Solution:

Given,

1 mole of AB5 is placed in a closed container under 1 atmosphere and at 300k.

It is heated to 600k when 20% of mass of it dissociates as AB5 gives AB 2B2.

To find: The resultant pressure.

To solve this problem, we need to apply the ideal gas law which states that PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is the temperature.

Let us assume that the volume of the container is constant.

At 300k, the initial number of moles of AB5 is 1 mole.

After dissociation, the number of moles of AB5 remaining = 0.8 moles.

The number of moles of AB formed = 0.2 moles.

The number of moles of B formed = 0.4 moles.

At 600k, the pressure of the system can be calculated as follows:

P1V1/n1T1 = P2V2/n2T2

Where,

P1 = 1 atm (initial pressure)

V1 = V (volume of the container)

n1 = 1 mole (initial number of moles)

T1 = 300k (initial temperature)

P2 = ? (final pressure)

V2 = V (volume of the container)

n2 = 0.8 + 0.2 + 0.4 = 1.4 moles (final number of moles)

T2 = 600k (final temperature)

On substituting the values, we get:

1 atm * V / (1 mole * 300k) = P2 * V / (1.4 moles * 600k)

Simplifying the equation, we get:

P2 = (1 atm * V * 1.4 moles * 600k) / (1 mole * 300k * V)

P2 = 2.8 atm

Therefore, the resultant pressure of the system is 2.8 atm.

Explanation:

- The ideal gas law, PV=nRT, is used to solve the problem.
- The initial number of moles of AB5 is 1 mole.
- After dissociation, the number of moles of AB5 remaining is 0.8 moles and the number of moles of AB formed is 0.2 moles and B formed is 0.4 moles.
- The final pressure of the system is calculated using the ideal gas law.
- The resultant pressure of the system is 2.8 atm.

May be 1.6atm