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Chapter 4: Problem 11

For a van der Waals gas, plot the isotherms in the \(\bar{P}-\bar{V}\) plane \((\bar{P}\) and \(\bar{V}\) are the reduced pressure and volume) for reduced temperatures \(\bar{T}=0.5, \bar{T}=1.0\), and \(\bar{T}=1.5\). For \(\bar{T}=0.5\), is \(\bar{P}=0.1\) the equilibrium pressure of the liquid-gas coexistence region?

Short Answer

Expert verified
Plot isotherms for \bar{T}=0.5, 1.0, 1.5; equilibrium pressure is determined from the critical point data.

Step by step solution

01

- Understand the van der Waals equation in reduced terms

The van der Waals equation in reduced form describes the behavior of a real gas. The equation is given by: The reduced form equation is then expressed as where: where
02

- Plot the isotherms for given reduced temperatures

For reduced temperatures plot the reduced pressure in plane. Use the van der Waals equation in reduced form. get the curves.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

isotherms
Isotherms are curves on a graph that represent the relationship between pressure and volume at constant temperature. For a van der Waals gas, this relationship is more complex compared to an ideal gas because it considers molecular size and interactions.
In the \(\bar{P} - \bar{V}\) plane, we plot the reduced pressure \(\bar{P}\) against the reduced volume \(\bar{V}\) for different reduced temperatures \(\bar{T}\).
Each curve on the plot corresponds to a specific temperature.
  • For higher temperatures, the isotherms resemble the hyperbolic shape typical of an ideal gas.
  • For lower temperatures, the isotherms show regions where liquid and gas coexist, known as the phase transition region.
This region is crucial for understanding the equilibrium between liquid and gas states in real gases.
reduced pressure
Reduced pressure \(\bar{P}\) is a dimensionless form of pressure, scaled by a characteristic pressure of the substance. It is given by:
\[\bar{P} = \frac{P}{P_c} \]
where:
  • \(P\) is the actual pressure of the gas.
  • \(P_c\) is the critical pressure of the gas.
Using reduced pressure allows us to compare different gases on a common scale, making it easier to study their behaviors.
Plotting \(\bar{P}\) versus reduced volume simplifies the analysis of phase transitions and other properties, revealing insights that would be obscured by using raw pressure values alone.
reduced temperature
Reduced temperature \(\bar{T}\) is another dimensionless quantity, which scales the actual temperature with the critical temperature of the gas. It is given by:
\[\bar{T} = \frac{T}{T_c} \]
where:
  • \(T\) is the actual temperature of the gas.
  • \(T_c\) is the critical temperature of the gas.
When \(\bar{T} = 1.0\), the gas is at its critical temperature.
For \(\bar{T} < 1.0\), the gas can exhibit phase transitions from gas to liquid.
For \(\bar{T} > 1.0\), the gas behaves more like an ideal gas, with less deviation seen in the van der Waals isotherms.
Understanding reduced temperature is crucial in predicting the behavior of gases under different thermal conditions.
van der Waals equation
The van der Waals equation accounts for real gas behavior by modifying the ideal gas law. It incorporates two factors:
  • Molecular volume: It adjusts the volume available to gas molecules.
  • Intermolecular forces: It accounts for the attraction between gas molecules.
In reduced form, the van der Waals equation is:
\[ ( \bar{P} + \frac{3} { \bar{V}^2 } ) ( 3\bar{V} - 1 ) = 8\bar{T} \]
Here, each variable is scaled by its critical value:
  • \(\bar{P} = \frac{P}{P_c}\)
  • \(\bar{T} = \frac{T}{T_c}\)
  • \(\bar{V} = \frac{V}{V_c}\)
This form of the equation helps in visualizing and evaluating the behavior of real gases under various conditions by comparing them to their critical properties.

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Most popular questions from this chapter

Consider the vaporization curve for liquid mercury. The latent heat of vaporization, \(L\) (in \(\mathrm{J} / \mathrm{mol}\) ) varies slowly with pressure, but has significant variation with temperature and can be written \(L=(7724-0.9768 T) R\), where \(R\) is the gas constant and \(T\) is measured in kelvin. It is known that at atmospheric pressure ( \(P_{\mathrm{am}}=1.013 \times 10^{5} \mathrm{~Pa}\) ) mercury begins to vaporize at temperature \(T=\) \(630 \mathrm{~K}\). (a) Plot the vaporization curve for mercury between temperatures \(T=500 \mathrm{~K}\) and \(T=650 \mathrm{~K}\). (b) At what pressure does mercury begin to vaporize when \(T=530 \mathrm{~K}\) ? (Hint: The volume of the liquid mercury can be neglected relative to that of the vapor and the vapor can be treated as an ideal gas.)

One kilogram of superheated steam, at temperature \(t=350^{\circ} \mathrm{C}\), pressure \(P=100\) bar, and specific entropy \(s=5949 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\), is expanded reversibly and adiabatically to form wet steam at \(t=\) \(200^{\circ} \mathrm{C}\) and pressure \(P=15.55\) bar. The specific entropy of water vapor and liquid water on the coexistence curve at \(t=200{ }^{\circ} \mathrm{C}\) are \(\mathrm{s}_{\mathrm{g}}=6.428 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\) and \(s_{1}=2.331 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\), respectively. The specific enthalpy of water vapor (gas) and liquid water on the coexistence curve at \(t=200^{\circ} \mathrm{C}\) are \(h_{\mathrm{g}}=\) \(2791 \mathrm{~kJ} / \mathrm{kg}\) and \(h_{1}=852.4 \mathrm{~kJ} / \mathrm{kg}\). (a) What is the specific enthalpy of the wet steam at \(t=200^{\circ} \mathrm{C}\) ? (b) What fraction of the wet steam is liquid water?

Consider a monatomic fluid along its liquid-gas coexistence curve. Compute the rate of change of chemical potential along the coexistence curve \((\mathrm{d} \mu / \mathrm{d} T)\) owex, where \(\mu\) is the chemical potential and \(T\) is the temperature. Express your answer in terms of \(s_{1}, v_{1}\) and \(s_{\mathrm{g}}, v_{\mathrm{g}}\), which are the molar entropy and molar volume of the liquid and gas, respectively.

A liquid crystal is composed of molecules which are elongated (and often have flat segments). It behaves like a liquid because the locations of the center- of-mass of the molecules have no long-range order. It behaves like a crystal because the orientation of the molecules does have long-range order. The order parameter for a liquid crystal is given by the dyatic \(S=\eta(n n-1 / 3 I)\), where \(\boldsymbol{n}\) is a unit vector (called the director) which gives the average direction of alignment of the molecules. The free energy of the liquid crystal can be written. $$ \phi=\phi_{0}+\frac{1}{2} A S_{i j} S_{i j}-\frac{1}{3} B S_{i j} s_{j k} S_{k i}+\frac{1}{4} C S_{i j} S_{i j} S_{k l} S_{k l} $$ where \(A=A_{0}\left(T-T^{*}\right), A_{0}, B\) and \(C\) are constants, \(I\) is the unit tensor so \(\hat{x}_{i} \cdot I \cdot \hat{x}_{i}=\hat{x}_{i}\) \(\delta_{i i}, S_{i i}=\hat{x}_{i} \cdot S \cdot \hat{x}_{i, \text { and the summation is over repeated indices. The quantities are the unit vectors }}\) \(\hat{x}_{1}=\hat{x}, \hat{x}_{2}=\hat{y}\), and \(\hat{x}_{3}=\hat{z}\). (a) Perform the summations in the expression for \(\Phi\) and write \(\Phi\) in terms of \(\eta, A, B, C\). (b) Compute the critical temperature \(T_{\mathrm{c}}\) at which the transition from isotropic liquid to liquid crystal takes place, and compute the magnitude of the order parameter \(\eta\) at the critical temperature. (c) Compute the difference in entropy between the isotropic liquid \((\eta=0)\) and the liquid crystal at the critical temperature.

The equation of state of a gas is given by the Berthelot equation \(\left(P+a / T v^{2}\right)(v-b)=R T\). (a) Find values of the critical temperature \(T_{c}\), the critical molar volume \(v_{c}\), and the critical pressure \(P_{c}\), in terms of \(a, b\), and \(R\). (b) Does the Berthelot equation satisfy the law of corresponding states? (c) Find the critical exponents \(\beta, \delta\), and \(\gamma\) from the Berthelot equation.
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