21 Jun 2025

Isothermal, Adiabatic, Reversible, and Irreversible Processes

Work and heat in ideal-gas processes, van der Waals isothermal work, reversibility, and mechanical equivalence.

bsc semester-iv mj-6 heat-and-thermodynamics thermodynamic-processes reversibility

Take heat into the gas and work done by the gas as positive, so $dU=\delta Q-\delta W$. For quasistatic boundary work,

\[W_{1\to2}=\int_{V_1}^{V_2}p_{\mathrm{ext}}\,dV.\]

A reversible process proceeds through equilibrium states and can be reversed by an infinitesimal change without leaving a net change in system and surroundings. During reversible boundary work, $p_{\mathrm{ext}}=p$ in the limiting sense. Finite pressure or temperature differences, friction, viscosity, free expansion, and unrestrained mixing make a process irreversible.

Reversible isothermal process

For $n$ moles of ideal gas at constant $T$,

\[p=\frac{nRT}{V}.\]

The work done by the gas is

\[\boxed{W_{\mathrm{iso}}=nRT\ln\!\frac{V_2}{V_1}}.\]

Ideal-gas internal energy depends only on $T$, so $\Delta U=0$. The first law therefore gives

\[Q_{\mathrm{iso}}=W_{\mathrm{iso}}.\]

For expansion, $V_2>V_1$, so $W$ and $Q$ are positive. For compression, both are negative.

For one mole of van der Waals gas at fixed $T$,

\[p=\frac{RT}{V_m-b}-\frac{a}{V_m^2},\]

and reversible isothermal work is

\[W=RT\ln\!\frac{V_{m2}-b}{V_{m1}-b} +a\left(\frac1{V_{m2}}-\frac1{V_{m1}}\right).\]

The attractive term reduces the expansion work relative to the excluded-volume contribution.

Reversible adiabatic process

An adiabatic boundary permits no heat transfer: $\delta Q=0$. For an ideal gas with temperature-independent molar heat capacity $C_{V,m}$,

\[nC_{V,m}\,dT=-p\,dV=-\frac{nRT}{V}\,dV.\]

Integration gives

\[C_{V,m}\ln\!\frac{T_2}{T_1}+R\ln\!\frac{V_2}{V_1}=0.\]

Using $\gamma=C_{P,m}/C_{V,m}$ and $C_{P,m}-C_{V,m}=R$,

\[\boxed{TV^{\gamma-1}=\text{constant}},\qquad \boxed{pV^\gamma=\text{constant}}.\]

The work can be written in equivalent forms:

\[\boxed{W_{\mathrm{ad}}=nC_{V,m}(T_1-T_2) =\frac{p_1V_1-p_2V_2}{\gamma-1}}.\]

Thus reversible adiabatic expansion cools the gas, while reversible adiabatic compression heats it. These power laws require an ideal gas, constant heat capacities, and a reversible process; $\delta Q=0$ alone is not sufficient.

Irreversible work and path dependence

If a gas expands against a constant external pressure,

\[W=p_{\mathrm{ext}}(V_2-V_1),\]

even though the gas pressure may be nonuniform during the change. In free expansion into vacuum, $p_{\mathrm{ext}}=0$, so $W=0$. If the container is also insulated, $Q=0$ and $\Delta U=0$; an ideal gas then has $T_2=T_1$. The same end states can be joined reversibly with nonzero $Q$ and $W$, showing that heat and work are path functions while $U$ is a state function.

Conversion of work and heat

Friction or electrical resistance can convert work completely into internal energy and heat. In an adiabatic paddle-wheel experiment on a closed fluid, work done on the system means $W<0$, so

\[\Delta U=-W>0.\]

Heat and work are measured in the same SI unit, the joule. The thermochemical calorie is

\[1\ \mathrm{cal}=4.184\ \mathrm{J}.\]

The first law fixes energy equivalence; the second law determines the limits on converting heat back into cyclic work.

Solved Problems

1. Compare reversible and irreversible isothermal expansion to twice the initial volume

Let one mole of ideal gas begin at $(T,V_1,p_1)$ and end at $V_2=2V_1$, so $p_2=p_1/2$. A reversible isothermal path gives

\[W_{\mathrm{rev}}=RT\ln2.\]

If the gas instead expands irreversibly against the constant final pressure $p_{\mathrm{ext}}=p_2$, then

\[W_{\mathrm{irr}}=p_2(V_2-V_1) =\frac{p_1}{2}V_1=\frac{RT}{2}.\]

Thus

\[W_{\mathrm{rev}}-W_{\mathrm{irr}} =RT\left(\ln2-\frac12\right)>0.\]

Both paths have $\Delta U=0$, so $Q=W$ on each path. The reversible path delivers the maximum expansion work because its external pressure is never lowered by a finite unnecessary amount.

2. Find the final temperature in an irreversible adiabatic expansion

An ideal gas with constant total heat capacity $C_V$ expands adiabatically from $(T_1,V_1)$ to $V_2$ against constant $p_{\mathrm{ext}}$. Since $Q=0$,

\[C_V(T_2-T_1)=-p_{\mathrm{ext}}(V_2-V_1).\]

Therefore

\[\boxed{T_2=T_1-\frac{p_{\mathrm{ext}}(V_2-V_1)}{C_V}}.\]

Expansion makes the second term positive before the minus sign, so $T_2<T_1$. This result uses the actual external pressure and must not be replaced by $pV^\gamma=\text{constant}$, which applies only to a reversible adiabat. Units check: $p\Delta V/C_V$ is kelvin.

Descriptive Questions

Give the mechanical and thermal conditions that distinguish a reversible process from a merely quasistatic one.
Why does an adiabatic process need not be isentropic?
Explain why free expansion can have $Q=W=0$ while still being irreversible.
State the assumptions behind $pV^\gamma=\text{constant}$ and identify one case where it fails.

Numerical Problems

One mole of a van der Waals gas expands reversibly and isothermally at $300\ \mathrm K$ from $V_{m1}=0.0100\ \mathrm{m^3\,mol^{-1}}$ to $V_{m2}=0.0200\ \mathrm{m^3\,mol^{-1}}$. Use $a=0.364\ \mathrm{Pa\,m^6\,mol^{-2}}$ and $b=4.27\times10^{-5}\ \mathrm{m^3\,mol^{-1}}$.

Final answer: $W=RT\ln[(V_{m2}-b)/(V_{m1}-b)]+a(1/V_{m2}-1/V_{m1})=1.716\ \mathrm{kJ}$.
A gas expands from $0.010$ to $0.020\ \mathrm{m^3}$ against a constant external pressure of $1.00\times10^5\ \mathrm{Pa}$.

Final answer: $W=p_{\mathrm{ext}}\Delta V=1.00\ \mathrm{kJ}$.
An ideal gas with $\gamma=1.40$ expands reversibly and adiabatically from $300\ \mathrm K$ to twice its volume.

Final answer: $T_2=T_1(V_1/V_2)^{\gamma-1}=227\ \mathrm K$.
A paddle wheel does $3.00\ \mathrm{kJ}$ of work on $2.00$ mol of a monatomic ideal gas in a rigid, adiabatic vessel. Find the internal-energy and temperature changes, taking $C_{V,m}=3R/2$ and work done by the gas as positive.

Final answer: $W=-3.00\ \mathrm{kJ}$, so $\Delta U=Q-W=+3.00\ \mathrm{kJ}$ and $\Delta T=\Delta U/(nC_{V,m})=120.3\ \mathrm K$.

The thermodynamic-processes Maxima worksheet checks reversible and irreversible work, the adiabatic temperature law, sign conventions, and every numerical answer.

References

Thermodynamic process, Wikipedia.
D. V. Schroeder, An Introduction to Thermal Physics, Addison-Wesley, 2000, chapter 1.
F. W. Sears and G. L. Salinger, Thermodynamics, Kinetic Theory, and Statistical Thermodynamics, 3rd ed., Addison-Wesley, 1975, chapters 4-5.