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## 物理代写|热力学代写thermodynamics代考|VAN DER WAAL’S EQUAIIUN OF SIAIE

van der Waal in 1873 developed an equation of state for real gases in an attempt to correct the equation of state for an ideal gas. At higher pressures, the volume occupied by the gas molecules is no longer negligible. Thus, the molar volume in the ideal gas law is replaced by $(\tilde{v}-b)$. To account for intermolecular attraction, the pressure is written as $p+\frac{a}{\tilde{v}^2}$. The correction term $\frac{a}{\tilde{v}^2}$ for the attractive intermolecular force is based on the fact that it depends on the number of molecules (i.e., the molar density $\rho=\frac{1}{\tilde{v}}$ ) and also the intermolecular distance (which also depends on density $\rho=\frac{1}{\tilde{v}}$ ). Thus, the pressure correction term varies as $\frac{1}{\tilde{v}^2}$ The van der Waal equation of state is thus written as
$$p+\frac{a}{\tilde{v}^2}=\frac{R_0 T}{\tilde{v}-b}$$

with $a$ and $b$ to be determined experimentally. The constants $a$ and $b$, however, actually depend on temperature and the values for $a$ and $b$ have to be determined for the particular regions of pressure and temperature of interest.

But since the critical constant temperature line (isotherm) on a $p-\tilde{v}$ diagram has zero slope and curvature at the critical point, we have the conditions
$$\left.\left.\frac{\partial p}{\partial \tilde{v}}\right){T_c}=0, \frac{\partial^2 p}{\partial \tilde{v}^2}\right){T_c}=0$$
where $T_C$ denotes the critical temperature, from which the van der Waal coefficients $a$ and $b$ can be evaluated. Using Eqs. $2.19$ and 2.18, we obtain
\begin{aligned} &\left.\frac{\partial p}{\partial \tilde{v}}\right){T_e}=-\frac{R_0 T_c}{\left(\tilde{v}_C-b\right)^2}+\frac{2 a}{\tilde{v}_C{ }^3}=0 \ &\left.\frac{\partial^2 p}{\partial \tilde{v}^2}\right){T_c}=\frac{2 R_0 T_c}{\left(\tilde{v}_C-b\right)^3}-\frac{6 a}{\tilde{v}_C{ }^4}=0 \end{aligned}
where the subscript ” $c$ ” refers to the critical state. Solving for $a$ and $b$ yields
$$a=\frac{27}{64} \frac{R_0{ }^2 T_C{ }^2}{p_C}, b=\frac{R_0 T_C}{8 p_C}, \frac{p_C \tilde{v}_c}{R_0 T_C}=\frac{3}{8}$$
Thus from the critical point data for a given gas, the van der Waal coefficients $a$ and $b$ can be obtained.

Note that Eq. $2.22$ gives $\frac{p_C \tilde{v}_C}{R_0 T_C}=\frac{3}{8}=0.375$, whereas experimentally it is found that $\frac{p_C \tilde{v}_C}{R_0 T_C}$ has values in the range of $0.2-0.3$. It would be more accurate to fit the experimental data in the region of $p$ and $T$ of interest to determine the coefficients $a$ and $b$ rather than using Eq. 2.22. The coefficients $a$ and $b$ for a few gases are given in Table $2.2$.

The van der Waal equation of state is of historical interest since it represents the first attempt to correct the equation of state for an ideal gas taking into account the real gas effects.

## 物理代写|热力学代写thermodynamics代考|Berthelot and Dieterici Equations of State

There are other two-parameter equations of state where the two constants can similarly be obtained in terms of the critical pressure $p_C$ and temperature $T_C$. Typical examples are the Berthelot and Dieterici equations, that is,
Berthelot : $p=\frac{R_0 T}{\tilde{v}-b}-\frac{a}{T \tilde{v}^2}$
Dieterici : $p=\frac{R_0 T}{\tilde{v}-b} \exp \left(-\frac{a}{R_0 T \tilde{v}}\right)$

The constants $a$ and $b$ in the above equations can be obtained using the critical point conditions given by Eq. $2.19$, to be
$$a=\frac{27 R_0{ }^2 T_C^2}{64 p_C}, b=\frac{R_0 T_C}{8 p_C}$$
for the Berthelot equation and
$$a=\frac{4 R_0{ }^2 T_C{ }^2}{p_C e^2}, b=\frac{R_0 T_C}{p_C e^2}$$
where $e=2.718$ for the Dieterici equation.
The Berthelot equation corrects for the attractive term in the van der Waal equation when the temperature is high and the kinetic energies of the molecules are large compared to the attractive potential energy. The correction term $\frac{a}{T^2{ }^2}$ would thus diminish with increase of temperature. The Dieterici equation was developed to give better agreement with the quantity $\frac{p_C \tilde{v}_C}{R_0 T_C}$ in Eq. $2.22$ which is in considerable error in the van der Waal equation when compared with experiments.

# 热力学代写

## 物理代写|热力学代写thermodynamics代考|VAN DER WAAL’S EQUAIIUN OF SIAIE

$$p+\frac{a}{\tilde{v}^2}=\frac{R_0 T}{\tilde{v}-b}$$

$$\left.\left.\frac{\partial p}{\partial \tilde{v}}\right) T_c=0, \frac{\partial^2 p}{\partial \tilde{v}^2}\right) T_c=0$$

$$\left.\left.\frac{\partial p}{\partial \tilde{v}}\right) T_e=-\frac{R_0 T_c}{\left(\tilde{v}_C-b\right)^2}+\frac{2 a}{\tilde{v}_C{ }^3}=0 \quad \frac{\partial^2 p}{\partial \tilde{v}^2}\right) T_c=\frac{2 R_0 T_c}{\left(\tilde{v}_C-b\right)^3}-\frac{6 a}{\tilde{v}_C{ }^4}=0$$

$$a=\frac{27}{64} \frac{R_0^2 T_C^2}{p_C}, b=\frac{R_0 T_C}{8 p_C}, \frac{p_C \tilde{v}_c}{R_0 T_C}=\frac{3}{8}$$

## 物理代写|热力学代写thermodynamics代考|Berthelot and Dieterici Equations of State

$$a=\frac{27 R_0{ }^2 T_C^2}{64 p_C}, b=\frac{R_0 T_C}{8 p_C}$$

$$a=\frac{4 R_0{ }^2 T_C{ }^2}{p_C e^2}, b=\frac{R_0 T_C}{p_C e^2}$$

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