Free PE Chemical practice questions.

Answer · B

Arrhenius: ln(k2/k1) = (Ea/R)(1/T1 − 1/T2). ln(10) = (Ea/8.314)(1/300 − 1/350) = (Ea/8.314)(0.003333 − 0.002857) = (Ea/8.314)(4.762×10⁻⁴). Ea = 2.303 × 8.314 / (4.762×10⁻⁴) = 40,200 J/mol = 40.2 kJ/mol.

Answer · C

First compute the hydraulic power transferred to the fluid: P_hyd = rho g Q H. Convert the flow rate and substitute: Q = 20 L/s = 0.020 m^3/s, so P_hyd = 1000 x 9.81 x 0.020 x 20 = 3924 W = 3.92 kW. Account for pump efficiency by dividing by eta: P_motor = P_hyd/eta = 3.92/0.75 = 5.23 kW. The hydraulic-power value alone ignores pump losses. A larger value usually comes from applying the efficiency incorrectly or using an incorrect efficiency value.

Answer · C

The ε–NTU method for counter-flow: ε = [1 − exp(−NTU(1−C*))] / [1 − C*·exp(−NTU(1−C*))]. Here NTU = 2.0 and C* = C_min/C_max = 0.5. Compute the argument: NTU(1−C*) = 2.0×0.5 = 1.0, so exp(−1.0) = 0.3679. Numerator: 1 − 0.3679 = 0.6321. Denominator: 1 − 0.5×0.3679 = 1 − 0.1840 = 0.8160. Therefore ε = 0.6321/0.8160 = 0.775. Choice A (0.632) is the effectiveness for a single-stream (C* = 0) heat exchanger at NTU = 1.0. This method is preferred over LMTD when outlet temperatures are unknown.

Answer · B

The Rachford-Rice equation Σ zᵢ(Kᵢ−1)/[1+ψ(Kᵢ−1)] = 0 is solved for vapor fraction ψ. Substituting K_A = 2.0 (K−1 = +1) and K_B = 0.5 (K−1 = −0.5): 0.40(1)/(1+ψ) + 0.60(−0.5)/(1−0.5ψ) = 0. Cross-multiplying: 0.40(1−0.5ψ) = 0.30(1+ψ). Solving: 0.40 − 0.20ψ = 0.30 + 0.30ψ → ψ = 0.10/0.50 = 0.20. Check: ψ = 0.20 is between 0 and 1 (two-phase), which is physically valid. Choice C (0.40) equals z_A, a common incorrect shortcut that ignores the equilibrium constraints.

Answer · C

y_out = 0.01 × (1−0.90) = 0.001. At minimum L/G, exiting liquid is in equilibrium with entering gas: x*_out = y_in/H = 0.01/300 = 3.33×10⁻⁵. (L/G)_min = (y_in − y_out)/(x*_out − x_in) = (0.01−0.001)/(3.33×10⁻⁵ − 0) = 0.009/3.33×10⁻⁵ = 270.

Answer · C

At steady state, all time-derivative terms vanish (s → 0 in Laplace domain), so G(0) = 3/(2·0+1) = 3 is the static (DC) gain. The new steady-state output change is: Δoutput = gain × Δinput = 3 × 5 = 15. Choice A (5) is wrong because it ignores the process gain entirely (assumes gain = 1). Choice D (20) results from incorrectly using the time constant (2) as a multiplier instead of setting s = 0. The transfer function G(s) = K/(τs+1) always has static gain K = G(0).

Answer · C

API 520 Part I defines maximum accumulated pressure during relieving for a single pressure relief valve (PRV): Non-fire case = 110% of MAWP; Fire case = 121% of MAWP; Multiple PRVs (non-fire) = 116% of MAWP. The fire case allowance is higher (21%) because fire is an abnormal, infrequent event with high heat input. The accumulation is the pressure rise above MAWP while the PRV is relieving. PRV set pressure is typically at or below MAWP. Exceeding 121% MAWP during a fire case would indicate undersized relief or blocked outlet.

Answer · B

The van der Waals equation P = nRT/(V−nb) − an²/V² corrects ideal gas for molecular volume (b) and intermolecular attraction (a). With n = 2 mol, V = 1 L: effective volume V−nb = 1 − 2(0.04) = 0.92 L; ideal pressure term nRT/(V−nb) = 2(0.08206)(400)/0.92 = 65.65/0.92 = 71.36 atm; attraction correction an²/V² = 3.6(4)/1 = 14.4 atm. P = 71.36 − 14.4 = 56.96 ≈ 57.0 atm. The ideal gas prediction would be P_ideal = nRT/V = 65.65 atm, so real intermolecular forces reduce P by about 13%.