Problem 1: Given x/m = 0.5 × P^(1/3), calculate x/m at P = 27 atm.

Solution: Step 1: Identify the given Freundlich isotherm: x/m = 0.5 × P^(1/3) Units: k = 0.5 (units of x/m), 1/n = 1/3, P in atm

Step 2: Substitute P = 27 atm: $\frac{x}{m} = 0.5 \times (27)^{1/3}$

Step 3: Calculate (27)^(1/3) = cube root of 27: $27^{1/3} = 3 \quad \text{(since } 3^3 = 27\text{)}$

Step 4: Multiply: $\frac{x}{m} = 0.5 \times 3 = \mathbf{1.5 \text{ units per gram}}$

Problem 2: From a log-log plot, slope = 0.5 and y-intercept = log 4. Write the Freundlich isotherm equation.

Solution: Step 1: Identify slope and intercept: slope = 1/n = 0.5 → n = 2 y-intercept = log k = log 4 → k = 4

Step 2: Write the isotherm: \frac{x}{m} = 4 \times $P^{0.5}$ = 4\sqrt{P}

Step 3: Verify 1/n is between 0 and 1: 0.5 ✓ (isotherm is valid)

Problem 3: Calculate amount adsorbed per gram at P = 64 atm using x/m = 0.5 × P^(1/3).

Solution: $\frac{x}{m} = 0.5 \times (64)^{1/3} = 0.5 \times 4 = \mathbf{2.0 \text{ units/g}}$ (64^(1/3) = 4, since 4^{3} = 64)

NEET Strategy: Always look for perfect cubes (8, 27, 64, 125) and perfect squares (4, 9, 16, 25) in NEET numerical problems on Freundlich isotherm. The exponent 1/n is usually 1/2 or 1/3 to make arithmetic clean.