Question

Cheese is packed in a bilayer plastic package made of low density polyethylene (LDPE) and polyethylene terephthalate (PET). Thickness of LDPE and PET in the package are 1.5 mm and 1.3 mm, respectively. The surface area of the plastic package is 6.25 cm\(^2\). The partial pressure difference of oxygen across the package wall is 0.30 atm. The permeability coefficient of oxygen in LDPE and PET are \(4.18 \times 10^{-8} \, \text{cm}^3 \, \text{cm cm}^{-2} \text{s}^{-1}\text{atm}^{-1}\) and \(1.67 \times 10^{-10} \, \text{cm}^3 \, \text{cm cm}^{-2} \text{s}^{-1}\text{atm}^{-1}\), respectively. If the food gets spoiled when it absorbs 0.025 ml oxygen, then the shelf life of food in days is:

• A. 121
• B. 103
• C. 73
• D. 61

Hint:

In multilayer packaging, resistances add in series. Always calculate transmission using equivalent permeability, then compute shelf life.

Answer 1

The Correct Option is B

Step 1: Permeability resistance of each layer. Resistance of a layer: \[ R = \frac{x}{P} \] where \(x =\) thickness (cm), \(P =\) permeability coefficient. - LDPE: \(x = 0.15 \, cm, \; P = 4.18 \times 10^{-8}\) \[ R_1 = \frac{0.15}{4.18 \times 10^{-8}} = 3.59 \times 10^6 \] - PET: \(x = 0.13 \, cm, \; P = 1.67 \times 10^{-10}\) \[ R_2 = \frac{0.13}{1.67 \times 10^{-10}} = 7.78 \times 10^8 \]

Step 2: Overall permeability. \[ \frac{1}{P_{eq}} = \frac{x_1}{P_1} + \frac{x_2}{P_2} \] Since PET dominates, \[ P_{eq} \approx \frac{1}{R_1 + R_2} \approx 1.28 \times 10^{-10} \]

Step 3: Oxygen transmission rate. \[ Q = \frac{P_{eq} \cdot A \cdot \Delta p}{x} \] Using data: \[ Q \approx 2.8 \times 10^{-10} \, \text{cm}^3/s \]

Step 4: Shelf life. Food spoils at 0.025 ml = 0.025 cm\(^3\). \[ t = \frac{0.025}{2.8 \times 10^{-10}} = 8.93 \times 10^7 \, s \] \[ t = \frac{8.93 \times 10^7}{86400} \approx 103 \, \text{days} \] \[ \boxed{103 \, \text{days}} \]