Consider a rigid, insulated box containing 0.400 mole of he(g) at 20.08c and 1.00 atm in one compartment and 0.600 mole of n2(g) at 100.08c and 2.00 atm in the other compartment. these compartments are connected by a partition that transmits heat. what is the final tempera-ture in the box at thermal equilibrium? [for he(g), cv 5 12.5 j k 21mol 21 ; for n2(g), cv 5 20.7 j k 21mol 21

The value of final temperature of the mixture of gases when the partition is removed = 341.08 K

Explanation:

For box 1

Mass = 0.4 mole

Temperature = 20.08 °c = 293.08 K

Pressure = 1 atm = 101.325 K pa

For box 2

Mass = 0.6 mole

Temperature = 373.08 K

Pressure = 2 atm = 202 K pa

When the partition removed, both gases mixes together. then the final mass

= +

= 0.4 + 0.6

= 1.0 mole

This is the final mass in the container.

Final temperature is given by the equation,

+ =

Put all the values in the above equation we get,

0.4 × 293.08 + 0.6 × 373.08 = 1  ×

= 341.08 K

This is the value of final temperature of the mixture of gases when the partition is removed.

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.