8.314$ J/mol·K)

Show solution

$$W = nRT \ln(V_2/V_1) = 1 \times 8.314 \times 300 \times \ln 2$$ $$= 2494.2 \times 0.693 \approx 1729 \text{ J}$$

Final Answer: $W \approx 1729$ J.

✎ Self-Check — 5 questions0 / 5
Q1.

The first law of thermodynamics is statement of:

Q2.

In an isothermal process, $\Delta U$ is:

Q3.

In an adiabatic process, the relation between $P$ and $V$ is:

Q4.

Work done by gas in cyclic process equals:

Q5.

For ideal gas, internal energy is function of:

Second Law, Entropy and Heat EnginesTopic 2

Second Law of Thermodynamics:

  • Kelvin-Planck statement: No process can have its sole effect that of converting heat from a reservoir into work.
  • Clausius statement: Heat cannot flow spontaneously from a colder to a hotter body.

Heat Engine: Converts heat into work cyclically. Absorbs $Q_1$ from hot reservoir at $T_1$, rejects $Q_2$ to cold reservoir at $T_2$, does work $W = Q_1 - Q_2$.

Efficiency: $\eta = \dfrac{W}{Q_1} = 1 - \dfrac{Q_2}{Q_1}$.

Carnot Engine: Reversible engine operating between two reservoirs. Maximum efficiency: $$\eta_{\text{Carnot}} = 1 - \frac{T_2}{T_1}$$ ($T$ in Kelvin). Carnot cycle: isothermal expansion → adiabatic expansion → isothermal compression → adiabatic compression.

Refrigerator (heat pump in reverse): Extracts $Q_2$ from cold reservoir, supplies $W$, rejects $Q_1$ to hot reservoir.

Coefficient of Performance (COP): $$\text{COP}_{\text{refrigerator}} = \frac{Q_2}{W} = \frac{Q_2}{Q_1-Q_2} = \frac{T_2}{T_1-T_2}$$

Entropy $S$: state function whose change is $\Delta S = \int dQ_{\text{rev}}/T$. For reversible process: $\Delta S_{\text{universe}} = 0$. For irreversible: $\Delta S_{\text{universe}} > 0$.

Worked Examples
1

A Carnot engine operates between $T_1 = 500$ K and $T_2 = 300$ K. Find efficiency and work done when $1000$ J of heat is absorbed.

Show solution

$$\eta = 1 - \frac{300}{500} = 0.4 = 40\%$$ $$W = \eta Q_1 = 0.4 \times 1000 = 400 \text{ J}$$

Final Answer: $\eta = 40\%$, $W = 400$ J.

2

A refrigerator works between $-3°$C and $27°$C. Find the COP.

Show solution

$T_2 = 270$ K, $T_1 = 300$ K. $$\text{COP} = \frac{T_2}{T_1 - T_2} = \frac{270}{30} = 9$$

Final Answer: COP = $9$.

✎ Self-Check — 5 questions0 / 5
Q1.

Maximum efficiency of any heat engine between $T_1$ and $T_2$:

Q2.

A Carnot engine has efficiency $25\%$. To double the efficiency, the sink temperature should be (source at $T_1 = 400$ K):

Q3.

Second law of thermodynamics implies:

Q4.

Coefficient of performance of refrigerator is greatest when:

Q5.

Entropy of isolated system in irreversible process:

2
Module 2

Kinetic Theory of Gases

Ideal Gas, Pressure and TemperatureTopic 1

Ideal Gas Equation: $PV = nRT = NkT$, where $n$ = moles, $N$ = molecules, $k = R/N_A = 1.38 \times 10^{-23}$ J/K (Boltzmann constant), $N_A = 6.022 \times 10^{23}$.

Pressure of Ideal Gas (kinetic interpretation): $$P = \frac{1}{3}\rho \overline{v^2} = \frac{1}{3}\frac{N}{V}m\overline{v^2} = \frac{2}{3}\frac{N}{V} \cdot \frac{1}{2}m\overline{v^2}$$

This gives mean translational KE per molecule: $$\overline{KE}_{\text{trans}} = \frac{3}{2}kT$$

Mean Speeds of gas molecules at temperature $T$:

Speed typeFormulaValue at $300$ K (N₂)
RMS speed $v_{\text{rms}}$$\sqrt{3RT/M} = \sqrt{3kT/m}$$517$ m/s
Mean speed $\bar v$$\sqrt{8RT/(\pi M)}$$476$ m/s
Most probable $v_p$$\sqrt{2RT/M}$$422$ m/s

Ratio: $v_p : \bar v : v_{\text{rms}} = \sqrt{2} : \sqrt{8/\pi} : \sqrt{3} \approx 1 : 1.128 : 1.225$.

Maxwell-Boltzmann Distribution: Speeds of molecules follow: $$f(v) = 4\pi n\left(\frac{m}{2\pi kT}\right)^{3/2} v^2 e^{-mv^2/(2kT)}$$

Gas Laws (special cases of ideal gas equation):

  • Boyle's law (T const): $PV = \text{const}$
  • Charles's law (P const): $V/T = \text{const}$
  • Gay-Lussac's law (V const): $P/T = \text{const}$
  • Avogadro's law: equal volumes contain equal molecules at same $P, T$
Worked Examples
1

Find the RMS speed of nitrogen molecules at $27°$C. ($M_{N_2} = 28$ g/mol, $R = 8.314$ J/mol·K)

Show solution

$T = 300$ K, $M = 28 \times 10^{-3}$ kg/mol. $$v_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.028}}$$ $$= \sqrt{\frac{7482.6}{0.028}} = \sqrt{267235} \approx 517 \text{ m/s}$$

Final Answer: $v_{\text{rms}} \approx 517$ m/s.

2

At what temperature will the RMS speed of hydrogen molecules be the same as nitrogen at $27°$C? ($M_{H_2} = 2$, $M_{N_2} = 28$)

Show solution

$v_{\text{rms}} = \sqrt{3RT/M}$. Equal speeds: $$\sqrt{T_H/M_H} = \sqrt{T_N/M_N}$$ $$T_H = T_N \cdot \frac{M_H}{M_N} = 300 \times \frac{2}{28} = \frac{600}{28} \approx 21.4 \text{ K}$$

Final Answer: $T_H \approx 21.4$ K ($\approx -252°$C).

✎ Self-Check — 5 questions0 / 5
Q1.

The kinetic energy per mole of a gas at temperature $T$ is:

Q2.

RMS speed of a gas molecule is proportional to:

Q3.

At constant temperature, doubling the pressure on a gas:

Q4.

The Boltzmann constant has units of:

Q5.

The ratio of most probable to RMS speeds is:

Specific Heats, Degrees of Freedom and Mean Free PathTopic 2

Degrees of Freedom $f$: independent ways a molecule can store energy.

  • Monatomic (e.g., He, Ar): $f = 3$ (translational only)
  • Diatomic (e.g., $O_2, N_2$) at moderate T: $f = 5$ (3 trans + 2 rot)
  • Diatomic at high T: $f = 7$ (+ 2 vibrational)
  • Polyatomic (non-linear): $f = 6$

Equipartition theorem: Each quadratic degree of freedom contributes $\dfrac{1}{2}kT$ to mean energy per molecule.

Internal energy: $U = \dfrac{f}{2}nRT$.

Molar Specific Heats:

Gas type$C_V = \dfrac{f}{2}R$$C_P = C_V + R = \dfrac{f+2}{2}R$$\gamma = C_P/C_V$
Monatomic$\dfrac{3}{2}R$$\dfrac{5}{2}R$$\dfrac{5}{3} \approx 1.67$
Diatomic (rigid)$\dfrac{5}{2}R$$\dfrac{7}{2}R$$\dfrac{7}{5} = 1.4$
Polyatomic$3R$$4R$$\dfrac{4}{3} \approx 1.33$

Mayer's Relation: $C_P - C_V = R$.

Mean Free Path: Average distance a molecule travels between collisions. $$\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$$ where $d$ = molecular diameter, $n$ = molecules per unit volume. $\lambda \propto 1/P$ at constant $T$.

Mixture of gases: Effective $C_V$ for mixture of $n_1$ moles ($C_{V_1}$) and $n_2$ moles ($C_{V_2}$): $$C_V^{\text{mix}} = \frac{n_1 C_{V_1} + n_2 C_{V_2}}{n_1 + n_2}$$

Worked Examples
1

Find $C_P, C_V, \gamma$ for diatomic gas ($f = 5$) in terms of $R$.

Show solution
  • $C_V = \dfrac{5}{2}R$
  • $C_P = C_V + R = \dfrac{7}{2}R$
  • $\gamma = C_P/C_V = 7/5 = 1.4$

Final Answer: $C_V = 5R/2$, $C_P = 7R/2$, $\gamma = 1.4$.

2

A mixture contains $2$ moles He ($\gamma = 5/3$) and $3$ moles N₂ ($\gamma = 7/5$). Find $\gamma$ of mixture.

Show solution

$C_{V,He} = 3R/2$, $C_{V,N_2} = 5R/2$. $$C_V^{\text{mix}} = \frac{2(3R/2) + 3(5R/2)}{5} = \frac{3R + 7.5R}{5} = \frac{10.5R}{5} = 2.1R$$ $C_P^{\text{mix}} = C_V + R = 3.1R$. So $\gamma = 3.1/2.1 \approx 1.476$.

Final Answer: $\gamma_{\text{mix}} \approx 1.48$.

✎ Self-Check — 5 questions0 / 5
Q1.

For a monatomic gas, $\gamma$ is:

Q2.

The value of $C_P - C_V$ for an ideal gas is:

Q3.

Degrees of freedom of a diatomic molecule at low temperature (rigid rotator):

Q4.

Mean free path depends on:

Q5.

According to equipartition theorem, each degree of freedom has average energy:

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