# Residual properties

Many books and articles give analytical expressions for the residual properties of the untranslated cubic equation of state. Such residual properties can be combined to the ideal gas properties of the same system under the same temperature and volume to derive the real fluid properties.

# 1 The translated cubic equation of state

The analytical expressions of the residual Tv properties are derived for the most general cubic equation of state, i.e. with universal parameters $r_1$ and $r_2$, an $\alpha$-function and a temperature dependent volume translation parameter.

The untranslated equation of state writes:

$$$P\left(T,v\right) = \frac{RT}{v-b_0} - \frac{a\left(T\right)}{\left(v-r_1b_0\right)\left(v-r_2b_0\right)} \tag{1.1}$$$

One can include a volume translation parameter $c\left(T\right)$ (abreviated to $c$ for simplicity) by substituing $v$ by $v+c$. The translated equation of state thus writes:

$$$P\left(T,v\right) = \frac{RT}{v+c-b_0} - \frac{a\left(T\right)}{\left(v+c-r_1b_0\right)\left(v+c-r_2b_0\right)} \tag{1.2}$$$

Because the covolume of the translated equation of state is no longer $b$ but $b = b_0-c$, the translated equation of state is recasted in the following form:

$$$P\left(T,v\right) = \frac{RT}{v-b} - \frac{a\left(T\right)}{\left[v-r_1b + c\left(1-r_1\right)\right]\left[v-r_2b + c\left(1-r_2\right)\right]} \tag{1.3}$$$

Because this writting is quite ugly, one can define the following variables in order to simplify the previous equation:

$$$\left\{ \begin{array}{l} s_1b = r_1b - c\left(1-r_1\right) \\ s_2b = r_2b - c\left(1-r_2\right) \end{array} \right. \Rightarrow \left\{ \begin{array}{l} s_1\left(T\right) = r_1 - \frac{c\left(T\right)}{b_0 - c\left(T\right)} \times \left(1-r_1\right) \\ s_2\left(T\right) = r_2 - \frac{c\left(T\right)}{b_0 - c\left(T\right)} \times \left(1-r_2\right) \end{array} \right. \tag{1.4}$$$

The translated cubic equation of state writes:

$$$P\left(T,v\right) = \frac{RT}{v-b} - \frac{a\left(T\right)}{\left(v-s_1b\right)\left(v-s_2b\right)} \tag{1.5}$$$

# 2 Residual molar Helmhotz energy $a^{res}$

This property is defined as:

$$$a^{res}\left(T,v\right) = - \int_{+\infty}^v \left[P\left(T,v\right) - \frac{RT}{v}\right] dv \tag{2.1}$$$

Injecting equation (1.5) into (2.1), one gets:

$$$a^{res}\left(T,v\right) = - \int_{+\infty}^v \left[\frac{RT}{v-b} - \frac{a}{\left(v-s_1b\right)\left(v-s_2b\right)} - \frac{RT}{v}\right] dv \tag{2.2}$$$

To integrate equation (2.2), one must perform the partial fraction decomposition of the second term under the integral:

$$$\frac{a}{\left(v-s_1b\right)\left(v-s_2b\right)} = \frac{a}{b\left(s_1 - s_2\right)} \left(\frac{1}{v-s_1b} - \frac{1}{v-s_2b}\right) \tag{2.3}$$$

Thus equation (2.2) rewrites:

$$$a^{res}\left(T,v\right) = - \int_{+\infty}^v \left[\frac{RT}{v-b} - \frac{a}{b\left(s_1 - s_2\right)}\left(\frac{1}{v-s_1b} - \frac{1}{v-s_2b}\right) - \frac{RT}{v}\right] dv \tag{2.4}$$$

After integration, one finds:

$$$a^{res}\left(T,v\right) = RT\ln\left(\frac{v}{v-b}\right) + \frac{a}{b\left(s_1-s_2\right)} \ln\left(\frac{v-s_1b}{v-s_2b}\right) \tag{2.5}$$$

One can notice that the quantity $b\left(s_1 - s_2\right)$ is temperature independant. Indeed:

$$$b\left(s_1 - s_2\right) = bs_1 - bs_2 = b\left(r_1 - r_2\right) + c\left(r_1 - r_2\right) = \left(b+c\right)\left(r_1 - r_2\right) = b_0 \left(r_1 - r_2\right) \tag{2.6}$$$

Because neither $b_0$, $r_1$ nor $r_2$ are temperature dependent, $b\left(s_1 - s_2\right)$ is temperature independent. Therefore, equation (2.5) becomes:

$$$a^{res}\left(T,v\right) = RT\ln\left(\frac{v}{v-b}\right) + \frac{a}{b_0\left(r_1-r_2\right)} \ln\left(\frac{v-s_1b}{v-s_2b}\right) \tag{2.7}$$$

For an untranslated equation of state, this equation becomes:

$$$a^{res}\left(T,v\right) = RT\ln\left(\frac{v}{v-b_0}\right) + \frac{a}{b_0\left(r_1-r_2\right)} \ln\left(\frac{v-r_1b_0}{v-r_2b_0}\right) \tag{2.8}$$$

# 3 Residual molar entropy $s^{res}$

The residual molar entropy is obtained by applying the following definition:

$$$s^{res}\left(T,v\right) = -\left(\frac{\partial a^{res}}{\partial T}\right)_v \tag{3.1}$$$

The derivative of equation (2.5) writes (remind that $b\left(s_1 - s_2\right)$ is temperature independant):

$$$\left(\frac{\partial a^{res}}{\partial T}\right)_v = R \ln\left(\frac{v}{v-b}\right) + \frac{RT}{v-b} \times \frac{db}{dT} + \frac{1}{b\left(s_1 - s_2\right)} \left[\frac{da}{dT}\ln\left(\frac{v-s_1b}{v-s_2b}\right) + a\left(-\frac{1}{v-s_1b} \frac{ds_1b}{dT} + \frac{1}{v-s_2b} \frac{ds_2b}{dT}\right)\right] \tag{3.2}$$$

Now we need some supplementary derivatives for $ds_1b/dT$ and $ds_2b/dT$. From equation (1.4), one can write:

$$$\begin{array}[ll] & \displaystyle \frac{ds_1b}{dT} & \displaystyle = r_1 \frac{db}{dT} - \left(1-r_1\right) \frac{dc}{dT} \\ & \displaystyle = -r_1 \frac{dc}{dT} - \left(1-r_1\right) \frac{dc}{dt} \\ & \displaystyle = -\frac{dc}{dT} \end{array} \tag{3.3}$$$

Obviously, the same result will be achieved for $ds_2b/dT$. Moreover, one can immediately derive that $db/dT = -dc/dT$

Equation (3.2) now writes:

$$$\left(\frac{\partial a^{res}}{\partial T}\right)_v = R \ln\left(\frac{v}{v-b}\right)-\frac{RT}{v-b} \times \frac{dc}{dT} + \frac{1}{b\left(s_1 - s_2\right)} \left[\frac{da}{dT}\ln\left(\frac{v-s_1b}{v-s_2b}\right) + a\times \frac{dc}{dT}\left(\frac{1}{v-s_1b} - \frac{1}{v-s_2b}\right)\right] \tag{3.4}$$$

By regrouping the terms involving $dc/dT$, one can similarely write:

$$$\left(\frac{\partial a^{res}}{\partial T}\right)_v = R \ln\left(\frac{v}{v-b}\right) + \frac{1}{bs_1 - bs_2} \times \frac{da}{dT}\ln\left(\frac{v-s_1b}{v-s_2b}\right) -\frac{dc}{dT} \times \left[\frac{RT}{v-b} - \frac{a}{b\left(s_1 - s_2\right)}\left(\frac{1}{v-s_1b} - \frac{1}{v-s_2b}\right)\right] \tag{3.5}$$$

The big term multiplying $dc/dT$ is nothing but the analytical expression of the cubic equation of state in its partial fraction form (compare to equation (2.3) to be convinced). So, one can write in a much simpler manner:

$$$\left(\frac{\partial a^{res}}{\partial T}\right)_v = R \ln\left(\frac{v}{v-b}\right) + \frac{1}{b\left(s_1 - s_2\right)} \times \frac{da}{dT}\ln\left(\frac{v-s_1b}{v-s_2b}\right) -\frac{dc}{dT} \times P\left(T,v\right) \tag{3.6}$$$

Therefore, the analytical expression for the residual molar entropy is:

$$$s^{res}\left(T,v\right) = -R \ln\left(\frac{v}{v-b}\right) - \frac{1}{b\left(s_1 - s_2\right)} \times \frac{da}{dT}\times\ln\left(\frac{v-s_1b}{v-s_2b}\right) +\frac{dc}{dT} \times P\left(T,v\right) \tag{3.7}$$$

For an untranslated equation of state, this equation becomes:

$$$s^{res}\left(T,v\right) = -R \ln\left(\frac{v}{v-b_0}\right) - \frac{1}{b_0\left(r_1 - r_2\right)} \times \frac{da}{dT}\times\ln\left(\frac{v-r_1b_0}{v-r_2b_0}\right) \tag{3.8}$$$

# 4 Residual interal energy $u^{res}$

This property can be calculated by the following formula:

$$$u^{res}\left(T,v\right) = a^{res}\left(T,v\right) + T s^{res}\left(T,v\right) \tag{4.1}$$$

Therefore, combining equations (2.5) and (3.7), one can write:

$$$u^{res}\left(T,v\right) = \frac{1}{b\left(s_1-s_2\right)} \left(a-T\frac{da}{dT}\right) \times \ln\left(\frac{v-s_1b}{v-s_2b}\right) + T\frac{dc}{dT} \times P\left(T,v\right) \tag{4.2}$$$

For an untranslated equation of state, this equation becomes:

$$$u^{res}\left(T,v\right) = \frac{1}{b_0\left(r_1-r_2\right)} \left(a-T\frac{da}{dT}\right) \times \ln\left(\frac{v-r_1b_0}{v-r_2b_0}\right) \tag{4.3}$$$

# 5 Residual molar enthalpy $h^{res}$

This property can be calculated by the following formula:

$$$h^{res}\left(T,v\right) = u^{res}\left(T,v\right) + P\left(T,v\right)v - RT \tag{5.1}$$$

Therefore, from equation (4.2) one can write:

$$$h^{res}\left(T,v\right) = \frac{1}{b\left(s_1-s_2\right)}\left(a-T\frac{da}{dT}\right)\times \ln\left(\frac{v-s_1b}{v-s_2b}\right) + P\left(T,v\right)\times \left(v+T \times \frac{dc}{dT}\right) - RT \tag{5.2}$$$

For an untranslated equation of state, this equation becomes:

$$$h^{res}\left(T,v\right) = \frac{1}{b_0\left(r_1-r_2\right)}\left(a-T\frac{da}{dT}\right)\times \ln\left(\frac{v-r_1b_0}{v-r_2b_0}\right) + P\left(T,v\right)v - RT \tag{5.3}$$$

# 6 Compressibility coefficient $\kappa_T$

By definition:

$$$\kappa_T = -\frac{1}{v}\left(\frac{\partial v}{\partial P}\right)_T$$$

The partial derivative analytical expression is:

$$$\left(\frac{\partial P}{\partial v}\right)_T = -\frac{RT}{\left(v-b\right)^2} + \frac{a \left(2v-s_1b-s_2b\right)}{\left[\left(v-s_1b\right)\left(v-s_2b\right)\right]^2} \tag{6.1}$$$

Therefore, coefficient $\kappa_T v$ writes:

$$$\kappa_T v = - \frac{1}{-\frac{RT}{\left(v-b\right)^2} + \frac{a \left(2v-s_1b-s_2b\right)}{\left[\left(v-s_1b\right)\left(v-s_2b\right)\right]^2}} \tag{6.2}$$$

# 7 Isochoric dilatation coefficient $\beta$

By definition:

$$$\beta = \frac{1}{P}\left(\frac{\partial P}{\partial T}\right)_v \tag{7.1}$$$

The partial derivative analytical expression is:

$$$\begin{array}[l] & \displaystyle \left(\frac{\partial P}{\partial T}\right)_v & = \displaystyle \frac{R\left(v-b\right)+RTb'}{\left(v-b\right)^2} - \frac{a'\left(v-s_1b\right)\left(v-s_2b\right)-a\left[-\left(s_1b\right)'\left(v-s_2b\right)-\left(s_2b\right)'\left(v-s_1b\right)\right]}{\left[\left(v-s_1b\right)\left(v-s_2b\right)\right]^2} \\ & = \displaystyle \frac{R\left(v-b\right)-RTc'}{\left(v-b\right)^2} - \frac{a'\left(v-s_1b\right)\left(v-s_2b\right)-a\left[c'\left(v-s_2b\right)+c'\left(v-s_1b\right)\right]}{\left[\left(v-s_1b\right)\left(v-s_2b\right)\right]^2} \\ & = \displaystyle \frac{R}{v-b}-\frac{RTc'}{\left(v-b\right)^2} - \frac{a'}{\left(v-s_1b\right)\left(v-s_2b\right)} + \frac{ac'\left[2v-s_1b-s_2b\right]}{\left[\left(v-s_1b\right)\left(v-s_2b\right)\right]^2} \\ & = \displaystyle \frac{R}{v-b} - \frac{a'}{\left(v-s_1b\right)\left(v-s_2b\right)} -c' \left[\frac{RT}{\left(v-b\right)^2}-\frac{a\left[2v-s_1b-s_2b\right]}{\left[\left(v-s_1b\right)\left(v-s_2b\right)\right]^2}\right] \end{array} \tag{7.2}$$$

Identification with equation (6.1) leads to:

$$$\beta P = \left(\frac{\partial P}{\partial T}\right)_v = \frac{R}{v-b} - \frac{a'}{\left(v-s_1b\right)\left(v-s_2b\right)} +c' \times \left(\frac{\partial P}{\partial v}\right)_T \tag{7.3}$$$

Although the calculation is complete, it is worth noting that equation (7.3) can rewrite:

$$$T\left[\left(\frac{\partial P}{\partial T}\right)_v - c' \times \left(\frac{\partial P}{\partial v}\right)_T\right] = \frac{RT}{v-b}-\frac{Ta'}{\left(v-s_1b\right)\left(v-s_2b\right)} \tag{7.4}$$$

# 8 Residual molar isochoric heat capacity $c_v^{res}$

By definition:

$$$c_v^{res} = \left(\frac{\partial u^{res}}{\partial T}\right)_v \tag{8.1}$$$

The calculation steps are the following:

$$$\begin{array}[ll] & \displaystyle c_v^{res} & = \displaystyle \frac{1}{b\left(s_1-s_2\right)}\left[\left(a'-a'-Ta''\right)\ln\left(\frac{v-s_1b}{v-s_2b}\right) + \left(a-T\frac{da}{dT}\right)\left[-\frac{(s_1b)'}{v-s_1b}+\frac{(s_2b)'}{v-s_2b}\right]\right] + Tc''P+Tc' \left(\frac{\partial P}{\partial T}\right)_v + c'P \\ & = \displaystyle -\frac{Ta''}{b\left(s_1-s_2\right)}\ln\left(\frac{v-s_1b}{v-s_2b}\right) + \frac{\left(a-Ta'\right)c'}{b\left(s_1-s_2\right)}\left(\frac{1}{v-s_1b} - \frac{1}{v-s_2b}\right)+Tc''P+Tc' \left(\frac{\partial P}{\partial T}\right)_v + c'P \\ & = \displaystyle -\frac{Ta''}{b\left(s_1-s_2\right)}\ln\left(\frac{v-s_1b}{v-s_2b}\right) + c'\left[\frac{a-Ta'}{\left(v-s_1b\right) \left(v-s_2b\right)} + T \left(\frac{\partial P}{\partial T}\right)_v + P\right] + Tc''P \end{array} \tag{8.2}$$$

Injecting equation (7.4), in equation (8.2), one finds:

$$$c_v^{res}\left(T,v\right) = -\frac{Ta''}{b\left(s_1-s_2\right)}\ln\left(\frac{v-s_1b}{v-s_2b}\right) + c'\left[T\left[\left(\frac{\partial P}{\partial T}\right)_v - c'\times \left(\frac{\partial P}{\partial v}\right)_T\right] + T \left(\frac{\partial P}{\partial T}\right)_v\right] + Tc''P \tag{8.3}$$$

Injecting equations (6.2) and (7.3), in equation (8.3), one finally finds:

$$$c_v^{res}\left(T,v\right) = -\frac{Ta''}{b\left(s_1-s_2\right)}\ln\left(\frac{v-s_1b}{v-s_2b}\right) + c'T \left(2 \beta P + \frac{c'}{\kappa_T v}\right) + Tc''P \tag{8.4}$$$