JEE Corner — Thermodynamic Processes, Cyclic Work and Advanced Applications

The Four Standard Processes in One Table

Almost every gas problem set at JEE level is one of four idealised processes, or a chain of them. Each process holds one variable fixed, and holding that variable fixed collapses the first law into a short expression you can write down without thinking.

Everything below uses the first law in the IUPAC form

$$\Delta U = q + w$$

where $\Delta U$ is the change in internal energy of the system, $q$ is the heat supplied to the system, and $w$ is the work done on the system. Expansion work against an external pressure $p_{ex}$ is $w = -p_{ex}\Delta V$, with $\Delta V = V_2 - V_1$.

Two results for an ideal gas do most of the work in this section. The internal energy of an ideal gas depends only on its temperature, because the molecules have no intermolecular potential energy to store. Enthalpy $H = U + pV = U + nRT$ then also depends only on temperature. So for $n$ moles of an ideal gas,

$$\Delta U = n\,C_{v,m}\,\Delta T \qquad \Delta H = n\,C_{p,m}\,\Delta T$$

where $C_{v,m}$ and $C_{p,m}$ are the molar heat capacities at constant volume and constant pressure.

Key Point: For an ideal gas, $\Delta U = nC_{v,m}\Delta T$ and $\Delta H = nC_{p,m}\Delta T$ hold on every path, not only at constant volume and constant pressure. The subscripts name where the heat capacities were measured, not where the formulae may be used. All you need is the temperature change.

The two molar heat capacities are tied together by

$$C_{p,m} - C_{v,m} = R$$

for one mole of an ideal gas. Now the four processes, for a fixed amount $n$ of an ideal gas doing only pressure-volume work.

Process	Held constant	$w$ (on the gas)	$\Delta U$	$q$	$\Delta H$
Isochoric	$V$	$0$	$nC_{v,m}\Delta T$	$q_v = \Delta U = nC_{v,m}\Delta T$	$nC_{p,m}\Delta T$
Isobaric	$p$	$-p\,\Delta V = -nR\,\Delta T$	$nC_{v,m}\Delta T$	$q_p = \Delta H = nC_{p,m}\Delta T$	$nC_{p,m}\Delta T$
Isothermal, reversible	$T$	$-nRT\ln\dfrac{V_2}{V_1}$	$0$	$-w = +nRT\ln\dfrac{V_2}{V_1}$	$0$
Adiabatic, reversible	$q = 0$	$nC_{v,m}\Delta T$	$nC_{v,m}\Delta T$	$0$	$nC_{p,m}\Delta T$

Four things in that table are worth spelling out.

In the isochoric row $\Delta V = 0$, so no expansion work is possible and every joule of heat goes into internal energy. $\Delta H$ is still non-zero, because the temperature changed; here $\Delta H = \Delta U + V\Delta p$.

In the isobaric row, $p\Delta V = nR\Delta T$ follows from writing $pV = nRT$ at both states and subtracting, which is legitimate only because $p$ is the same at both ends.

In the isothermal row $\Delta T = 0$, so $\Delta U$ and $\Delta H$ are both zero and the first law reduces to $q = -w$. Whatever work the gas does on the surroundings is replaced exactly by heat drawn in. The logarithmic expression for $w$ is the reversible one; for a one-step expansion against a constant $p_{ex}$ the work is the smaller $-p_{ex}\Delta V$ instead, while $\Delta U$ and $\Delta H$ stay zero.

In the adiabatic row $q = 0$ by definition, so $\Delta U = w$ exactly. Work done on the gas raises its temperature; work done by the gas cools it. This row is the one that carries most of the marks, and the next block takes it apart.

The four processes are one family

All four curves are special cases of a single reversible path called a polytropic process, defined by

$$pV^{k} = \text{constant}$$

for some fixed number $k$ called the polytropic index, written with a letter of its own so that it is never confused with the number of moles $n$. Reading off the four values of $k$ makes the family obvious.

$k$	The condition $pV^k =$ constant becomes	Process
$0$	$p =$ constant	isobaric
$1$	$pV =$ constant, which for an ideal gas means $T$ constant	isothermal
$\gamma$	$pV^{\gamma} =$ constant	reversible adiabatic
$k \rightarrow \infty$	$p^{1/k}V =$ constant, that is $V =$ constant	isochoric

For any $k \ne 1$ the reversible work integrates to a single expression,

$$w = \frac{p_2V_2 - p_1V_1}{k - 1}$$

Putting $k = \gamma$ gives the adiabatic work formula used later in this section. Putting $k = 0$ gives $w = (pV_2 - pV_1)/(-1) = -p\,\Delta V$, the isobaric result. The case $k = 1$ has to be handled separately because the denominator vanishes, and integrating $-\int p\,dV$ with $pV$ constant gives the logarithm instead. Seeing the four processes as one family stops them feeling like four unrelated formulae to memorise.

[JEE Main] A question that gives you $\Delta T$ and asks for $\Delta H$ is answered in one line by $nC_{p,m}\Delta T$, whatever the path was, provided the gas is ideal and no phase change or reaction occurs.

Adiabatic Change in an Ideal Gas

An adiabatic process is one in which no heat crosses the boundary: $q = 0$. It is achieved either by insulating the system or by running the change so fast that heat has no time to flow. The first law then gives

$$\Delta U = w$$

An adiabatic expansion cools the gas; an adiabatic compression heats it.

Deriving the relation between $p$ and $V$

Take one infinitesimal step of a reversible adiabatic change, so that the gas pressure $p$ and the external pressure are equal throughout. For an infinitesimal step,

$$dU = dq + dw = 0 + (-p\,dV) = -p\,dV$$

For $n$ moles of an ideal gas $dU = nC_{v,m}\,dT$, and $p = nRT/V$. Substituting both,

$$nC_{v,m}\,dT = -\frac{nRT}{V}\,dV$$

Divide through by $nT$ to separate the variables:

$$C_{v,m}\,\frac{dT}{T} = -R\,\frac{dV}{V}$$

Integrate from state 1 to state 2, treating $C_{v,m}$ as constant over the range:

$$C_{v,m}\ln\frac{T_2}{T_1} = -R\ln\frac{V_2}{V_1}$$

Now define the adiabatic exponent

$$\gamma = \frac{C_{p,m}}{C_{v,m}}$$

Since $C_{p,m} - C_{v,m} = R$ for an ideal gas, dividing that relation by $C_{v,m}$ gives $R/C_{v,m} = \gamma - 1$. Divide the integrated equation by $C_{v,m}$:

$$\ln\frac{T_2}{T_1} = -(\gamma - 1)\ln\frac{V_2}{V_1} = \ln\left(\frac{V_1}{V_2}\right)^{\gamma-1}$$

Removing the logarithms,

$$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \qquad \text{so} \qquad TV^{\gamma-1} = \text{constant}$$

Replace $T$ by $pV/nR$ in that result. The constant factor $1/nR$ cancels from both sides and leaves

$$pV^{\gamma} = \text{constant}$$

Eliminating $V$ instead gives the third member of the family:

$$p^{1-\gamma}T^{\gamma} = \text{constant} \qquad \text{equivalently} \qquad \frac{T_2}{T_1} = \left(\frac{p_2}{p_1}\right)^{(\gamma-1)/\gamma}$$

Key Point: For a reversible adiabatic change of an ideal gas with constant heat capacities, $pV^{\gamma} =$ constant, $TV^{\gamma-1} =$ constant and $p^{1-\gamma}T^{\gamma} =$ constant, where $\gamma = C_{p,m}/C_{v,m}$. Drop any one of those three conditions and the relations no longer hold.

Values of $\gamma$

For an ideal monatomic gas such as He, Ne or Ar, kinetic theory gives three translational degrees of freedom and

$$C_{v,m} = \frac{3}{2}R \qquad C_{p,m} = C_{v,m} + R = \frac{5}{2}R \qquad \gamma = \frac{5/2}{3/2} = \frac{5}{3} \approx 1.67$$

For an ideal diatomic gas such as $\mathrm{H_2}$, $\mathrm{N_2}$ or $\mathrm{O_2}$ at ordinary temperatures, two rotational degrees of freedom are also active and

$$C_{v,m} = \frac{5}{2}R \qquad C_{p,m} = \frac{7}{2}R \qquad \gamma = \frac{7/2}{5/2} = \frac{7}{5} = 1.4$$

Numerically, $C_{v,m} = 12.47\ \mathrm{J\,K^{-1}\,mol^{-1}}$ for a monatomic gas and $20.79\ \mathrm{J\,K^{-1}\,mol^{-1}}$ for a diatomic one.

These values assume the vibrational modes are not excited. At high temperature the vibrations of a diatomic molecule start to absorb energy, $C_{v,m}$ rises above $5R/2$ and $\gamma$ falls below $1.4$. For a linear triatomic or a bent triatomic molecule the count changes again. Exam problems say "monatomic" or "diatomic" precisely so that you may use $5/3$ and $7/5$ without worrying about this.

Work in an adiabatic process

Since $q = 0$, the work follows directly from the internal energy change:

$$w = \Delta U = nC_{v,m}(T_2 - T_1)$$

Replacing $C_{v,m} = R/(\gamma - 1)$ gives the form most useful when temperatures are known,

$$w = \frac{nR(T_2 - T_1)}{\gamma - 1}$$

and using $pV = nRT$ at each end, $nRT_2 = p_2V_2$ and $nRT_1 = p_1V_1$, gives the form most useful when pressures and volumes are known:

$$w = \frac{p_2V_2 - p_1V_1}{\gamma - 1}$$

All three expressions are the same equation. In an expansion $T_2 < T_1$, so $w$ is negative: the gas does work on the surroundings and pays for it out of its own internal energy. In a compression $T_2 > T_1$ and $w$ is positive.

One warning that costs marks every year is worth stating flatly. The three $pV^\gamma$-type relations were derived for a reversible adiabatic path. An adiabatic expansion against a constant external pressure is irreversible, and those relations do not apply to it. What still applies is $q = 0$ and

$$nC_{v,m}(T_2 - T_1) = w = -p_{ex}(V_2 - V_1)$$

which is solved for $T_2$ directly. Question 6 below works through exactly this case.

The two adiabatic paths differ in every quantity except $q$:

	Reversible adiabatic	Irreversible adiabatic (constant $p_{ex}$)
$q$	$0$	$0$
$pV^{\gamma} =$ constant	applies	does not apply
Route to $T_2$	$T_2 = T_1(V_1/V_2)^{\gamma-1}$	solve $nC_{v,m}(T_2-T_1) = -p_{ex}(V_2-V_1)$
$w$	$nC_{v,m}(T_2-T_1)$	$-p_{ex}(V_2-V_1)$, equal to $nC_{v,m}(T_2-T_1)$
Cooling on expansion	larger	smaller
$\Delta S$ of the gas	$0$	greater than $0$

The last row is worth holding on to. Both processes have $q = 0$, but only the reversible one has $\Delta S = 0$. Adiabatic and isentropic mean the same thing only when the path is also reversible.

A third adiabatic case sits at the far end of the same scale. In a free expansion the gas expands into an evacuated space, so $p_{ex} = 0$ and $w = 0$. If the container is also insulated, $q = 0$ as well, and the first law gives $\Delta U = 0$. For an ideal gas, zero $\Delta U$ means zero $\Delta T$: a free expansion of an ideal gas is both adiabatic and isothermal, and neither $pV^\gamma$ nor the cooling of the ordinary adiabatic case occurs. The entropy still rises, by $nR\ln(V_2/V_1)$, because the final state is the same as that reached by an isothermal expansion.

Isothermal and Adiabatic Curves Compared

Draw both processes on a $p$-$V$ diagram starting from the same state $(p_1, V_1)$ and let the gas expand to the same final volume $V_2$. The two curves separate immediately, and the reason is temperature.

Along the isotherm the gas is held at $T_1$ by contact with a thermostat, so heat flows in to replace the energy spent as work. Along the adiabat no heat enters, so the energy spent as work comes out of the internal energy and the gas cools. A cooler gas at a given volume has a lower pressure. The adiabat therefore lies below the isotherm at every volume beyond the common starting point.

The slopes

Differentiate the isothermal condition $pV =$ constant:

$$p\,dV + V\,dp = 0 \qquad \Rightarrow \qquad \left(\frac{dp}{dV}\right)_{T} = -\frac{p}{V}$$

Differentiate the adiabatic condition $pV^{\gamma} =$ constant by the product rule:

$$\gamma p V^{\gamma-1}\,dV + V^{\gamma}\,dp = 0 \qquad \Rightarrow \qquad \left(\frac{dp}{dV}\right)_{ad} = -\frac{\gamma p}{V}$$

Both slopes are negative, and at any common point $(p, V)$ they differ by exactly the factor $\gamma$.

Key Point: For an ideal gas, at any point on a $p$-$V$ diagram the reversible adiabat is steeper than the isotherm through that point by the factor $\gamma = C_{p,m}/C_{v,m}$. Since $\gamma > 1$ always, the adiabat always falls away more sharply. The statement belongs to reversible paths only: an irreversible adiabatic expansion against a constant external pressure does not obey $pV^{\gamma} =$ constant and is not a curve of this kind at all.

For a monatomic gas the adiabat is $5/3$ times as steep as the isotherm through the same point; for a diatomic gas, $1.4$ times as steep.

Numbers make the separation concrete. Take 1.0 mol of an ideal monatomic gas starting at 300 K in 10.0 L, where $p_1 = nRT/V = (0.08314)(300)/10.0 = 2.49\ \mathrm{bar}$, and follow both paths outward.

$V$ / L	$p$ on the isotherm / bar	$p$ on the adiabat / bar	Gas temperature on the adiabat / K
10.0	2.49	2.49	300
15.0	1.66	1.27	229
20.0	1.25	0.79	189
30.0	0.83	0.40	144

The isothermal pressures come from $p = p_1V_1/V$ and the adiabatic ones from $p = p_1(V_1/V)^{5/3}$. By the time the volume has tripled, the adiabatic pressure has fallen to less than half the isothermal one, and the gas has cooled to 144 K.

Which process does more work

Work done by the gas is the area under the curve on a $p$-$V$ diagram. Between the same two volumes, starting from the same state, the isotherm lies above the adiabat everywhere, so the area under it is larger.

$$\lvert w_{isothermal} \rvert > \lvert w_{adiabatic} \rvert \qquad \text{(same initial state, same } V_2 > V_1)$$

The physical reading is short: in the isothermal expansion the gas is being fed heat continuously, and that borrowed energy is delivered as extra work. In the adiabatic expansion the gas spends only what it already had.

The comparison flips for a compression. Compressing from the same initial state down to the same final volume, the adiabat rises more steeply, so the area under it is larger and more work must be done on the gas adiabatically than isothermally. An adiabatic compression also leaves the gas hotter, which is why a bicycle pump warms up.

A useful ordering for a single expansion between the same two volumes, from most work delivered to least:

1. Isothermal reversible expansion — largest.
2. Adiabatic reversible expansion — smaller, because the gas cools on the way.
3. Any single-step irreversible expansion against a constant external pressure — smaller still, because the opposing pressure never falls.
4. Free expansion into a vacuum — zero, because $p_{ex} = 0$.

That ordering carries a condition worth stating plainly: all four paths must start from the same state and finish at the same volume. If instead the paths are made to finish at the same final pressure, the volumes reached differ, the areas are no longer being compared over the same interval, and the ranking can change. Any comparison of work has to fix what the two paths have in common before it means anything.

The same caution applies to the heat. The isothermal expansion absorbs $\lvert w \rvert$ of heat, the adiabatic one absorbs none, and the irreversible one absorbs whatever its smaller work demands. Only $\Delta U$ and $\Delta H$ are indifferent to the path, and even they are equal across these cases only when the final temperatures agree, which they do not here.

[JEE Main] A question that asks only "which is greater" can be answered from the diagram in seconds. Sketch the two curves from the common point and compare areas; no arithmetic is needed.

Cyclic Processes and Work as an Area

A cyclic process returns the system to its exact starting state. Every state function has the same value at the end as at the start, so for one complete cycle

$$\Delta U = 0 \qquad \Delta H = 0 \qquad \Delta S = 0 \qquad \Delta G = 0$$

Feeding $\Delta U = 0$ into the first law leaves the single most useful statement about cycles:

$$q_{cycle} = -w_{cycle}$$

Net heat absorbed over a cycle equals net work done by the gas. Neither $q$ nor $w$ is a state function, so neither is zero in general; only their sum is.

Reading the area

For any step, the work done on the gas is $w = -\int p_{ex}\,dV$, and on a reversible path $p_{ex} = p$, so the magnitude of the work is the area under that step of the curve. Around a closed loop the areas under the outward and return paths partly cancel, and what survives is the area enclosed by the loop.

Key Point: For a cyclic process the magnitude of the net work equals the area enclosed by the loop on the $p$-$V$ diagram. A clockwise loop means net work done by the gas, so $w < 0$ and $q > 0$ in the IUPAC convention. An anticlockwise loop means net work done on the gas, so $w > 0$ and $q < 0$.

The sign rule follows from the geometry. In a clockwise loop the expansion half of the cycle happens along the upper, higher-pressure branch and the compression half along the lower branch. The gas therefore pushes out at high pressure and is pushed back in at low pressure, and it wins on the exchange. This is exactly what a heat engine does: absorb heat, deliver work, reject the rest. An anticlockwise loop is that engine run backwards, which is a refrigerator or a heat pump.

Working a cycle in practice

The reliable method is a small table. List the steps down the side and $w$, $\Delta U$, $q$ across the top, fill in whichever two entries each process makes easy, and get the third from $\Delta U = q + w$. Then check that the $\Delta U$ column sums to zero. That check catches almost every sign error.

Step type	Which entry is easiest first	Then
Isochoric	$w = 0$	$\Delta U = nC_{v,m}\Delta T$, and $q = \Delta U$
Isobaric	$w = -p\Delta V = -nR\Delta T$	$q = \Delta H = nC_{p,m}\Delta T$, and $\Delta U = q + w$
Isothermal reversible	$\Delta U = 0$	$w = -nRT\ln(V_2/V_1)$, and $q = -w$
Adiabatic	$q = 0$	$w = \Delta U = nC_{v,m}\Delta T$

Two conversions make the arithmetic painless when pressures are in bar and volumes in litres:

$$1\ \mathrm{bar\,L} = 100\ \mathrm{J} \qquad 1\ \mathrm{atm\,L} = 101.3\ \mathrm{J}$$

A rectangular cycle built from two isobars and two isochores has an area that is simply $\Delta p \times \Delta V$, which is the fastest area to evaluate. For a cycle containing an isothermal step the logarithmic work expression is needed for that step, and for one containing an adiabatic step the work is $nC_{v,m}\Delta T$ for that step.

One caution about efficiency. The net work of a cycle is not the whole heat absorbed, because some steps of the cycle reject heat. The efficiency is

$$\eta = \frac{\lvert w_{net} \rvert}{q_{absorbed}}$$

where $q_{absorbed}$ counts only the steps in which $q$ is positive, not the algebraic sum. Using the net heat in the denominator gives $\eta = 1$ for every cycle, which is the giveaway that the wrong quantity was used.

The upper limit on $\eta$ is set by the second law. A cycle built from two reversible isotherms at $T_{hot}$ and $T_{cold}$ joined by two reversible adiabats — the Carnot cycle — has

$$\eta_{max} = 1 - \frac{T_{cold}}{T_{hot}}$$

with both temperatures in kelvin. No cycle working between the same two temperatures can beat it, and every real cycle falls short because its steps are irreversible. The efficiency depends only on the two temperatures, not on the gas or on how much of it there is.

Where the state functions help

The reason cycles are worth setting up carefully is that they turn hard questions into easy ones. Any quantity that is a state function can be obtained by whichever route through the cycle is simplest, because the answer does not depend on the route. If a problem asks for $\Delta U$ from A to C and the direct path is awkward, going A to B to C gives the same number. The same freedom is what Hess's law exploits for reaction enthalpies, and it is the same principle wearing different clothes.

What cannot be moved between routes is $q$ and $w$. Each of those belongs to the particular path taken, which is exactly why the enclosed area of a cycle is not zero even though $\Delta U$ is.

Cycles drawn on other axes

A cycle is sometimes given on $p$-$T$ or $V$-$T$ axes instead, and the first job is to identify each step before any arithmetic starts. Reading the graph is a matter of knowing what each process looks like on those axes.

Process	On $p$-$V$ axes	On $p$-$T$ axes	On $V$-$T$ axes
Isochoric	vertical line	straight line through the origin	horizontal line
Isobaric	horizontal line	horizontal line	straight line through the origin
Isothermal	rectangular hyperbola	vertical line	vertical line

The straight lines through the origin come from $p \propto T$ at fixed $V$ and $V \propto T$ at fixed $p$, both direct readings of $pV = nRT$. Once each step is named, the work is computed on $p$-$V$ thinking as usual, since the area rule belongs to those axes alone. An area enclosed on a $p$-$T$ or $V$-$T$ plot has no thermodynamic meaning at all.

Kirchhoff's Equation — How $\Delta_r H$ Changes with Temperature

Tables of standard enthalpies of formation are quoted at 298 K. Industrial reactions are run at 700 K or 1200 K, and the reaction enthalpy is not the same there. Kirchhoff's equation supplies the correction.

Start from the definition of heat capacity at constant pressure applied to a single substance. Warming a substance from $T_1$ to $T_2$ at constant pressure changes its enthalpy by

$$H(T_2) = H(T_1) + C_p (T_2 - T_1)$$

when $C_p$ is constant over the interval. Apply this to the products and to the reactants of a reaction separately, then subtract. The enthalpy difference between products and reactants is $\Delta_r H$, and the heat capacity difference is

$$\Delta C_p = \sum C_{p}(\text{products}) - \sum C_{p}(\text{reactants})$$

each term weighted by its stoichiometric coefficient. Subtracting gives

$$\Delta_r H(T_2) - \Delta_r H(T_1) = \Delta C_p\,(T_2 - T_1)$$

Key Point (Definition): Kirchhoff's equation is $\Delta_r H(T_2) = \Delta_r H(T_1) + \Delta C_p (T_2 - T_1)$, where $\Delta C_p$ is the stoichiometric sum of the molar heat capacities of the products minus that of the reactants. It assumes $\Delta C_p$ is effectively constant across the temperature range and that no substance changes phase within that range.

Both conditions matter. Heat capacities themselves drift with temperature, so over a range of many hundreds of kelvin the constant-$\Delta C_p$ form is an approximation. The exact statement is a derivative,

$$\left(\frac{\partial \Delta_r H}{\partial T}\right)_p = \Delta C_p$$

whose integrated form is

$$\Delta_r H(T_2) = \Delta_r H(T_1) + \int_{T_1}^{T_2} \Delta C_p\,dT$$

and the linear equation above is what that integral becomes when $\Delta C_p$ is pulled out as a constant. If a reactant or product melts or boils between $T_1$ and $T_2$, the latent heat of that transition has to be added separately, because the simple linear form cannot see it. For a reaction of water, for instance, a range that straddles 373 K needs the enthalpy of vaporisation inserted at that point and the heat capacity of steam used above it.

The sign of $\Delta C_p$ decides the direction of the drift.

• $\Delta C_p > 0$: the products absorb heat faster than the reactants, so $\Delta_r H$ becomes more positive (less exothermic) as $T$ rises.
• $\Delta C_p < 0$: $\Delta_r H$ becomes more negative (more exothermic) as $T$ rises.
• $\Delta C_p = 0$: the reaction enthalpy is independent of temperature over that range.

There is a matching statement for internal energy at constant volume,

$$\Delta_r U(T_2) = \Delta_r U(T_1) + \Delta C_v\,(T_2 - T_1)$$

with $\Delta C_v$ built the same way from constant-volume heat capacities. The two forms are consistent with each other. Differentiating $\Delta_r H = \Delta_r U + \Delta n_g RT$ with respect to temperature gives

$$\Delta C_p = \Delta C_v + \Delta n_g R$$

which is just $C_{p,m} - C_{v,m} = R$ applied to every gaseous species and summed with its stoichiometric sign. Solids and liquids contribute nothing to $\Delta n_g$, so only the gases matter in that correction.

A worked feel for the size of the effect: for the ammonia synthesis $\Delta C_p$ is about $-45\ \mathrm{J\,K^{-1}\,mol^{-1}}$, so raising the temperature by 200 K shifts $\Delta_r H$ by about $-9\ \mathrm{kJ\,mol^{-1}}$ against a reaction enthalpy of $-92.4\ \mathrm{kJ\,mol^{-1}}$. The correction is a tenth of the value — small enough to ignore in a rough estimate, far too large to ignore in a plant design.

[JEE Main] Watch the units. Reaction enthalpies come in $\mathrm{kJ\,mol^{-1}}$ and heat capacities in $\mathrm{J\,K^{-1}\,mol^{-1}}$. The correction term $\Delta C_p(T_2-T_1)$ lands in joules and must be divided by 1000 before it is added.

Entropy Changes for an Ideal Gas

Entropy is a state function defined through a reversible path by $dS = dq_{rev}/T$. For an ideal gas that definition can be turned into a formula in $T$ and $V$ that works for any change of state, reversible or not, because $S$ depends only on the endpoints.

For an infinitesimal reversible step, the first law gives $dq_{rev} = dU + p\,dV$. For an ideal gas $dU = nC_{v,m}\,dT$ and $p = nRT/V$, so

$$dS = \frac{dq_{rev}}{T} = \frac{nC_{v,m}\,dT}{T} + \frac{nR\,dV}{V}$$

Integrating between the two states, with $C_{v,m}$ taken as constant,

$$\Delta S = nC_{v,m}\ln\frac{T_2}{T_1} + nR\ln\frac{V_2}{V_1}$$

The same change expressed in $T$ and $p$, obtained by substituting $V = nRT/p$ and using $C_{p,m} = C_{v,m} + R$, is

$$\Delta S = nC_{p,m}\ln\frac{T_2}{T_1} - nR\ln\frac{p_2}{p_1}$$

Key Point: For $n$ moles of an ideal gas, $\Delta S = nC_{v,m}\ln(T_2/T_1) + nR\ln(V_2/V_1)$. Because entropy is a state function, this holds for any process joining those two states, however irreversible the actual path was. The reversible path was used only to evaluate the integral.

Three special cases fall straight out.

Isothermal change. $T_2 = T_1$, so the first term vanishes:

$$\Delta S = nR\ln\frac{V_2}{V_1} = 2.303\,nR\log\frac{V_2}{V_1} = 2.303\,nR\log\frac{p_1}{p_2}$$

Expansion raises entropy; compression lowers it. For an isothermal reversible expansion the surroundings lose exactly what the system gains, so $\Delta S_{total} = 0$, which is the signature of a reversible process. For an isothermal irreversible expansion $\Delta S_{sys}$ is the same number, but the surroundings give up less heat, so $\Delta S_{total} > 0$.

Constant volume. $\Delta S = nC_{v,m}\ln(T_2/T_1)$. Heating raises entropy. At constant pressure the matching result is $\Delta S = nC_{p,m}\ln(T_2/T_1)$, larger for the same temperature rise because the gas also expands.

Reversible adiabatic change. Substituting $T_2/T_1 = (V_1/V_2)^{\gamma-1}$ into the general formula gives

$$\Delta S = nC_{v,m}(\gamma-1)\ln\frac{V_1}{V_2} + nR\ln\frac{V_2}{V_1} = -nR\ln\frac{V_2}{V_1} + nR\ln\frac{V_2}{V_1} = 0$$

using $C_{v,m}(\gamma-1) = R$. A reversible adiabatic process is isentropic. An irreversible adiabatic expansion, by contrast, has $q = 0$ but $\Delta S > 0$, so "adiabatic" and "constant entropy" are not synonyms.

The surroundings and the total

Entropy decides spontaneity only through the total, $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}$. The surroundings are treated as a reservoir so large that heat entering or leaving does not change its temperature, and any heat it exchanges is therefore exchanged reversibly:

$$\Delta S_{surr} = \frac{q_{surr}}{T} = -\frac{q_{sys}}{T}$$

Two isothermal expansions of the same gas between the same two states show what this buys. Take 1.0 mol expanding at 300 K from 10.0 L to 20.0 L. In both cases $\Delta S_{sys} = R\ln 2 = 5.76\ \mathrm{J\,K^{-1}}$, because entropy is a state function.

Carried out reversibly, the gas absorbs $q_{rev} = 1729\ \mathrm{J}$, so $\Delta S_{surr} = -1729/300 = -5.76\ \mathrm{J\,K^{-1}}$ and $\Delta S_{total} = 0$. Carried out against a constant external pressure equal to the final pressure, the gas does less work and absorbs only $q = 1247\ \mathrm{J}$, so $\Delta S_{surr} = -1247/300 = -4.16\ \mathrm{J\,K^{-1}}$ and

$$\Delta S_{total} = 5.76 - 4.16 = +1.60\ \mathrm{J\,K^{-1}} > 0$$

The system's entropy change is identical on the two paths; the surroundings' is not, because $q$ is a path function. That difference is the whole content of the second law for this process.

For a phase change at its transition temperature, where the change is reversible by construction,

$$\Delta S_{trans} = \frac{\Delta H_{trans}}{T_{trans}}$$

with $T_{trans}$ the melting or boiling point in kelvin.

Entropy of mixing

Two ideal gases at the same temperature and pressure, separated by a partition, mix spontaneously when the partition is removed. No heat is exchanged and no work is done, yet the process runs one way only. Entropy explains it.

Each gas expands from the volume it occupied alone into the whole container. For gas A, $V_2/V_1 = V_{total}/V_A = 1/x_A$, where $x_A$ is its mole fraction. Its entropy rise is $-n_A R\ln x_A$. Adding the contribution of every component,

$$\Delta S_{mix} = -R\sum_i n_i \ln x_i \qquad \text{or per mole of mixture} \qquad \Delta S_{mix} = -R\sum_i x_i \ln x_i$$

Every mole fraction is less than 1, so every logarithm is negative and $\Delta S_{mix}$ is always positive. For one mole each of two gases, $\Delta S_{mix} = -8.314(\ln 0.5 + \ln 0.5) = 11.53\ \mathrm{J\,K^{-1}}$.

Since mixing ideal gases has $\Delta H_{mix} = 0$, the Gibbs energy of mixing is $\Delta G_{mix} = -T\Delta S_{mix}$, which is negative at every temperature. Mixing of ideal gases is spontaneous and never reverses on its own.

Gibbs Energy Away from Standard Conditions

$\Delta_r G^{\circ}$ is a single fixed number for a balanced equation at a stated temperature, defined with every reactant and product in its standard state — pure, at 1 bar, and at unit concentration for solutions. A real reaction vessel almost never contains that mixture. The quantity that decides which way a real mixture will move is $\Delta_r G$, and the two are connected by

$$\Delta_r G = \Delta_r G^{\circ} + RT\ln Q$$

Here $Q$ is the reaction quotient: the same combination of partial pressures or concentrations that appears in the equilibrium constant, but evaluated with whatever amounts are present at that moment rather than with equilibrium amounts. For a gas reaction

$$a\mathrm{A}(g) + b\mathrm{B}(g) \rightleftharpoons c\mathrm{C}(g) + d\mathrm{D}(g) \qquad Q_p = \frac{(p_{\mathrm{C}}/p^{\circ})^{c}\,(p_{\mathrm{D}}/p^{\circ})^{d}}{(p_{\mathrm{A}}/p^{\circ})^{a}\,(p_{\mathrm{B}}/p^{\circ})^{b}}$$

with $p^{\circ} = 1\ \mathrm{bar}$, which makes $Q$ dimensionless so that its logarithm is defined. In practice the partial pressures are simply entered in bar and the $p^{\circ}$ divisions are silent.

Two checks confirm that the equation is sensible. If the mixture happens to be in standard states, every ratio is 1, $Q = 1$, $\ln Q = 0$ and $\Delta_r G = \Delta_r G^{\circ}$. And at equilibrium the mixture has no tendency to move either way, so $\Delta_r G = 0$ while $Q$ has reached its equilibrium value $K$. Putting both facts in,

$$0 = \Delta_r G^{\circ} + RT\ln K \qquad \Rightarrow \qquad \Delta_r G^{\circ} = -RT\ln K = -2.303\,RT\log K$$

Key Point: $\Delta_r G = \Delta_r G^{\circ} + RT\ln Q$ gives the driving force of a mixture at its actual composition. Setting $\Delta_r G = 0$ and $Q = K$ recovers $\Delta_r G^{\circ} = -RT\ln K$. The first equation describes a particular vessel; the second is a property of the reaction alone.

Substituting $\Delta_r G^{\circ} = -RT\ln K$ back into the first equation collapses both into one statement:

$$\Delta_r G = RT\ln\frac{Q}{K}$$

which reads off the direction of change immediately.

Condition	Sign of $\Delta_r G$	What the mixture does
$Q < K$	negative	runs forward, making more product
$Q = K$	zero	at equilibrium, composition fixed
$Q > K$	positive	runs backward, remaking reactant

This settles a confusion that costs marks. A reaction with a large positive $\Delta_r G^{\circ}$ is not forbidden. It has a small $K$, so it proceeds only a little way — but if the vessel starts with almost no product, $Q$ is smaller still, $\ln(Q/K)$ is negative, and the reaction does run forward until $Q$ climbs to $K$. What $\Delta_r G^{\circ}$ fixes is the stopping point, not whether anything happens at all.

The temperature dependence sits inside $\Delta_r G^{\circ} = \Delta_r H^{\circ} - T\Delta_r S^{\circ}$. Combining it with $\Delta_r G^{\circ} = -RT\ln K$ and dividing by $-RT$,

$$\ln K = -\frac{\Delta_r H^{\circ}}{RT} + \frac{\Delta_r S^{\circ}}{R}$$

A plot of $\ln K$ against $1/T$ is a straight line of slope $-\Delta_r H^{\circ}/R$ and intercept $\Delta_r S^{\circ}/R$, provided $\Delta_r H^{\circ}$ and $\Delta_r S^{\circ}$ are roughly constant over the range. An exothermic reaction has a positive slope, so $K$ falls as $T$ rises — the quantitative form of the qualitative rule that heating shifts an exothermic equilibrium backwards.

Writing that line at two temperatures and subtracting eliminates the entropy term and gives the two-point form,

$$\ln\frac{K_2}{K_1} = \frac{\Delta_r H^{\circ}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

which converts a value of $K$ at one temperature into a value at another using only the reaction enthalpy. The same caution attaches: $\Delta_r H^{\circ}$ must be effectively constant across the interval, which is the assumption Kirchhoff's equation exists to correct when it is not.

What $\Delta G$ measures

At constant temperature and pressure, $-\Delta G$ is the maximum work other than expansion work that a process can deliver:

$$w_{non-expansion,\ max} = \Delta G$$

in the IUPAC convention, where work done on the system is positive, so a negative $\Delta G$ corresponds to work delivered by the system. That maximum is reached only on a reversible path; any real process delivers less, with the shortfall appearing as entropy generated. This is the quantity a galvanic cell converts into electrical work, and the reason $\Delta_r G^{\circ} = -nFE^{\circ}$ appears in electrochemistry rather than $\Delta_r H^{\circ}$.

[JEE/NEET] A negative $\Delta_r G^{\circ}$ means $K > 1$, and a positive $\Delta_r G^{\circ}$ means $K < 1$. Neither says anything about how fast the reaction goes: thermodynamics fixes the destination, kinetics fixes the travelling time.

Question 1: The isochoric row, filled in

2.0 mol of an ideal monatomic gas is heated from 300 K to 500 K in a sealed rigid vessel. Find $w$, $\Delta U$, $q$ and $\Delta H$. Take $R = 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}$.

Answer:

The vessel is rigid, so $\Delta V = 0$ and no expansion work is possible.

$$w = -p_{ex}\Delta V = 0$$

The gas is monatomic, so $C_{v,m} = \tfrac{3}{2}R = 12.47\ \mathrm{J\,K^{-1}\,mol^{-1}}$ and $C_{p,m} = \tfrac{5}{2}R = 20.79\ \mathrm{J\,K^{-1}\,mol^{-1}}$. The temperature rise is $\Delta T = 500 - 300 = 200\ \mathrm{K}$.

$$\Delta U = nC_{v,m}\Delta T = 2.0 \times 12.47 \times 200 = 4988\ \mathrm{J}$$

From the first law with $w = 0$, the heat is the whole of the internal energy change.

$$q_v = \Delta U = +4.99\ \mathrm{kJ}$$

For the enthalpy I use the ideal-gas result that works on any path.

$$\Delta H = nC_{p,m}\Delta T = 2.0 \times 20.79 \times 200 = 8316\ \mathrm{J} = +8.32\ \mathrm{kJ}$$

Ans: $w = 0$, $\Delta U = q = +4.99\ \mathrm{kJ}$, $\Delta H = +8.32\ \mathrm{kJ}$ Watch out: $\Delta H$ is not zero just because the volume is fixed. The temperature changed, and for an ideal gas that alone fixes $\Delta H$. Here $\Delta H = \Delta U + V\Delta p$, and the pressure did rise.

Question 2: The isobaric row, filled in

3.0 mol of an ideal diatomic gas is heated from 300 K to 400 K at a constant pressure of 1.0 bar. Find $w$, $\Delta U$, $q$ and $\Delta H$.

Answer:

For a diatomic gas $C_{v,m} = \tfrac{5}{2}R = 20.79\ \mathrm{J\,K^{-1}\,mol^{-1}}$ and $C_{p,m} = \tfrac{7}{2}R = 29.10\ \mathrm{J\,K^{-1}\,mol^{-1}}$, and $\Delta T = +100\ \mathrm{K}$.

The pressure is constant at both ends, so I can convert $p\Delta V$ into $nR\Delta T$ using the gas equation at each state.

$$w = -p\Delta V = -nR\Delta T = -3.0 \times 8.314 \times 100 = -2494\ \mathrm{J}$$

The sign is negative because the gas expanded on heating and pushed the surroundings back.

$$\Delta H = q_p = nC_{p,m}\Delta T = 3.0 \times 29.10 \times 100 = 8730\ \mathrm{J} = +8.73\ \mathrm{kJ}$$

$$\Delta U = nC_{v,m}\Delta T = 3.0 \times 20.79 \times 100 = 6237\ \mathrm{J} = +6.24\ \mathrm{kJ}$$

I check with the first law: $q + w = 8730 - 2494 = 6236\ \mathrm{J}$, which matches $\Delta U$.

Ans: $w = -2.49\ \mathrm{kJ}$, $\Delta U = +6.24\ \mathrm{kJ}$, $q = \Delta H = +8.73\ \mathrm{kJ}$

Question 3: The isothermal row, filled in

2.0 mol of an ideal gas expands reversibly and isothermally at 300 K from 5.0 L to 25.0 L. Find $w$, $q$, $\Delta U$ and $\Delta H$.

Answer:

The temperature is constant, so for an ideal gas both $\Delta U$ and $\Delta H$ are zero straight away.

$$\Delta U = nC_{v,m}\Delta T = 0 \qquad \Delta H = nC_{p,m}\Delta T = 0$$

For the reversible isothermal work,

$$w = -2.303\,nRT\log\frac{V_2}{V_1}$$

$$2.303\,nRT = 2.303 \times 2.0 \times 8.314 \times 300 = 11488\ \mathrm{J}$$

$$\log\frac{25.0}{5.0} = \log 5 = 0.699$$

$$w = -11488 \times 0.699 = -8030\ \mathrm{J} = -8.03\ \mathrm{kJ}$$

With $\Delta U = 0$ the first law gives $q = -w$.

$$q = +8.03\ \mathrm{kJ}$$

Suppose the same expansion is instead carried out in one step against a constant external pressure equal to the final pressure of the gas. The final pressure is

$$p_2 = \frac{nRT}{V_2} = \frac{2.0 \times 0.08314 \times 300}{25.0} = 2.00\ \mathrm{bar}$$

$$w = -p_{ex}\Delta V = -(2.00)(25.0-5.0) = -40.0\ \mathrm{bar\,L} = -4.00\ \mathrm{kJ}$$

$\Delta U$ and $\Delta H$ are still zero, because the end states are the same, and $q$ is still $-w$, now $+4.00\ \mathrm{kJ}$.

Ans: Reversible: $w = -8.03\ \mathrm{kJ}$, $q = +8.03\ \mathrm{kJ}$. Irreversible one-step: $w = -4.00\ \mathrm{kJ}$, $q = +4.00\ \mathrm{kJ}$. In both, $\Delta U = \Delta H = 0$ Watch out: $\Delta U = 0$ does not mean nothing happened. It also does not fix $w$: the same pair of end states gave two different amounts of work, because work is a path function while internal energy is not.

Question 4: Reversible adiabatic expansion of a monatomic gas

1.0 mol of an ideal monatomic gas at 300 K expands reversibly and adiabatically until its volume doubles. Find the final temperature, $w$, $\Delta U$ and $\Delta H$.

Answer:

The gas is monatomic, so $\gamma = 5/3$ and $\gamma - 1 = 2/3$. The path is a reversible adiabatic one for an ideal gas, so $TV^{\gamma-1}$ is constant.

$$T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} = 300 \times \left(\frac{1}{2}\right)^{2/3}$$

$$\left(\frac{1}{2}\right)^{2/3} = 2^{-0.667} = 0.630$$

$$T_2 = 300 \times 0.630 = 189\ \mathrm{K}$$

The gas cooled by 111 K, which is what an adiabatic expansion must do. With $q = 0$ the work equals the internal energy change.

$$w = \Delta U = nC_{v,m}(T_2 - T_1) = 1.0 \times 12.47 \times (-111) = -1384\ \mathrm{J}$$

The same number from the other form, as a check:

$$w = \frac{nR(T_2-T_1)}{\gamma-1} = \frac{1.0 \times 8.314 \times (-111)}{2/3} = -1384\ \mathrm{J}$$

$$\Delta H = nC_{p,m}\Delta T = 1.0 \times 20.79 \times (-111) = -2308\ \mathrm{J}$$

Ans: $T_2 = 189\ \mathrm{K}$, $q = 0$, $w = \Delta U = -1.38\ \mathrm{kJ}$, $\Delta H = -2.31\ \mathrm{kJ}$ Watch out: The exponent in $TV^{\gamma-1}$ is $\gamma - 1$, not $\gamma$. Using $\gamma$ by mistake gives $300 \times (0.5)^{5/3} = 94.5\ \mathrm{K}$, which looks like a perfectly reasonable answer and is not one.

Question 5: Adiabatic compression using pressures and volumes

1.0 mol of an ideal diatomic gas at 1.00 bar occupying 20.0 L is compressed reversibly and adiabatically to 5.0 L. Find the final pressure and the work done. Take $1\ \mathrm{bar\,L} = 100\ \mathrm{J}$.

Answer:

For a diatomic gas $\gamma = 1.4$. The path is reversible and adiabatic, so $p_1V_1^{\gamma} = p_2V_2^{\gamma}$.

$$p_2 = p_1\left(\frac{V_1}{V_2}\right)^{\gamma} = 1.00 \times \left(\frac{20.0}{5.0}\right)^{1.4} = 4^{1.4}$$

$$\log(4^{1.4}) = 1.4 \times 0.602 = 0.843 \qquad 4^{1.4} = 6.96$$

$$p_2 = 6.96\ \mathrm{bar}$$

For the work I use the form built from pressures and volumes.

$$w = \frac{p_2V_2 - p_1V_1}{\gamma - 1} = \frac{(6.96)(5.0) - (1.00)(20.0)}{0.4}\ \mathrm{bar\,L}$$

$$= \frac{34.8 - 20.0}{0.4} = \frac{14.8}{0.4} = 37.0\ \mathrm{bar\,L} = +3.70\ \mathrm{kJ}$$

The positive sign is correct: this is a compression, so work is done on the gas.

The temperatures come from the gas equation at each end, with $R = 0.08314\ \mathrm{L\,bar\,K^{-1}\,mol^{-1}}$.

$$T_1 = \frac{p_1V_1}{nR} = \frac{20.0}{1.0 \times 0.08314} = 241\ \mathrm{K} \qquad T_2 = \frac{p_2V_2}{nR} = \frac{34.8}{0.08314} = 419\ \mathrm{K}$$

Checking the work by the temperature form, with $C_{v,m} = \tfrac{5}{2}R = 20.79\ \mathrm{J\,K^{-1}\,mol^{-1}}$ for a diatomic gas:

$$w = nC_{v,m}(T_2-T_1) = 1.0 \times 20.79 \times (419-241) = +3700\ \mathrm{J}$$

which agrees with the pressure-volume form to the precision of the rounding.

Ans: $p_2 = 6.96\ \mathrm{bar}$, $T_2 = 419\ \mathrm{K}$, $w = \Delta U = +3.70\ \mathrm{kJ}$, $q = 0$ Watch out: $\gamma - 1$ in the denominator is 0.4, not 1.4. Dividing by 1.4 gives 1.06 kJ, and the answer is out by a factor of three and a half.

Question 6: Adiabatic expansion against a constant external pressure

1.0 mol of an ideal monatomic gas at 300 K and 5.0 bar expands adiabatically against a constant external pressure of 1.0 bar until mechanical equilibrium is reached. Find the final temperature and the work done.

Answer:

The path is adiabatic but irreversible, because the gas pressure is 5.0 bar while the opposing pressure is 1.0 bar. The relation $pV^{\gamma} =$ constant was derived for a reversible path and cannot be used here. What survives is $q = 0$, so $\Delta U = w$, together with $w = -p_{ex}(V_2 - V_1)$.

The gas stops expanding when its pressure equals the external pressure, so $p_2 = 1.0\ \mathrm{bar}$.

$$nC_{v,m}(T_2 - T_1) = -p_{ex}\left(\frac{nRT_2}{p_2} - \frac{nRT_1}{p_1}\right)$$

Putting in $n = 1$, $C_{v,m} = \tfrac{3}{2}R$, $p_{ex} = p_2 = 1.0\ \mathrm{bar}$, $p_1 = 5.0\ \mathrm{bar}$ and dividing both sides by $R$:

$$\frac{3}{2}(T_2 - 300) = -\left(T_2 - \frac{300}{5}\right) = -(T_2 - 60)$$

$$1.5T_2 - 450 = -T_2 + 60$$

$$2.5T_2 = 510 \qquad T_2 = 204\ \mathrm{K}$$

$$w = \Delta U = nC_{v,m}(T_2-T_1) = 1.0 \times 12.47 \times (204 - 300) = -1197\ \mathrm{J}$$

The entropy change is worth adding, because it shows what "irreversible" costs. Using the pressure form for an ideal gas,

$$\Delta S = nC_{p,m}\ln\frac{T_2}{T_1} - nR\ln\frac{p_2}{p_1} = 20.79\ln\frac{204}{300} - 8.314\ln\frac{1.0}{5.0}$$

$$= 20.79 \times (-0.386) - 8.314 \times (-1.609) = -8.02 + 13.38 = +5.36\ \mathrm{J\,K^{-1}}$$

No heat crossed the boundary, so $\Delta S_{surr} = 0$ and $\Delta S_{total} = +5.36\ \mathrm{J\,K^{-1}}$, which is positive as the second law requires for a change that happens on its own. A reversible adiabatic expansion between a different pair of end states would have given exactly zero.

Ans: $T_2 = 204\ \mathrm{K}$, $w = \Delta U = -1.20\ \mathrm{kJ}$, $q = 0$, $\Delta H = 20.79 \times (-96) = -2.00\ \mathrm{kJ}$, $\Delta S = +5.36\ \mathrm{J\,K^{-1}}$ Watch out: The reversible formula $T_2 = T_1(p_2/p_1)^{(\gamma-1)/\gamma}$ would give 158 K here. That number is wrong for this problem, and it is the single most common error in adiabatic questions: the word "adiabatic" alone does not license $pV^{\gamma}$.