Chemical Engineering - MCQ Practice Questions

In adiabatic reactors, Q = 0, so ΔT depends on heat of reaction (ΔH_r) and heat capacity of reactants.

Space-time τ = V/v₀ = $\frac{100}{10}$ = 10 min.

Conversion X = (C_A0 - C_A)/C_A0 = (2 - 0.5)/2 = 0.75 or 75%.

Temperature is an intensive property as it does not depend on the quantity of substance. Volume, mass, and internal energy are extensive properties that depend on the amount of material.

In the convention ΔU = Q - W, W is work done by the system. This is the IUPAC recommended convention used in most thermodynamics calculations.

Using ΔU = Q - W, where Q = +500 J (absorbed) and W = +200 J (work done by system). Therefore, ΔU = 500 - 200 = 300 J.

For ideal gas, Z = PV/nRT = 1 always. For real gases, Z ≠ 1 due to intermolecular forces and molecular volume.

Q = m × Cₚ × ΔT = 1 × 4.18 × (75-25) = 1 × 4.18 × 50 = 209 kJ. At constant pressure, Q = ΔH.

On a T-S diagram, horizontal line means constant temperature (T = const), which is an isothermal process. A vertical line would represent isentropic process (S = const).

Partial pressure = Total pressure × Mole fraction. P_O₂ = 1 atm × 0.60 = 0.6 atm (Dalton's Law)

For an isothermal process of ideal gas, entropy depends only on volume ratio: ΔS = nR ln(V₂/V₁), regardless of reversibility

Internal energy (U), Enthalpy (H), Gibbs energy (G) are state functions. Work and heat are path functions

At constant volume, W = P×ΔV = 0. Therefore, ΔU = Q - 0 = Q

At equilibrium, ΔG = 0, which means ΔG° = -RT ln(K), and when ΔG° = 0, K = 1 (equilibrium)

η_Carnot = 1 - (T_C/T_H) = 1 - ($\frac{300}{600}$) = 0.5 = 50%

This is the van der Waals equation where 'a' accounts for intermolecular forces and 'b' for molecular volume

ΔU = Q - W = 100 - 60 = 40 J (First Law of Thermodynamics)

A reversible adiabatic process has constant entropy (dS = 0), which defines an isentropic process. This is a fundamental concept in thermodynamics.

At constant pressure, ΔH = qₚ (heat at constant pressure). This is the definition of enthalpy and its primary application.

For constant volume: W = 0, so Q = ΔU from first law (ΔU = Q - W). This is isochoric process where all heat goes to internal energy change.