IPhO 2014 · Theory problem 2

Van der Waals equation of state

Download PDFIPhO 2014

In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered.

Part A. Non-ideal gas equation of state (2 points)

Taking into account the finite size of the molecules, the gaseous equation of state takes the form

\[ P(V-b)=RT, \tag{1} \]

where \(P,V,T\) stands for the gas pressure, its volume per mole and temperature, respectively, \(R\) denotes the universal gas constant, and \(b\) is a specific constant extracting some volume.

A1 Estimate \(b\) and express it in terms of the diameter of the molecules \(d\). (0.3 points)

With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter

\[ \left(P+\frac{a}{V^2}\right)(V-b)=RT. \tag{2} \]

where \(a\) is another specific constant.

At temperatures \(T\) below a certain critical value \(T_c\) the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment \(AB\) drawn at some constant pressure \(P_{LG}\). This straight segment is located between the volumes \(V_L\) and \(V_G\), and corresponds to the equilibrium of the liquid phase (indicated by \(L\)) and the gaseous phase (referred to by \(G\)). From the second law of thermodynamics J. Maxwell showed that the pressure \(P_{LG}\) must be chosen such that the areas \(I\) and \(II\) shown in Figure 1 must be equal.

Figure 1

Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2).

Figure 2

Figure 2. Several isotherms for van der Waals equation of state.

With increasing temperature the straight segment \(AB\) on the isotherm shrinks to a single point when the temperature and the pressure reaches some values \(T_c\) and \(P_{LG}=P_c\), respectively. The parameters \(P_c\) and \(T_c\) are called critical and can be measured experimentally with high degree of accuracy.

A2 Express the van der Waals constants \(a\) and \(b\) in terms of \(T_c\) and \(P_c\). (1.3 points)
A3 For water \(T_c=647\ \mathrm{K}\) and \(P_c=2.2\cdot10^7\ \mathrm{Pa}\). Calculate \(a_w\) and \(b_w\) for water. (0.2 points)
A4 Estimate the diameter of water molecules \(d_w\). (0.2 points)

Part B. Properties of gas and liquid (6 points)

This part of the problem deals with the properties of water in the gaseous and liquid states at temperature \(T=100\,^\circ\mathrm{C}\). The saturated vapor pressure at this temperature is known to be \(p_{LG}=p_0=1.0\cdot10^5\ \mathrm{Pa}\), and the molar mass of water is \(\mu=1.8\cdot10^{-2}\ \frac{\mathrm{kg}}{\mathrm{mole}}\).

Gaseous state

It is reasonable to assume that the inequality \(V_G\gg b\) is valid for the description of water properties in a gaseous state.

B1 Derive the formula for the volume \(V_G\) and express it in terms of \(R,T,p_0,\) and \(a\). (0.8 points)

Almost the same volume \(V_{G0}\) can be approximately evaluated using the ideal gas law.

B2 Evaluate in percentage the relative decrease in the gas volume due to intermolecular forces, \(\frac{\Delta V_G}{V_{G0}}=\frac{V_{G0}-V_G}{V_{G0}}\). (0.3 points)

If the system volume is reduced below \(V_G\), the gas starts to condense. However, thoroughly purified gas can remain in a mechanically metastable state (called supercooled vapor) until its volume reaches a certain value \(V_{G\min}\).

The condition of mechanical stability of supercooled gas at constant temperature is written as: \(\frac{dP}{dV}<0\).

B3 Find and evaluate how many times the volume of water vapor can be reduced and still remains in a metastable state. In other words, what is \(V_G/V_{G\min}\)? (0.7 points)

Liquid state

For the van der Waals’ description of water in a liquid state it is reasonable to assume that the following inequality holds \(P\ll a/V^2\).

B4 Express the volume of liquid water \(V_L\) in terms of \(a,b,R,\) and \(T\). (1.0 points)

Assuming that \(bRT\ll a\), find the following characteristics of water. Do not be surprised if some of the data evaluated do not coincide with the well-known tabulated values!

B5 Express the liquid water density \(\rho_L\) in some of the terms of \(\mu,a,b,R\) and evaluate it. (0.5 points)
B6 Express the volume thermal expansion coefficient \(\alpha=\frac{1}{V_L}\frac{\Delta V_L}{\Delta T}\) in terms of \(a,b,R\), and evaluate it. (0.6 points)
B7 Express the specific heat of water vaporization \(L\) in terms of \(\mu,a,b,R\) and evaluate it. (1.1 points)
B8 Considering the monomolecular layer of water, estimate the surface tension \(\sigma\) of water. (1.2 points)

Part C. Liquid-gas system (3 points)

From Maxwell’s rule (equalities of areas, by applying trivial integration) and the van der Waals’ equation of state together with the approximations made in Part B, it can be shown that the saturated vapor pressure \(p_{LG}\) depends on temperature \(T\) as follows

\[ \ln p_{LG}=A+\frac{B}{T}, \tag{3} \]

where \(A\) and \(B\) are some constants, that can be expressed in terms of \(a\) and \(b\) as \(A=\ln\left(\frac{a}{b^2}\right)-1\); \(B=-\frac{a}{bR}\)

W. Thomson showed that the pressure of saturated vapor depends on the curvature of the liquid surface. Consider a liquid that does not wet the material of a capillary (contact angle \(180^\circ\)). When the capillary is immersed into the liquid, the liquid in the capillary drops to a certain level because of the surface tension (see Figure 3).

C1 Find a small change in pressure \(\Delta p_T\) of the saturated vapor over the curved surface of liquid and express it in terms of the vapor density \(\rho_s\), the liquid density \(\rho_L\), the surface tension \(\sigma\) and the radius of surface curvature \(r\). (1.3 points)
Figure 3

Figure 3. Capillary immersed in a liquid that does not wet its material

Metastable states, considered in part B3, are widely used in real experimental setups, such as the cloud chamber designed for registration of elementary particles. They also occur in natural phenomena such as the formation of morning dew. Supercooled vapor is subject to condensation by forming liquid droplets. Very small droplets evaporate quickly but large enough ones can still grow.

C2 Suppose that at the evening temperature of \(t_e=20\,^\circ\mathrm{C}\) the water vapor in the air was saturated, but in the morning the ambient temperature has fallen by a small amount of \(\Delta t=5.0\,^\circ\mathrm{C}\). Assuming that the vapor pressure has remained unchanged, estimate the minimum radius of droplets that can grow. Use the tabulated value of water surface tension \(\sigma=7.3\cdot10^{-2}\ \mathrm{N/m}\). (1.7 points)

Official English text of the problem.

Answers

Numerical calculations use \(R=8.31\ \mathrm{J\,mol^{-1}K^{-1}}\), \(N_A=6.02\times10^{23}\ \mathrm{mol^{-1}}\), and \(T=373\ \mathrm K\) in Part B. Volumes are molar volumes. The molecular estimates below use the standard hard-sphere excluded-volume convention; their model dependence is discussed in the solutions of A1, A4 and B8.

  • A1: \(b\sim N_A d^3\); the hard-sphere estimate is \[ \boxed{b=\frac{2\pi}{3}N_A d^3.} \]

  • A2: \[ \boxed{a=\frac{27R^2T_c^2}{64P_c},\qquad b=\frac{RT_c}{8P_c}.} \]

  • A3: \[ \boxed{a_w\simeq0.55\ \mathrm{Pa\,m^6\,mol^{-2}},\qquad b_w\simeq3.1\times10^{-5}\ \mathrm{m^3\,mol^{-1}}.} \]

  • A4: \[ \boxed{d_w=\left(\frac{3b_w}{2\pi N_A}\right)^{1/3} \simeq2.9\times10^{-10}\ \mathrm m.} \]

  • B1: \[ \boxed{V_G=\frac{RT+\sqrt{R^2T^2-4ap_0}}{2p_0} \simeq3.08\times10^{-2}\ \mathrm{m^3\,mol^{-1}}.} \]

  • B2: \[ \boxed{\frac{\Delta V_G}{V_{G0}} =\frac{1-\sqrt{1-\frac{4ap_0}{R^2T^2}}}{2} \simeq0.58\%.} \]

  • B3: \[ \boxed{V_{G\min}=\frac{2a}{RT}\simeq3.6\times10^{-4}\ \mathrm{m^3\,mol^{-1}}, \qquad \frac{V_G}{V_{G\min}}\simeq86.} \]

  • B4: \[ \boxed{V_L=\frac{a-\sqrt{a^2-4abRT}}{2RT} =\frac{2b}{1+\sqrt{1-\frac{4bRT}{a}}}.} \]

  • B5: \[ \boxed{\rho_L\simeq\frac{\mu}{b}\simeq5.9\times10^2\ \mathrm{kg\,m^{-3}}.} \]

  • B6: \[ \boxed{\alpha\simeq\frac{bR}{a}\simeq4.6\times10^{-4}\ \mathrm{K^{-1}}.} \]

  • B7: \[ \boxed{L\simeq\frac{a}{\mu b}\simeq1.0\times10^6\ \mathrm{J\,kg^{-1}}.} \]

  • B8: A monolayer estimate with half the bulk cohesive contribution missing gives \[ \boxed{\sigma\sim\frac12\rho_L Ld_w =\frac{ad_w}{2b^2}\simeq8.6\times10^{-2}\ \mathrm{N\,m^{-1}}.} \] Its reliable precision is an order-of-magnitude estimate, \(\sigma\sim10^{-1}\ \mathrm{N\,m^{-1}}\).

  • C1: For the convex liquid surface shown, \[ \boxed{\Delta p_T=\frac{2\sigma\rho_s}{r(\rho_L-\rho_s)} \simeq\frac{2\sigma\rho_s}{r\rho_L}>0.} \]

  • C2: With \(T_e=293\ \mathrm K\) and \(\Delta T=5.0\ \mathrm K\), \[ \boxed{r_{\min}\simeq\frac{2\sigma T_e}{\rho_L L\Delta T} =\frac{2\sigma b^2T_e}{a\Delta T} \simeq1.4\times10^{-8}\ \mathrm m=14\ \mathrm{nm}.} \] Droplets larger than this critical radius can grow.

Solution

A1. Molecular size and excluded volume

Model the molecules as hard spheres of diameter \(d\). The center of one molecule cannot enter a sphere of radius \(d\) around the center of another. The excluded volume associated with a pair is therefore

\[ v_{\mathrm{ex}}=\frac{4\pi d^3}{3}. \]

In the standard dilute hard-sphere excluded-volume calculation, each pair is counted once, giving an excluded-volume parameter per molecule of \(v_{\mathrm{ex}}/2\). For one mole,

\[ \boxed{b=\frac12N_Av_{\mathrm{ex}} =\frac{2\pi}{3}N_A d^3.} \]

Thus the usual van der Waals \(b\) is four times the total geometrical volume of the spherical molecules. Independently of this geometric prefactor, the requested scale is \(b\sim N_A d^3\).

A2. Critical-point conditions

Write the equation of state as

\[ P(V,T)=\frac{RT}{V-b}-\frac{a}{V^2}. \]

At the critical point, the two extrema of the subcritical isotherm merge into a horizontal inflection point. Hence

\[ \left(\frac{\partial P}{\partial V}\right)_{T_c,V_c}=0, \qquad \left(\frac{\partial^2P}{\partial V^2}\right)_{T_c,V_c}=0. \]

These conditions give

\[ \frac{RT_c}{(V_c-b)^2}=\frac{2a}{V_c^3}, \qquad \frac{RT_c}{(V_c-b)^3}=\frac{3a}{V_c^4}. \]

Dividing the second equation by the first,

\[ \frac{1}{V_c-b}=\frac{3}{2V_c}, \]

so

\[ V_c=3b. \]

Substitution into the first derivative condition and the equation of state yields

\[ RT_c=\frac{8a}{27b}, \qquad P_c=\frac{a}{27b^2}. \]

Solving for the constants,

\[ \boxed{b=\frac{RT_c}{8P_c},\qquad a=\frac{27R^2T_c^2}{64P_c}.} \]

A3. Constants for water

Using the given critical data,

\[ b_w=\frac{8.31(647)}{8(2.2\times10^7)} =3.055\times10^{-5}\ \mathrm{m^3\,mol^{-1}}, \]

and

\[ a_w=\frac{27(8.31)^2(647)^2}{64(2.2\times10^7)} =0.5543\ \mathrm{Pa\,m^6\,mol^{-2}}. \]

To two significant figures,

\[ \boxed{a_w\simeq0.55\ \mathrm{Pa\,m^6\,mol^{-2}},\qquad b_w\simeq3.1\times10^{-5}\ \mathrm{m^3\,mol^{-1}}.} \]

The unrounded values are retained in subsequent calculations.

A4. Molecular diameter

From A1,

\[ d_w=\left(\frac{3b_w}{2\pi N_A}\right)^{1/3}. \]

Consequently,

\[ \boxed{d_w\simeq2.9\times10^{-10}\ \mathrm m=0.29\ \mathrm{nm}.} \]

This is an effective diameter within the hard-sphere model.

B1. Saturated vapor volume

Apply the prescribed approximation \(V_G\gg b\):

\[ \left(p_0+\frac{a}{V_G^2}\right)V_G=RT. \]

Multiplication by \(V_G\) gives

\[ p_0V_G^2-RTV_G+a=0. \]

The gaseous state corresponds to the larger root, which approaches \(RT/p_0\) as \(a\to0\):

\[ \boxed{V_G=\frac{RT+\sqrt{R^2T^2-4ap_0}}{2p_0}.} \]

At \(T=373\ \mathrm K\),

\[ V_G=0.0308164\ \mathrm{m^3\,mol^{-1}} \simeq3.08\times10^{-2}\ \mathrm{m^3\,mol^{-1}}. \]

For weak attraction, expansion of the square root also gives

\[ V_G\simeq\frac{RT}{p_0}-\frac{a}{RT}. \]

B2. Relative volume decrease

The ideal-gas volume is

\[ V_{G0}=\frac{RT}{p_0} =0.0309963\ \mathrm{m^3\,mol^{-1}}. \]

Using B1,

\[ \frac{V_{G0}-V_G}{V_{G0}} =\frac12\left(1-\sqrt{1-\frac{4ap_0}{R^2T^2}}\right). \]

Numerically,

\[ \frac{\Delta V_G}{V_{G0}} =5.803\times10^{-3}, \]

so

\[ \boxed{\frac{\Delta V_G}{V_{G0}}\simeq0.58\%.} \]

To first order, this fraction is \(ap_0/(R^2T^2)\).

B3. Limit of metastable compression

Within the same gas approximation,

\[ P(V)=\frac{RT}{V}-\frac{a}{V^2}. \]

Mechanical stability requires

\[ \frac{dP}{dV} =-\frac{RT}{V^2}+\frac{2a}{V^3} =\frac{2a-RTV}{V^3}<0. \]

Thus the gas remains mechanically stable against small volume perturbations when

\[ V>\frac{2a}{RT}. \]

The limiting, marginally stable volume is therefore

\[ \boxed{V_{G\min}=\frac{2a}{RT} =3.58\times10^{-4}\ \mathrm{m^3\,mol^{-1}}.} \]

The compression factor is

\[ \frac{V_G}{V_{G\min}} =\frac{RT}{2a} \frac{RT+\sqrt{R^2T^2-4ap_0}}{2p_0} \simeq86.2. \]

Hence

\[ \boxed{\frac{V_G}{V_{G\min}}\simeq86.} \]

At the limiting volume, \(V_{G\min}/b\simeq12\), so the result is an estimate within the prescribed approximation that neglects \(b\).

B4. Liquid volume

For the liquid, neglect \(P\) relative to \(a/V_L^2\):

\[ \frac{a}{V_L^2}(V_L-b)=RT. \]

Therefore

\[ RTV_L^2-aV_L+ab=0. \]

The liquid is the smaller-volume root, which approaches \(b\) as \(T\to0\):

\[ \boxed{ V_L=\frac{a-\sqrt{a^2-4abRT}}{2RT} =\frac{2b}{1+\sqrt{1-4bRT/a}}. } \]

At \(373\ \mathrm K\), this expression gives approximately

\[ V_L=3.91\times10^{-5}\ \mathrm{m^3\,mol^{-1}}. \]

For the remaining questions, introduce the stipulated small parameter

\[ \varepsilon=\frac{bRT}{a}\ll1. \]

Expanding the liquid root,

\[ \boxed{V_L=b\left(1+\frac{bRT}{a} +O(\varepsilon^2)\right).} \]

For the supplied water parameters, \(\varepsilon\simeq0.17\), so the following leading-order results are approximate.

B5. Liquid density

To leading order, \(V_L\simeq b\). Since one mole has mass \(\mu\),

\[ \boxed{\rho_L=\frac{\mu}{V_L}\simeq\frac{\mu}{b}.} \]

Numerically,

\[ \rho_L\simeq \frac{1.8\times10^{-2}}{3.055\times10^{-5}} =\boxed{5.9\times10^2\ \mathrm{kg\,m^{-3}}.} \]

This is the density predicted by the specified approximation.

B6. Thermal expansion coefficient

Retaining the first temperature-dependent term from B4,

\[ V_L\simeq b+\frac{b^2RT}{a}. \]

Thus, at fixed external pressure within the approximation used,

\[ \frac{\partial V_L}{\partial T}\simeq\frac{b^2R}{a}. \]

Using \(V_L\simeq b\) in the denominator,

\[ \boxed{\alpha=\frac{1}{V_L} \left(\frac{\partial V_L}{\partial T}\right)_P \simeq\frac{bR}{a}.} \]

Numerically,

\[ \boxed{\alpha\simeq4.6\times10^{-4}\ \mathrm{K^{-1}}.} \]

B7. Specific latent heat of vaporization

The thermodynamic identity

\[ \left(\frac{\partial U}{\partial V}\right)_T =T\left(\frac{\partial P}{\partial T}\right)_V-P \]

relates the internal energy to the equation of state. For the van der Waals fluid,

\[ T\left(\frac{\partial P}{\partial T}\right)_V-P =\frac{RT}{V-b} -\left(\frac{RT}{V-b}-\frac{a}{V^2}\right) =\frac{a}{V^2}. \]

Hence the molar internal energy has the form

\[ U(T,V)=U_0(T)-\frac{a}{V}. \]

At the same temperature, the internal-energy increase upon vaporization is

\[ U_G-U_L=a\left(\frac1{V_L}-\frac1{V_G}\right). \]

Vaporization at constant pressure also requires expansion work, so the heat supplied per mole is

\[ \mu L =a\left(\frac1{V_L}-\frac1{V_G}\right) +p_0(V_G-V_L). \]

Since \(V_G\gg V_L\simeq b\), and \(RT\ll a/b\), the leading contribution is \(a/b\). Therefore

\[ \boxed{L\simeq\frac{a}{\mu b}.} \]

Numerically,

\[ \boxed{L\simeq1.0\times10^6\ \mathrm{J\,kg^{-1}}.} \]

The omission of thermal corrections is consistent: retaining \(V_L\simeq b(1+bRT/a)\) gives \(a/V_L\simeq a/b-RT\), whose \(-RT\) correction cancels the leading expansion-work term \(p_0V_G\simeq RT\).

B8. Surface tension from a monomolecular layer

Surface tension is the surface free energy per unit area. For an energetic estimate, neglect the surface entropy contribution and use the cohesive energy found in B7.

The bulk cohesive energy per molecule is approximately

\[ E_{\mathrm{coh}}\simeq\frac{\mu L}{N_A} \simeq\frac{a}{bN_A}. \]

A molecule near a free surface lacks neighbors on the vapor side. In a simple estimate, its assigned cohesive binding is reduced by about one half, giving a surface excess energy of order \(E_{\mathrm{coh}}/2\).

A surface layer of area \(S\) and thickness \(d_w\) contains approximately

\[ N_{\mathrm{layer}}\simeq \frac{\rho_L S d_w}{\mu}N_A \]

molecules. Its excess energy is therefore

\[ E_{\mathrm{surface}} \sim \frac12E_{\mathrm{coh}}N_{\mathrm{layer}} =\frac12\rho_L Ld_wS. \]

Identifying this with \(\sigma S\),

\[ \boxed{\sigma\sim\frac12\rho_L Ld_w =\frac{ad_w}{2b^2}.} \]

Using the estimates above,

\[ \boxed{\sigma\sim8.6\times10^{-2}\ \mathrm{N\,m^{-1}}.} \]

The monolayer construction fixes the scale, \(\sigma\sim10^{-1}\ \mathrm{N\,m^{-1}}\); its numerical prefactor depends on the assumed molecular arrangement and missing interactions.

C1. Saturation pressure above a curved surface

Let \(p_s\) be the saturated vapor pressure above the flat liquid surface. The nonwetting meniscus is convex toward the vapor and lies a depth \(h\) below that surface.

For a small height difference, hydrostatic equilibrium gives the pressures at the meniscus level:

\[ p_{\ell}=p_s+\rho_Lgh, \qquad p_v=p_s+\rho_sgh. \]

The spherical meniscus obeys the Laplace pressure relation

\[ p_{\ell}-p_v=\frac{2\sigma}{r}. \]

Consequently,

\[ (\rho_L-\rho_s)gh=\frac{2\sigma}{r}. \]

The change in the equilibrium vapor pressure is

\[ \Delta p_T=p_v-p_s=\rho_sgh. \]

Eliminating \(h\),

\[ \boxed{\Delta p_T =\frac{2\sigma\rho_s}{r(\rho_L-\rho_s)} \simeq\frac{2\sigma\rho_s}{r\rho_L}.} \]

The last step uses \(\rho_s\ll\rho_L\). The change is positive: a convex liquid surface requires a higher equilibrium vapor pressure than a flat surface.

C2. Minimum radius of a growing droplet

Let \(p_s(T)\) denote the saturation pressure over a flat surface. Initially the actual vapor pressure is

\[ p=p_s(T_e),\qquad T_e=293\ \mathrm K. \]

After cooling by \(\Delta T=5.0\ \mathrm K\), the actual pressure stays fixed while the flat-surface saturation pressure falls. To first order,

\[ \Delta p_{\mathrm{cool}} =p_s(T_e)-p_s(T_e-\Delta T) \simeq\left.\frac{dp_s}{dT}\right|_{T_e}\Delta T. \]

From equation (3),

\[ \frac1{p_s}\frac{dp_s}{dT} =-\frac{B}{T^2} =\frac{a}{bRT^2}. \]

Thus

\[ \Delta p_{\mathrm{cool}} \simeq p_s\frac{a\Delta T}{bRT_e^2}. \]

A droplet can grow if the ambient vapor pressure exceeds its curvature-dependent equilibrium pressure. At the critical radius, the two pressure changes are equal:

\[ p_s\frac{a\Delta T}{bRT_e^2} =\frac{2\sigma\rho_s}{\rho_Lr_{\min}}. \]

Treating the dilute vapor as ideal,

\[ \rho_s\simeq\frac{\mu p_s}{RT_e}. \]

Substituting and cancelling \(p_s\),

\[ r_{\min} \simeq\frac{2\sigma\mu bT_e}{\rho_La\Delta T}. \]

Using the Part B results \(\rho_L\simeq\mu/b\) and \(L\simeq a/(\mu b)\),

\[ \boxed{ r_{\min}\simeq \frac{2\sigma b^2T_e}{a\Delta T} =\frac{2\sigma T_e}{\rho_LL\Delta T}. } \]

With the supplied tabulated surface tension,

\[ r_{\min}\simeq \frac{2(7.3\times10^{-2})(3.055\times10^{-5})^2(293)} {(0.5543)(5.0)} =\boxed{1.4\times10^{-8}\ \mathrm m\approx14\ \mathrm{nm}.} \]

At this radius a droplet is in unstable equilibrium. Larger droplets can grow; smaller droplets tend to evaporate.

The solution was checked against the official solution.