For a first-order reaction A(g) → 2B(g) + C(g) at constant volume and 300 K, the total pressure at the beginning (t=0) and at time t are P0 and Pt respectively. Initially, only A is present with concentration [A]0, and t1/3 is the time required for the partial pressure of A to reach 1/3rd of its initial value. The correct option(s) is/are: (Assume that all these gases behave as ideal gases).

Let the initial pressure of A be P0. At time t, the partial pressure of A is P0/3. The reaction is A(g) → 2B(g) + C(g).

At t = 0, the total pressure is P0.
At time t, the partial pressure of A is P0/3. The moles of A that have reacted are (2/3)P0/RT. According to stoichiometry, 2 moles of B and 1 mole of C are formed for every mole of A reacted.

Therefore, partial pressure of B = 2 * (2/3)P0/3 = (4/3)P0/3
Partial pressure of C = (2/3)P0/3

Total pressure at time t (Pt) = P0/3 + (4/3)P0/3 + (2/3)P0/3 = (7/3)P0/3 = 7P0/9

For a first-order reaction, we have:
ln(P0/P) = kt
ln(P0/(P0/3)) = kt1/3
ln3 = kt1/3

k = ln3/t1/3

The rate constant k is given by k = ln3/t1/3. The correct option would depend on the specific options given (A, B, C, D) as they are missing from the input. However, the above calculations are the necessary steps to determine the correct answer. The relationship between P0, Pt and t1/3 can be determined using the integrated rate law for first-order reactions and the stoichiometry of the reaction.