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An insulator container contains 4 moles of an ideal diatomic gas at temperature T. Heat Q is supplied to this gas, due to which 2 moles of the gas are dissociated into atoms but temperature of the gas remains constant. Then
A) $Q = 2RT$
B) $Q = RT$
C) $Q = 3RT$
D) $Q = 4RT$

Answer
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Hint:We will be using the concept of First law of thermodynamics. For monoatomic gas the degrees of freedom $f = 3$ and for diatomic gas $f = 5$ . We know that Internal energy of moles of ideal gas $E = \dfrac{f}{2}RT$

Complete Solution
It is given that an insulator container contains $4$ moles of an ideal gas. $2$ Moles of the gas are dissociated into atoms hence heat is supplied to gas, temperature remains constant.

The internal energy is the total amount of kinetic energy and potential energy of all the particles in the system. There is no inter- particle interaction in ideal gases. The internal energy is equal to the heat of the system
For the diatomic gas, the internal energy of moles of an ideal gas of degrees of freedom is given by
$E = \dfrac{f}{2}RT$
Where, f is the degrees of freedom
For the diatomic gas, the degrees of freedom (f) is equal to $5$
The amount of heat supplied ‘Q’ is given by
$Q = \Delta E$
$\Delta E$ is the change in internal energy
$Q = {E_f} - {E_{_i}}$
${E_f}$ is the internal energy of diatomic gas
${E_i}$ is the internal energy of monoatomic gas and internal energy of diatomic gas
Suppose $2$ moles of diatomic gas combines together to form $4$ moles of monatomic gas.
For $4$ molecules of monoatomic gas and $2$ moles of diatomic gas
$ {E_f} = 4 \times \dfrac{3}{2}RT + 2 \times \dfrac{5}{2}RT $
Simplifying,
$ \Rightarrow 6RT + 5RT$
$ \Rightarrow {E_f} = 11RT$
For $4$ molecules of diatomic gas
${E_i} = 4 \times \dfrac{5}{2}RT$
$ = 10RT$
$Q = {E_f} - {E_i}$
$ = 11RT - 10RT$
$Q$ $ = RT$

Answer; Option B: RT

NOTES:When gas dissociated into atoms $2$ moles of diatomic gas becomes $4$ moles of a monatomic gas. The degrees of freedom is defined as the number of ways a molecule in the gas phase may move or vibrate.