Thermodynamics

Before you read anything else:

1 mole of a monatomic ideal gas goes from 300 K to 400 K at constant pressure. Find ΔU. Take $R=8.31$.

Most candidates use $C_P$ because the pressure is constant. That gives $\frac{5}{2}(8.31)(100)=2078\text{J}$ — wrong.

Correct: $\Delta U=n{C}_V\Delta T=\frac{3}{2}(8.31)(100)=1247\text{J}$. The other 831 J became expansion work.

Section 4 proves why $C_V$ is right even here. That proof is the chapter.

1. Thermal equilibrium and the zeroth law

Thermal equilibrium — two systems at the same temperature, with no net heat flow between them.

Thermodynamic equilibrium — mechanical + chemical + thermal equilibrium, all at once.

Adiabatic wall — allows no heat exchange. Example: thermos flask.

Diathermic wall — allows heat exchange. Example: metal partition.

Zeroth law — if A is in thermal equilibrium with C, and B with C, then A is in equilibrium with B.

Why it matters: it makes temperature well-defined, and it is the reason a thermometer works — the thermometer is C.

Trap. Same temperature does not mean same internal energy. Two systems in thermal equilibrium can hold wildly different $U$.

2. Heat, work, internal energy

A gas contains internal energy. It does not contain heat or work — those exist only while crossing the boundary.

Sign convention (Physics). $Q>0$ heat into the system. $W>0$ work done by the system.

Chemistry uses $\Delta U=Q+W$ with $W$ done on the system. Write your convention on the rough sheet before question 1.

3. First law

$\Delta U=Q-W$

Valid for every process — reversible or not, quasi-static or violent.

Differential form: $dQ=dU+PdV$.

Illustration 1

A system absorbs 500 J and does 200 J of work. A second process between the same two states absorbs 350 J. Find its work.

$\Delta U=500-200=300\text{J}$

Same two states $\Rightarrow$ same $\Delta U$ $\Rightarrow$ $300=350-W$

$W=50\text{J}$

Neither path was described, and neither needed to be.

Illustration 2

A gas goes from A $(2\text{atm},1\text{L})$ to B $(1\text{atm},2\text{L})$ by two routes. Path 1: expand at constant $2$ atm to $2$ L, then drop the pressure at constant volume. Path 2: drop to $1$ atm at constant volume first, then expand at constant $1$ atm. Find $W$, $\Delta U$ and $Q$ for each.

Internal energy first, as always. For an ideal gas $T\propto PV$, and

$P_A{V}_A=2\text{L atm}=P_B{V}_B$

The temperature is unchanged, so $\Delta U=0$ on both paths.

Work is the area under each path, and only the isobaric legs contribute:

Heat, from the first law with $\Delta U=0$, so $Q=W$: $203$ J on path 1 and $101$ J on path 2.

This is the whole point of the chapter in one table. Identical endpoints force identical $\Delta U$; $Q$ and $W$ are free to differ, and here they differ by a factor of two. Path 1 does more work because it expands while the pressure is still high — the area under it is larger, and you can see that directly in the figure.

4. The theorem: ΔU = nCᵥΔT in every process

Three lines, and it removes half the chapter's memory load.

1. Ideal gas $\Rightarrow$ $U$ depends on $T$ alone $\Rightarrow$ $\Delta U$ is fixed by $T_1$ and $T_2$, whatever the path.
2. So evaluate it along a constant-volume path between those same two temperatures: there $W=0$, so $\Delta U=Q=n{C}_V\Delta T$.
3. The path cannot matter $\Rightarrow$ the same value holds for every process.

$\Delta U=n{C}_V\Delta T$

The subscript $V$ labels the coefficient, not the process.

5. The four processes

All four leave the same state. The adiabatic is the steeper of the two curves — section 9 shows it is steeper by exactly $\gamma$.

Work is the area under the path:

$W={\int }_{V_1}^{V_2}PdV$

Isochoric — constant volume

$dV=0\Rightarrow W=0\Rightarrow \Delta U=Q=n{C}_V\Delta T$

Isobaric — constant pressure

$P$ leaves the integral: $W=P\Delta V=nR\Delta T$. Heat: $Q=n{C}_P\Delta T$. And $\Delta U$ is still $n{C}_V\Delta T$.

Isothermal — constant temperature

$\Delta U=0\Rightarrow Q=W$. Substitute $P=nRT/V$:

$W={\int }_{V_1}^{V_2}\frac{nRT}{V}dV=nRT[\ln V{]}_{V_1}^{V_2}$

$W=nRT\ln \frac{V_2}{V_1}$

Needs a reservoir and slow motion — heat must flow in continuously to hold $T$ steady.

Illustration 3

Two moles of an ideal gas expand isothermally at $300\text{K}$ from $10\text{L}$ to $30\text{L}$. Find $W$, $Q$ and $\Delta U$. Then compare with the work if the same expansion were adiabatic, with $\gamma =1.4$. Take $R=8.31$.

Isothermal. $\Delta U=0$ because $T$ never changes, so the first law gives $Q=W$ immediately:

$W=nRT\ln \frac{V_2}{V_1}=2(8.31)(300)\ln 3=4986(1.0986)=5478\text{J}$

$Q=5478\text{J}$ in, $\Delta U=0$. Every joule that entered left again as work.

Adiabatic, same expansion. Now $Q=0$ and the gas must pay for the work out of its own internal energy, so it cools:

$T_2=300{\left(\frac{1}{3}\right)}^{0.4}=300(0.644)=193\text{K}$

$W=\frac{nR(T_1-T_2)}{\gamma -1}=\frac{2(8.31)(107)}{0.4}=4446\text{J}$

Read the comparison. The adiabatic expansion does less work for the same volume change — about 19% less — because its pressure falls away faster as it goes. That is the same fact as "the adiabatic curve is steeper", seen from the energy side rather than the geometry side.

Adiabatic — no heat exchange

$Q=0\Rightarrow \Delta U=-W$. Expansion always cools.

Derivation, from $dQ=0$:

$n{C}_{V}dT+PdV=0$

Put $P=nRT/V$, divide by $nT$:

$C_V\frac{dT}{T}+R\frac{dV}{V}=0$

Integrate: $C_V\ln T+R\ln V=\text{const}$. Divide by $C_V$, and use $\frac{R}{C_V}=\frac{C_P-C_V}{C_V}=\gamma -1$:

$T{V}^{\gamma -1}=\text{constant}$

Replace $T$ by $PV/nR$ for one more power of $V$:

$P{V}^{\gamma }=\text{constant}$

Third form: $P^{1-\gamma }{T}^{\gamma }=$ constant.

Work: put $P=K/V^{\gamma }$, and use $K{V}^{1-\gamma }=PV$ at each end:

$W=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=\frac{nR(T_1-T_2)}{\gamma -1}$

Adiabatic needs insulation or speed. A fast process leaves no time for heat to flow — which is why the compression stroke of an engine and a sound wave are both adiabatic.

Read down the $\Delta U$ column. Fill that column in first, always.

Illustration 4

Air at 300 K is compressed adiabatically to one-eighth of its volume. $\gamma =1.4$. Find $T_2$.

$T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}=300\times 8^{0.4}$

$8^{0.4}=2^{1.2}\approx 2.30\Rightarrow T_2\approx 690\text{K}$

More than double, with zero heat added. This is how a diesel engine ignites fuel without a spark plug.

Illustration 5

A gas at 2 atm, 3 L expands adiabatically to 12 L. $\gamma =1.5$. Find $W$ in L·atm.

$P_2=2{\left(\frac{3}{12}\right)}^{1.5}=2(0.125)=0.25\text{atm}$

$W=\frac{(2)(3)-(0.25)(12)}{0.5}=\frac{6-3}{0.5}=6\text{L atm}$

Check: $P_2{V}_2=3<P_1{V}_1=6$, so $T$ fell — as an adiabatic expansion requires.

6. Mayer's relation

One mole, constant pressure, temperature change $\Delta T$:

• $Q=C_P\Delta T$ — definition of $C_P$
• $W=P\Delta V=R\Delta T$ — gas law
• $\Delta U=C_V\Delta T$ — section 4, and this is the step that needs the theorem

Into $\Delta U=Q-W$:

$C_P-C_V=R$

$R$ is exactly the per-mole expansion work per degree. That is why $C_P$ is the larger.

7. Gamma and degrees of freedom

Equipartition: $\frac{1}{2}RT$ per degree of freedom per mole $\Rightarrow C_V=\frac{f}{2}R$.

$\gamma =\frac{C_P}{C_V}=1+\frac{2}{f}$

Illustration 6

Mixture: 2 mol He + 3 mol ${\text{O}}_2$. Find $\gamma$.

Specific heats add by moles. Gammas do not.

$C_V=\frac{2\left(\frac{3}{2}R\right)+3\left(\frac{5}{2}R\right)}{5}=2.1R$

$C_P=C_V+R=3.1R\Rightarrow \gamma =\frac{3.1}{2.1}\approx 1.48$

Averaging the gammas gives 1.51 — wrong, and sitting in the options as a distractor.

8. Polytropic processes: one formula for all four

The four standard processes look like four separate cases. They are one case.

Any process obeying

$P{V}^n=\text{constant}$

is called polytropic, and its molar heat capacity is

$C=C_V+\frac{R}{1-n}$

Where it comes from. Along $P{V}^n=$ const the work is $W=\frac{nR\Delta T}{1-n}$ per mole (the adiabatic derivation with $\gamma$ replaced by $n$), and the first law $Q=\Delta U+W$ divided by $\Delta T$ gives the result.

Now feed in the four values of $n$:

Four rows of the summary table, recovered from one line.

A negative $C$ is possible and is not an error. For $1<n<\gamma$ the gas absorbs heat and cools, because it does more work than the heat supplied. Nothing forbids it — $C$ is not a state function.

Illustration 7

A monatomic ideal gas is taken along $P{V}^2=\text{constant}$. Find its molar heat capacity, and say whether heat enters or leaves during an expansion.

$C_V=\frac{3}{2}R$ and $n=2$:

$C=\frac{3}{2}R+\frac{R}{1-2}=\frac{3}{2}R-R=\frac{1}{2}\mathbf{R}$

Direction of the heat flow. Since $T\propto PV$ and $P{V}^2$ is fixed, $T\propto \frac{1}{V}$ — so expanding cools this gas, and $\Delta T<0$.

With $C=+\frac{1}{2}R$ positive and $\Delta T$ negative, $Q=nC\Delta T$ is negative: heat leaves the gas even as it expands.

Cross-check from the first law. Per mole, with $\Delta T=T_2-T_1<0$:

$W=\frac{R\Delta T}{1-2}=-R\Delta T>0\text{(expansion, so positive)}$

$\Delta U=\frac{3}{2}R\Delta T$

$Q=\Delta U+W=\frac{3}{2}R\Delta T-R\Delta T=\frac{1}{2}R\Delta T$

which is exactly $C\Delta T$ with $C=\frac{1}{2}R$, and negative. The two routes agree.

9. Slopes, and cyclic processes

Isothermal: $PV=c\Rightarrow PdV+VdP=0\Rightarrow \frac{dP}{dV}=-\frac{P}{V}$

Adiabatic: $P{V}^{\gamma }=c\Rightarrow V^{\gamma }dP+\gamma P{V}^{\gamma -1}dV=0\Rightarrow \frac{dP}{dV}=-\gamma \frac{P}{V}$

${\left(\frac{dP}{dV}\right)}_{\text{adia}}=\gamma {\left(\frac{dP}{dV}\right)}_{\text{iso}}$

$\gamma >1$ always $\Rightarrow$ the adiabatic is steeper. That is how you label two unlabelled curves in a figure question.

Cyclic process

Return to the start $\Rightarrow$ $\Delta U=0\Rightarrow Q=W=$ area enclosed.

Clockwise → engine. Anticlockwise → refrigerator or heat pump.

Illustration 8

Rectangular cycle between $P$ and $3P$, $V$ and $3V$, run clockwise. Find net $W$ and $Q$.

Area $=(3P-P)(3V-V)=4PV$. Clockwise $\Rightarrow$ $W=+4PV$.

$\Delta U=0$ over the cycle $\Rightarrow$ $Q=W=4PV$. It is a heat engine.

10. Free expansion — the chapter's favourite trap

Nothing to push against $\Rightarrow$ $W=0$. Insulated $\Rightarrow$ $Q=0$.

$\Rightarrow \Delta U=0\Rightarrow \Delta T=0$ for an ideal gas.

Trap. Free expansion is adiabatic — but it is not quasi-static and not reversible. The gas has no single well-defined $P$ during it, so there is no $P$ to put in $\int PdV$. $P{V}^{\gamma }$ does not apply.

Illustration 9

A gas doubles its volume by free expansion. Compare with a reversible adiabatic doubling, $\gamma =1.4$.

Free: $\frac{T_2}{T_1}=1$. No change.

Reversible adiabatic: $\frac{T_2}{T_1}={\left(\frac{1}{2}\right)}^{0.4}\approx 0.76$ — a 24% drop.

Both adiabatic. Only one is reversible.

11. Second law and entropy

Kelvin–Planck — no process whose sole result is complete conversion of heat into work.

Clausius — no process whose sole result is heat flowing from colder to hotter.

The two are equivalent: violate either and you can build a violation of the other.

A refrigerator moves heat cold → hot, but not as its sole result — it also consumes work. That phrase is what excludes it.

Reversible process — can be run backwards leaving no trace on system or surroundings. Requires quasi-static + zero dissipation. No real process qualifies.

$\Delta S=\frac{Q_{\text{rev}}}{T}$

Entropy of an isolated system never decreases. It can fall locally — that is what a refrigerator does — provided the surroundings gain more.

Illustration 10

1 kg of ice melts at 0 °C. $L_f=3.34\times 10^5{\text{J kg}}^{-1}$. Find $\Delta S$.

$T$ constant at 273 K $\Rightarrow$ $\Delta S=\frac{3.34\times 10^5}{273}\approx 1223{\text{J K}}^{-1}$

Positive, as melting must be.

Illustration 11

One mole of an ideal gas doubles its volume at $300\text{K}$, (a) by reversible isothermal expansion, (b) by free expansion into a vacuum. Find $\Delta S$ in each case.

(a) Reversible isothermal. $\Delta U=0$, so $Q=W=RT\ln 2$:

$\Delta S=\frac{Q_{rev}}{T}=R\ln 2=8.31(0.693)=5.76{\text{J K}}^{-1}$

(b) Free expansion. Here $Q=0$ — and the tempting conclusion is $\Delta S=0$. It is wrong.

Entropy is a state function. The free expansion starts and ends at exactly the same two states as part (a): same temperature, same doubled volume. So

$\Delta S=5.76{\text{J K}}^{-1}$, identical.

The resolution of the apparent contradiction is the subscript on $Q_{rev}$. Entropy change is computed along any reversible path joining the two states, not along the actual one. Free expansion is irreversible, so its own $Q$ tells you nothing about $\Delta S$ — you must invent a reversible route, and part (a) is that route.

And the second law is satisfied: the system is isolated, its entropy rose by 5.76 J/K, and it will never spontaneously go back.

12. Heat engines and Carnot — JEE Advanced only

Removed from JEE Main in the 2023 syllabus revision; still in the Advanced syllabus. Skip if you are sitting Main only.

$\eta =1-\frac{T_C}{T_H}$

Also $\eta =W/Q_H=1-Q_C/Q_H$. No engine between the same two reservoirs beats the Carnot value, and every reversible engine between them ties it, whatever the working substance.

Refrigerator: $\text{COP}=Q_C/W=\frac{T_C}{T_H-T_C}$. It routinely exceeds 1, which is why it is not called an efficiency.

Illustration 12

Carnot engine, 500 K to 300 K, absorbs 1000 J per cycle. Find $W$, then the COP when reversed.

$\eta =1-\frac{300}{500}=0.4\Rightarrow W=400\text{J}$, rejecting 600 J.

Reversed: $\text{COP}=\frac{300}{200}=1.5$.

Check: 600 J moved for 400 J of work, and $600/400=1.5$. Same numbers, read backwards.

Summary

• $U$ is a state function; $Q$ and $W$ are not. Same endpoints $\Rightarrow$ same $\Delta U$.
• $\Delta U=n{C}_V\Delta T$ in every process. Write this line first, every time.
• First law $\Delta U=Q-W$, with $Q$ in and $W$ done by the gas. Valid even for irreversible processes.
• Work: isochoric 0, isobaric $nR\Delta T$, isothermal $nRT\ln (V_2/V_1)$, adiabatic $\frac{nR(T_1-T_2)}{\gamma -1}$.
• Adiabatic: $P{V}^{\gamma }=T{V}^{\gamma -1}=$ const, from integrating $C_{V}dT/T+RdV/V=0$.
• $C_P-C_V=R$; $\gamma =1+2/f$ → 1.67, 1.40, 1.33. Mixtures: combine $C_V$, never $\gamma$.
• Adiabatic slope $=\gamma \times$ isothermal slope at any shared point.
• Polytropic $P{V}^n$: $C=C_V+\frac{R}{1-n}$ reproduces all four standard processes from $n=0,1,\gamma ,\infty$. Negative $C$ is legal.
• Cycle: $\Delta U=0$, $Q=W=$ enclosed area. Clockwise engine, anticlockwise fridge.
• Free expansion: $Q=W=\Delta U=0$, no temperature change, and $P{V}^{\gamma }$ is invalid.
• Entropy is a state function: free expansion has $Q=0$ but $\Delta S=nR\ln 2$, computed along an invented reversible path.
• Carnot and $\eta =1-T_C/T_H$: Advanced only since 2023.