The amount of phenol adsorbed by the adsorbent can be calculated by comparing the initial amount of phenol in the solution and the final residual concentration after adsorption.
Step 1: Initial amount of phenol in the solution
The initial concentration of phenol in the solution is 100 mg/L, and the volume of the solution is 100 mL (0.1 L). Therefore, the initial amount of phenol in the solution is:
\[ \text{Initial phenol} = 100 \, \text{mg/L} \times 0.1 \, \text{L} = 10 \, \text{mg} \]
Step 2: Final amount of phenol in the solution
The residual concentration of phenol after adsorption is 30 mg/L. Therefore, the final amount of phenol in the solution is:
\[ \text{Final phenol} = 30 \, \text{mg/L} \times 0.1 \, \text{L} = 3 \, \text{mg} \]
Step 3: Amount of phenol adsorbed
The amount of phenol adsorbed is the difference between the initial and final amounts of phenol:
\[ \text{Phenol adsorbed} = 10 \, \text{mg} - 3 \, \text{mg} = 7 \, \text{mg} \]
Step 4: Amount of phenol adsorbed per gram of dry adsorbent
The mass of the dry adsorbent is 500 mg, which is 0.5 g. The amount of phenol adsorbed per gram of adsorbent is:
\[ \frac{\text{Phenol adsorbed}}{\text{Mass of adsorbent}} = \frac{7 \, \text{mg}}{0.5 \, \text{g}} = 14 \, \text{mg/g} \]
Therefore, the correct answer is: 14 mg per gram of dry adsorbent.
Consider a process with transfer function: \[ G_p = \frac{2e^{-s}}{(5s + 1)^2} \] A first-order plus dead time (FOPDT) model is to be fitted to the unit step process reaction curve (PRC) by applying the maximum slope method. Let \( \tau_m \) and \( \theta_m \) denote the time constant and dead time, respectively, of the fitted FOPDT model. The value of \( \frac{\tau_m}{\theta_m} \) is __________ (rounded off to 2 decimal places).
Given: For \( G = \frac{1}{(\tau s + 1)^2} \), the unit step output response is: \[ y(t) = 1 - \left(1 + \frac{t}{\tau}\right)e^{-t/\tau} \] The first and second derivatives of \( y(t) \) are: \[ \frac{dy(t)}{dt} = \frac{t}{\tau^2} e^{-t/\tau} \] \[ \frac{d^2y(t)}{dt^2} = \frac{1}{\tau^2} \left(1 - \frac{t}{\tau}\right) e^{-t/\tau} \]
Choose the transfer function that best fits the output response to a unit step input change shown in the figure:

An electrical wire of 2 mm diameter and 5 m length is insulated with a plastic layer of thickness 2 mm and thermal conductivity \( k = 0.1 \) W/(m·K). It is exposed to ambient air at 30°C. For a current of 5 A, the potential drop across the wire is 2 V. The air-side heat transfer coefficient is 20 W/(m²·K). Neglecting the thermal resistance of the wire, the steady-state temperature at the wire-insulation interface __________°C (rounded off to 1 decimal place).

GIVEN:
Kinematic viscosity: \( \nu = 1.0 \times 10^{-6} \, {m}^2/{s} \)
Prandtl number: \( {Pr} = 7.01 \)
Velocity boundary layer thickness: \[ \delta_H = \frac{4.91 x}{\sqrt{x \nu}} \]
The first-order irreversible liquid phase reaction \(A \to B\) occurs inside a constant volume \(V\) isothermal CSTR with the initial steady-state conditions shown in the figure. The gain, in kmol/m³·h, of the transfer function relating the reactor effluent \(A\) concentration \(c_A\) to the inlet flow rate \(F\) is:
