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by Pravegaa Education
Chapter 11

Phase Transitions and Low-Temperature Physics

Theory, worked examples and 7 previous-year questions with full solutions — IIT-JAM, GATE, JEST and CSIR NET-JRF, 2020 onwards. Click Show solution under any question.

Theory

11.1 The third law and low temperatures

(a) Statements of the third law

The third law is usually stated in two equivalent forms. In the Nernst–Planck form, as \(T\to0\) the entropy of every system in internal equilibrium approaches a constant \(S_0\), independent of pressure, volume, field and so on; for a non-degenerate ground state, \(S_0=k_B\ln1=0\). In the unattainability form, no process can reduce a system to \(T=0\) in a finite number of steps.

Consequences. Since \(S(T)=S_0+\displaystyle\int_0^T\frac{C}{T'}dT'\) must stay finite, it follows that \(C_P,C_V\to0\) as \(T\to0\) (examples are the Debye law \(C\propto T^3\) and the electronic \(C\propto T\)). From the Maxwell relation \(\Big(\dfrac{\partial V}{\partial T}\Big)_P=-\Big(\dfrac{\partial S}{\partial P}\Big)_T\to0\), the expansion coefficient \(\alpha\to0\); similarly, \((\partial P/\partial T)_V\to0\). The classical ideal gas, with \(C_V=\frac32Nk_B\) and \(S\propto\ln T\), violates the third law, a defect that quantum statistics cures (Ch. 10).

(b) Joule–Kelvin (throttling) cooling

In a throttling process (porous plug or valve), gas passes steadily from pressure \(P_1\) to \(P_2\) with no heat exchange. The work balance \(U_1+P_1V_1=U_2+P_2V_2\) shows the process is isenthalpic: \(H_1=H_2\).

The Joule–Kelvin coefficient. Using \(dH=T\,dS+V\,dP\) and \(T\,dS=C_PdT-T(\partial V/\partial T)_PdP\):

\[\mu_{JK}=\Big(\frac{\partial T}{\partial P}\Big)_H=\frac1{C_P}\Big[T\Big(\frac{\partial V}{\partial T}\Big)_P-V\Big]=\frac{T^2}{C_P}\Big[\frac{\partial}{\partial T}\Big(\frac VT\Big)\Big]_P .\]

In a throttling expansion the pressure falls (\(dP<0\)). Hence when \(\mu_{JK}>0\) (below the inversion temperature \(T_i\)), \(dT<0\) and the gas cools on expansion, whereas when \(\mu_{JK}<0\) (above \(T_i\)) the gas heats on expansion. For an ideal gas, \(T(\partial V/\partial T)_P=V\), so \(\mu_{JK}=0\) and there is no temperature change.

van der Waals gas. To first order in \(a\) and \(b\), \(V\simeq\dfrac{RT}{P}+b-\dfrac{a}{RT}\). Then

\[T\Big(\frac{\partial V}{\partial T}\Big)_P-V=\frac{2a}{RT}-b\quad\Rightarrow\quad \mu_{JK}\simeq\frac{1}{C_P}\Big(\frac{2a}{RT}-b\Big),\qquad T_i=\frac{2a}{Rb}.\]

Attraction (\(a\)) thus favours cooling, while the finite molecular size (\(b\)) favours heating. In terms of the critical temperature \(T_c=\dfrac{8a}{27Rb}\), we have \(T_i=\dfrac{27}{4}T_c\). H₂ and He have very low inversion temperatures (about 200 K and 40 K). At room temperature they warm on throttling, so they must be pre-cooled (with liquid N₂, or liquid H₂ for He) before Joule–Kelvin liquefaction.

11.2 Phase equilibrium

(a) Conditions for coexistence

Two phases in contact reach equilibrium when \(T_1=T_2\) (thermal), \(P_1=P_2\) (mechanical) and \(\mu_1=\mu_2\) (chemical, i.e. no net particle flow). For a pure substance, \(\mu=g\), the Gibbs energy per particle. So along a coexistence line,

\[g_1(T,P)=g_2(T,P).\]

Gibbs phase rule. With \(C\) components and \(P\) phases, the number of independent intensive variables is

\[F=C-P+2 .\]

For a pure substance (\(C=1\)), a single phase has \(F=2\) (a region of the \(P\)–\(T\) plane), two phases have \(F=1\) (a line), and three phases have \(F=0\) (the triple point).

(b) The Clausius–Clapeyron equation

Derivation. Move along the coexistence line from \((T,P)\) to \((T+dT,P+dP)\). Equality of \(g\) is maintained, so \(dg_1=dg_2\). Using \(dg=-s\,dT+v\,dP\):

\[-s_1dT+v_1dP=-s_2dT+v_2dP\quad\Rightarrow\quad \boxed{\frac{dP}{dT}=\frac{s_2-s_1}{v_2-v_1}=\frac{L}{T\,\Delta v}} ,\]

where \(L=T(s_2-s_1)\) is the latent heat per particle (or per unit mass, with \(v\) the specific volume).

Vapour pressure. For liquid–vapour, \(v_g\gg v_l\) and the vapour is ideal, \(v_g=RT/P\) per mole. Then

\[\frac{d\ln P}{dT}=\frac{L}{RT^2}.\]

For constant \(L\), integration gives \(\ln P=-\dfrac{L}{RT}+C\), i.e. \(P=P_0e^{-L/RT}\). As a function of \(T\), \(\ln P\) rises from \(-\infty\) at \(T\to0^+\) and levels off to the constant \(C\) at high \(T\); plotted against \(1/T\), it is a straight line with slope \(-L/R\). If instead \(L=a-bT\), integrating \(\dfrac{a-bT}{RT^2}\) gives

\[\ln\frac{P}{P_c}=\frac aR\Big(\frac1{T_c}-\frac1T\Big)+\frac bR\ln\frac{T_c}{T}.\]

Similarly, for \(L=a-bT^{1/2}\) the second term integrates as \(-\dfrac bR\displaystyle\int T^{-3/2}dT=\dfrac{2b}{R}T^{-1/2}\), giving

\[\ln\frac{P}{P_c}=\frac aR\Big(\frac1{T_c}-\frac1T\Big)+\frac{2b}{R}\Big(\frac{1}{T^{1/2}}-\frac{1}{T_c^{1/2}}\Big).\]

Melting of ice. Ice is less dense than water, so \(v_{\rm liq}<v_{\rm ice}\) and \(\Delta v<0\) on melting while \(L>0\). So \(dP/dT<0\): increasing the pressure lowers the melting point. (This is also why water expands on freezing.) For most substances, \(\Delta v>0\) and the melting curve has positive slope.

Triple point. Where two coexistence curves meet. For vapour-pressure curves given as \(\ln P=A_i-B_i/T\), set them equal: \(T_3=\dfrac{B_1-B_2}{A_1-A_2}\).

Andrews’ experiments on CO₂. Above the critical temperature \(T_c=31.1\) °C, no liquefaction occurs at any pressure. Below it, isotherms show a flat two-phase region, which shrinks to a point (an inflection) at \(T_c\).

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