Ecuación de Virial para Fluidos Puros

Volumetric Properties of Pure Fluids

Virial Equation of State

• The virial equation of state expresses pressure (P) as a function of molar volume (V) and temperature (T).
• The equation is: $P=\frac{RT}{V}+\frac{B(T)}{V^2}+\frac{C(T)}{V^3}+\frac{D(T)}{V^4}+...$
• $B$, $C$, $D$ are temperature-dependent virial coefficients
• An alternative form of the virial equation: $P=\frac{RT}{V}+B'\frac{P}{RT}+C'(\frac{P}{RT})^2+D'(\frac{P}{RT})^3+...$
• For practical engineering calculations, the virial equation is often truncated to two or three terms.

Truncated Virial Equation

• The truncated virial equation uses the compressibility factor Z: $Z=1+\frac{B}{V}$
• B is the second virial coefficient.
• B can be estimated with: $B=\frac{RT_c}{P_c}(B^0+\omega B^1)$
• $T_c$ is critical temperature, $P_c$ is critical pressure, and $\omega$ is the acentric factor
• $B^0$ and $B^1$ are dimensionless parameters dependent on reduced temperature $T_r = T/T_c$
• Equations to calculate $B^0$ and $B^1$ are:
  • $B^0=0.083-\frac{0.422}{T_r^{1.6}}$
  • $B^1=0.139-\frac{0.172}{T_r^{4.2}}$

Example 1

• Problem: Determine the molar volume of $n$-butane at $500 K$ and $25 bar$ using the virial equation. Compare with the experimental value of $0.15325 m^3/mol$ given $T_c = 425.2 K$, $P_c = 37.96 bar$, $\omega = 0.200$.
• Solution:
  • Calculate reduced temperature: $T_r=\frac{T}{T_c}=\frac{500}{425.2}=1.176$
  • Calculate $B^0$: $B^0=0.083-\frac{0.422}{T_r^{1.6}}=0.083-\frac{0.422}{(1.176)^{1.6}}=-0.243$
  • Calculate $B^1$: $B^1=0.139-\frac{0.172}{T_r^{4.2}}=0.139-\frac{0.172}{(1.176)^{4.2}}=0.036$
  • Calculate B: $B=\frac{RT_c}{P_c}(B^0+\omega B^1)=\frac{83.14425.2}{37.96}(-0.243+0.20.036)=-224.2 cm^3/mol$
  • Calculate Z: $Z=1+\frac{BP}{RT}=1+\frac{-224.225}{83.14500}=0.865$
  • Calculate V: $V=\frac{ZRT}{P}=\frac{0.86583.14500}{25}=1438 cm^3/mol = 0.001438 m^3/mol$
• The virial equation provides and estimate with an error of 6.2%.

The Ideal Gas Model

Equation of State

• The Ideal Gas Law is: $PV=RT$

Internal Energy

• Internal energy (U) depends only on temperature.
• $U=U(T)$
• $dU = C_v dT$
• $\Delta U = \int_{T_1}^{T_2} C_v dT = C_v (T_2 - T_1)$

Enthalpy

• Enthalpy (H) depends only on temperature.
• $H = H(T)$
• $dH = C_p dT$
• $\Delta H = \int_{T_1}^{T_2} C_p dT = C_p (T_2 - T_1)$

Entropy

• Entropy (S) depends on temperature and pressure.
• $dS = C_p \frac{dT}{T} - R\frac{dP}{P}$
• $\Delta S = \int_{T_1}^{T_2} C_p \frac{dT}{T} - Rln\frac{P_2}{P_1}$
• If $C_p$ is constant: $\Delta S = C_p ln\frac{T_2}{T_1} - Rln\frac{P_2}{P_1}$

Example 2

• Problem: Calculate the change in entropy of an ideal gas undergoing a reversible process from $25^\circ C$ and $1 bar$ to $75^\circ C$ and $4 bar$, given $C_p = 29.1 J/mol \cdot K$.
• Solution:
  • $\Delta S = C_p ln\frac{T_2}{T_1} - Rln\frac{P_2}{P_1} = 29.1 ln\frac{348}{298} - 8.314ln\frac{4}{1} = -2.32 J/mol \cdot K$