Chemical Engineering Thermodynamics Tutoring

Chemical engineering thermodynamics is not the same course mechanical engineers take. Engines and cycles give way to mixtures, phase equilibrium, and chemical reactions that do not run to completion. The ideas also get more abstract: chemical potential, fugacity, and activity coefficients. Many students can follow each derivation in lecture and still cannot say which model fits a new problem. STEM Tutor 360 builds the course around that decision. Is the vapor ideal? Is the liquid? Which property do you need, and which equation supplies it?

Sessions cover PVT behavior and equations of state, departure functions, partial molar properties, fugacity and activity, vapor-liquid equilibrium, and reaction equilibrium. We follow your textbook, such as Smith, Van Ness, Abbott, and Swihart, Koretsky, or Elliott and Lira, and we use your professor's problem sets and exams to set the difficulty. If you want the first-law-and-cycles course that mechanical engineers take, the thermodynamics tutoring page is the better fit.

A typical vapor-liquid equilibrium problem, worked

A liquid is 40 mol% benzene and 60 mol% toluene. Assume ideal behavior and find the bubble-point temperature at 760 mmHg, and the composition of the first vapor. Use the Antoine equation, $\log_{10} P^{\text{sat}} = A - B/(T + C)$, with $P^{\text{sat}}$ in mmHg and $T$ in °C. Benzene: $A = 6.90565$, $B = 1211.033$, $C = 220.79$. Toluene: $A = 6.95464$, $B = 1344.8$, $C = 219.482$.

At the bubble point the partial pressures add up to the total: $x_B P_B^{\text{sat}} + x_T P_T^{\text{sat}} = 760$. Try 95°C. Then $P_B^{\text{sat}} = 1176.8\ \text{mmHg}$ and $P_T^{\text{sat}} = 476.9\ \text{mmHg}$, so the sum is $0.40 \times 1176.8 + 0.60 \times 476.9 = 756.9\ \text{mmHg}$, a little low. At 96°C the sum is 779.2 mmHg, a little high. Interpolating puts the bubble point at 95.14°C, about 95.1°C.

At that temperature the vapor pressures are about 1181 and 479 mmHg. The vapor composition is $y_B = \dfrac{0.40 \times 1181}{760} = 0.622$, so $y_T = 0.378$. The first vapor is richer in benzene because benzene is the more volatile component. The ratio of vapor pressures, about 2.5, is the relative volatility your separations course will use for distillation.

Where students usually get stuck

• Using Raoult's law where it does not apply: polar or hydrogen-bonding mixtures, high pressure, or near an azeotrope. Modified Raoult's law adds an activity coefficient.
• Mixing up fugacity with the fugacity coefficient, and activity with the activity coefficient.
• Mixing up bubble-point and dew-point setups. A bubble point starts from a known liquid composition, and a dew point from a known vapor composition.
• Antoine constants in the wrong units: mmHg or kPa, °C or K, $\log_{10}$ or natural log.
• Departure functions: leaving out the ideal-gas part, or evaluating the departure at only one of the two end states.
• Getting the sign wrong between the standard Gibbs energy change and $\ln K$.