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# Chemical Engineering Thermodynamics ## Chapter 3 ### Volumetric Properties of Pure Fluids #### 3.1 PVT Behavior of Pure Substances ##### 3.1.1 The Ideal Gas ###### The Ideal-Gas Model - Ideal gas assumptions: - Intermolecular forces are negligible - Molecules are point masses - The idea...

# Chemical Engineering Thermodynamics ## Chapter 3 ### Volumetric Properties of Pure Fluids #### 3.1 PVT Behavior of Pure Substances ##### 3.1.1 The Ideal Gas ###### The Ideal-Gas Model - Ideal gas assumptions: - Intermolecular forces are negligible - Molecules are point masses - The ideal-gas equation: $PV = RT$ or $P \hat{V} = RT$ - Where: - $P$ is the absolute pressure - $V$ is the volume - $\hat{V}$ is the molar volume - $T$ is the absolute temperature - $R$ is the gas constant - Limitations: - Valid only at low pressures and/or high temperatures ###### Example 3.1: A cylinder with a volume of $0.1 m^3$ contains $12 kg$ of ethane at $296.15 K$. Compare the pressure predicted by the ideal-gas equation of state with the experimental value of $68.04 \times 10^5 Pa$. **Solution:** $$ \begin{aligned} P & =\frac{n R T}{V}=\frac{m R T}{M V} \\ & =\frac{12 \mathrm{~kg} \times 8.314 \times 10^{-3} \frac{\mathrm{m}^{3} \mathrm{kPa}}{\mathrm{mol} \mathrm{K}} \times 296.15 \mathrm{~K}}{0.03007 \frac{\mathrm{kg}}{\mathrm{mol}} \times 0.1 \mathrm{~m}^{3}} \\ & =9.77 \times 10^{3} \mathrm{kPa}=97.7 \times 10^{5} \mathrm{~Pa} \end{aligned} $$ The ideal-gas equation of state predicts a pressure $43.6 \%$ higher than the experimental value. ##### 3.1.2 Virial Equation of State - An equation of state in the form of a series: $\frac{P \hat{V}}{R T}=1+\frac{B}{\hat{V}}+\frac{C}{\hat{V}^{2}}+\frac{D}{\hat{V}^{3}}+\cdots$ or $P \hat{V}=R T+B^{\prime} P+C^{\prime} P^{2}+D^{\prime} P^{3}+\cdots$ - Where: - $B, C, D,$ etc. are virial coefficients - $B^{\prime}, C^{\prime}, D^{\prime},$ etc. are virial coefficients - Virial coefficients are temperature-dependent. - The virial equation of state can be derived from statistical mechanics. - The two-term virial equation of state: $\frac{P \hat{V}}{R T}=1+\frac{B}{\hat{V}}$ or $Z=1+\frac{B}{\hat{V}}$ - Where: - $Z$ is the compressibility factor - $B$ can be estimated using the following equation: $B=\frac{R T_{c}}{P_{c}}\left(B_{0}+\omega B_{1}\right)$ $B_{0}=0.083-\frac{0.422}{T_{r}^{1.6}}$ $B_{1}=0.139-\frac{0.172}{T_{r}^{4.2}}$ Where: - $T_{r}=\frac{T}{T_{c}}$ is the reduced temperature - $\omega$ is the acentric factor ##### Example 3.2: For the conditions of Example 3.1, estimate the pressure using the virial equation. Experimental value of pressure is $68.04 \times 10^5 Pa$. **Solution:** For ethane: $T_{c}=305.4 \mathrm{~K}, P_{c}=48.72 \times 10^{5} \mathrm{~Pa}, \omega=0.099$ $T_{r}=\frac{T}{T_{c}}=\frac{296.15}{305.4}=0.9697$ $B_{0}=0.083-\frac{0.422}{T_{r}^{1.6}}=0.083-\frac{0.422}{(0.9697)^{1.6}}=-0.3526$ $B_{1}=0.139-\frac{0.172}{T_{r}^{4.2}}=0.139-\frac{0.172}{(0.9697)^{4.2}}=-0.0446$ $B=\frac{R T_{c}}{P_{c}}\left(B_{0}+\omega B_{1}\right)$ $=\frac{8.314 \times 10^{-3} \frac{\mathrm{m}^{3} \mathrm{kPa}}{\mathrm{mol} \mathrm{K}} \times 305.4 K}{48.72 \times 10^{3} \mathrm{kPa}}(-0.3526+0.099 \times(-0.0446))=-1.88 \times 10^{-4} \frac{\mathrm{m}^{3}}{\mathrm{~mol}}$ $\hat{V}=\frac{V}{n}=\frac{M V}{m}=\frac{0.03007 \frac{\mathrm{kg}}{\mathrm{mol}} \times 0.1 \mathrm{~m}^{3}}{12 \mathrm{~kg}}=2.506 \times 10^{-4} \frac{\mathrm{m}^{3}}{\mathrm{~mol}}$ $Z=1+\frac{B}{\hat{V}}=1+\frac{-1.88 \times 10^{-4} \frac{\mathrm{m}^{3}}{\mathrm{~mol}}}{2.506 \times 10^{-4} \frac{\mathrm{m}^{3}}{\mathrm{~mol}}}=0.249$ $P=\frac{Z R T}{\hat{V}}=\frac{0.249 \times 8.314 \times 10^{-3} \frac{\mathrm{m}^{3} \mathrm{kPa}}{\mathrm{mol} \mathrm{K}} \times 296.15 \mathrm{~K}}{2.506 \times 10^{-4} \frac{\mathrm{m}^{3}}{\mathrm{~mol}}}=24.5 \times 10^{3} \mathrm{kPa}=24.5 \times 10^{5} \mathrm{~Pa}$ The predicted value deviates significantly from the experimental value.