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For a fixed amount of substance and composition of classical ideal gas, internal energy depends only on temperature: $U=U(T,n)$, with no independent volume dependence. This does not require constant heat capacity at constant volume. Between equilibrium endpoints,Here $C_V$ is total heat capacity at constant volume, $C_{V,\mathrm{m}}$ is molar heat capacity, and $U/m$ for sample mass $m$ is specific internal energy, whose change is ${\int }_{T_1}^{T_2}{c}_V(T)dT$. These endpoint relations apply even when the intervening process is not at fixed volume; they do not say that transferred heat always equals $\Delta U$. With constant $C_V$, the safe statement is $U(T)=U(T_{\ast })+C_V(T-T_{\ast })$. Writing $U=C_{V}T$ additionally chooses the reference energy; it does not prove that a physical gas remains classical and ideal down to zero temperature. ^[ref]

A thermodynamic proof outline uses the first law of thermodynamics in equilibrium form, $dU=TdS-pdV$, and the Maxwell relation $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$:The last step uses the ideal gas law and the molar gas constant $R$, with fixed composition and amount of substance. Microscopically, separated ideal gas particles store no intermolecular interaction energy; their translational kinetic energy and any internal molecular energy are determined by temperature. Negligible intermolecular interactions still allow a molecule to have internal vibrational potential energy.

An original analytical model makes the distinction between an ideal gas and constant heat capacity concrete. Assume classical translation, two fully active rotational terms, and one independent quantum harmonic oscillator per diatomic molecule, with fixed angular frequency $\omega$. Define the vibrational temperature scale $\Theta =\hslash \omega /k_B$, using the reduced Planck constant $\hslash =h/(2\pi )$ and Boltzmann constant $k_B$. Summing its thermal occupations givesThe constant $U_{\ast }$ includes the chosen reference and zero-point energy, whose temperature derivative vanishes. When $T\ll \Theta$ the vibrational contribution to heat capacity is exponentially small; when $T\gg \Theta$ it tends to $R$ per mole. This model assumes rotations remain active throughout the interval and excludes dissociation, electronic excitation and real gas corrections. Its low-$T/\Theta$ edge is not a model of rotational freezing or of an actual substance at absolute zero.^[ref]

For this stipulated model, heating from $T_1=0.2\Theta$ to $T_2=\Theta$ gives $\Delta U/(nR\Theta )\simeq 2.57519$. Holding the frozen-vibration value $C_{V,\mathrm{m}}=5R/2$ throughout would give $2$, underestimating the required internal energy increase by about $22.3\%$ of the model's full value. The integral measures all thermal storage activated along the interval. At fixed volume with only heat input, this is also the supplied heat; at fixed pressure, expansion work must be added through Mayer's relation.

Joule's free-expansion experiment tests the independence from volume in a different way: an insulated rigid vessel allows gas into an evacuated compartment, so no external work is delivered and ideally no heat is exchanged. The first law of thermodynamics gives $\Delta U=0$; positive $C_V$ then predicts equal initial and final equilibrium temperatures for the ideal gas. The apparatus, thermal-capacity controls and limitations of this inference are described in Joule's free-expansion experiment and temperature sensitivity of a free-expansion experiment. This endpoint test differs from the reversible adiabatic expansion of a loaded piston, which delivers work and lowers internal energy.

Across a horizontal material interface with continuous background mass density but a velocity jump, an internal gravity wave preserves laboratory frequency and horizontal wavenumber. The matching conditions are continuity of displacement $w/\hat{\omega }$ and pressure, not continuity of $w$. For wave amplitudes and upward phase-line angles measured from the vertical ${\theta }_1,{\theta }_2$, define $r={\hat{\omega }}_2/{\hat{\omega }}_1$. Then the transmission coefficient isThe displacement condition is $A_i+A_r=A_t\cos {\theta }_2/(r\cos {\theta }_1)$, while pressure continuity gives $A_i-A_r=r{A}_t\sin {\theta }_2/\sin {\theta }_1$. Adding eliminates the reflected amplitude and gives the coefficient. It assumes nonzero intrinsic frequencies and propagating outgoing branches.

The intrinsic planetary luminosity $L_{\mathrm{int}}$ is the power supplied by interior cooling, contraction, nuclear reactions if present, or internal heating, excluding reflected or passively reradiated stellar light. For fixed mass and no accretion or nuclear source, global conservation of energy gives $L_{\mathrm{int}}=Q_{\mathrm{deep}}-d(U+\Omega )/dt$, where $Q_{\mathrm{deep}}$ is any additional power deposited in the interior, such as tidal heating or Joule heating. For $Q_{\mathrm{deep}}=0$ and a gas-supported quasi-static sphere of fixed specific-heat ratio $\gamma$ and negligible surface pressure, the stellar virial theorem gives $\Omega =-3(\gamma -1)U$, henceFor a dynamically stable gas-supported model take $\gamma >4/3$. For $\gamma =5/3$ this is half the gravitational release. Degenerate giants require a different internal energy reservoir; the half-factor is not universal.

A steady incompressible jet at atmospheric pressure conserves Bernoulli's equation and volume flux. Its radius contracts as $R=a[u/u_0{]}^{-1/2}$. After impact, a thin nearly horizontal axisymmetric film has $2\pi rhv=\pi a^2{u}_0$. Evaluating Bernoulli's equation on its free surface gives $v^2=u_0^2+2g(H-h)$, hence $a^4/(4r^2{h}^2)=1+2g(H-h)/u_0^2$. This inviscid thin-film description applies away from the impact region and does not describe a dissipative hydraulic jump.

With fixed pressure $p$, let $A=\mu p/\mathcal{R}$, so $\rho =A/T$. A material mass coordinate has ${\partial }_x=\rho {\partial }_m$ and the material time derivative becomes ${\partial }_t{|}_m$. The energy and continuity equations then give the displayed equation with $\tau =Ct$ and $C=(\gamma -1)(\mu /\mathcal{R}{)}^{2}p/\gamma$. A mass integral from a fixed origin is material only when the mass flux there vanishes; otherwise subtract its time-integrated flux to label parcels. Uniform pressure is a low-Mach approximation when sound crossing is faster than thermal evolution.

Isothermal compressibility is the compressibility measured at fixed temperature, particle number and composition. A slowly compressed sealed piston in contact with a thermal reservoir can approach this constraint. For an ideal gas, differentiating $V=N{k}_{B}T/P$ at fixed temperature gives $(\partial V/\partial P{)}_{T,N}=-V/P$, hence ${\kappa }_T=1/P$. A small pressure increase $\Delta P$ therefore gives $\Delta V/V\simeq -\Delta P/P$. For a finite change, use the whole isothermal process curve $V_2/V_1=P_1/P_2$ instead.

Take one mole of classical monatomic gas, with stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, initially at $T_0=300\mathrm{K}$ and $P_0=100\mathrm{kPa}$. The initial volume is $V_0=nR{T}_0/P_0=0.024942{\mathrm{m}}^3$. Compressing to $V_0/2$ givesThe isothermal compression releases heat to the thermal reservoir; the reversible adiabatic compression retains the input work as internal energy. Reversing either curve gives the corresponding expansion, with adiabatic expansion cooling the gas.

The equilibrium curvature of Gibbs free energy provides another measurement of isothermal compressibility:This follows from $V=(\partial G/\partial P{)}_{T,N}$, with $G$ denoting Gibbs free energy. Within a stable single thermodynamic phase, increasing pressure reduces volume and makes this curvature negative. At phase coexistence, changing the phase fraction can change the volume at constant temperature and pressure; the usual finite single-phase isothermal compressibility then need not exist. Near a critical point, it can diverge. ^[ref]

For an isothermal gas slab, a horizontal Fourier mode has $\delta \rho =-ik\rho {\xi }_x-(\rho {\xi }_z{)}_z$, $W=c_s^2\delta \rho /\rho +\varphi$, ${\omega }^2{\xi }_x=ikW$, ${\omega }^2{\xi }_z=W_z$ and $({\partial }_z^2-k^2)\varphi =4\pi G\delta \rho$. Multiply momentum by $\rho {\mathbf{\xi }}^{\ast }$ and integrate by parts, then use the Poisson equation for Newtonian gravity. Vanishing pressure-work and gravitational boundary terms give the displayed real quadratic identity, with $I=\int \rho |\mathbf{\xi }{|}^{2}dz>0$. Thus every admissible squared eigenfrequency is real; a negative value gives gravitational growth rather than an oscillatory instability.

An isothermal expansion increases the volume of a thermodynamic system at fixed temperature. For a reversible ideal gas path, the gas delivers positive work and absorbs positive heat, with $Q=W_{\mathrm{by},\mathrm{rev}}=nRT\ln (V_f/V_i)$ and $\Delta U=0$. The heat bath supplies the expansion energy; without sufficient heat transfer, doing work instead lowers the temperature. This transfer is the hot step of a Carnot cycle, and it is why an expanding gas need not cool when its surroundings continuously replenish its internal energy.

An insulated free expansion into an evacuated compartment is different. In the idealization with no heat exchange, no external work and no container energy change, $Q=W_{\mathrm{by}}=0$. Since internal energy of an ideal gas depends only on temperature, its final equilibrium temperature equals its initial value. The transient need not have a uniform temperature or pressure, so equal endpoint temperatures do not make it a reversible isotherm. Do not apply the logarithmic isothermal ideal-gas work formula to this irreversible process. Its entropy still increases by $\Delta S=nR\ln (V_f/V_i)$; entropy change in free expansion evaluates this state function difference using a hypothetical reversible path.^[ref]

For a real gas, even equal-temperature equilibrium endpoints need not have equal internal energy. An insulated free expansion can therefore change the temperature, while a controlled isothermal expansion must provide the heat needed for both its internal energy change and its delivered work. The internal energy at constant temperature article gives the Van der Waals equation example.^[ref]

A gas syringe makes the distinction between a constant-temperature path and equal-temperature endpoints observable. The Ideal Gas Law Apparatus (TD-8596A), Instruction Manual 012-08420B by PASCO scientific specifies a TD-8596A gas syringe with a mechanical stop, an internal thermistor, an attached pressure sensor and data acquisition. Record the stop volume with the coupling disconnected, then set $40\mathrm{mL}$ and seal the coupling. Record pressure and temperature, compress quickly to the stop near $20\mathrm{mL}$, and hold until both readouts settle. The manual gives roughly $30\mathrm{s}$ for settling and about $0.5\mathrm{s}$ for the thermistor response. Releasing the plunger gives a second transient and a final volume to record.^[ref]

The manual describes a pressure overshoot and temperature rise during rapid compression, followed by cooling at the held volume; the settled pressure remains higher than initially. This illustrates approximately adiabatic compression followed by heat rejection, rather than an isothermal compression throughout. Use the settled initial and compressed readings to test Boyle's law, correcting both syringe readings by the unmarked tubing volume $V_d$:Use absolute pressure, not gauge pressure. A transient comparison of $pV/T$ must pair simultaneous pressure and temperature readings: the sensor lag can separate their peaks.^[ref]

For an additional test of the path, repeat with small compression steps, wait for stable temperature and pressure at every step, and repeat in reverse. Keep the sealed amount of substance, connecting tubing and ambient temperature fixed; prevent hand warming of the barrel and check for leaks by watching for continued pressure drift at held volume after thermal equilibrium. These are proposed controls, rather than reported measurements. Agreement of equilibrated points with Boyle's law supports the fixed-temperature ideal gas law approximation, but does not prove that the rapid segments between them were isothermal processes. A single local temperature sensor can miss temperature gradients; plunger friction can produce direction-dependent readings. The pressure and volume data estimate reversible work only insofar as the path approaches a reversible thermodynamic process; they do not directly measure heat. After motion stops, cooling at fixed volume has zero boundary work and $Q=\Delta U<0$, even though the complete equal-endpoint-temperature ideal gas run has $\Delta U=0$.