Question

Heat$Q=\left( {}^{3}/{}_{2} \right)RT$ is supplied to $4$ moles of an ideal diatomic gas at temperature $\text{T}$. How many moles the gas are dissociated into atoms if the temperature of gas is constant?
(A) $\text{3}$
(B) $\text{2}$
(C) $\text{1}$
(D) Data insufficient

Answer 1

Hint Find the change in the internal energy for initial 4 moles of diatomic gas and final dissociations into atoms. One mole of diatomic gas becomes two when it gets dissociated into monoatomic gas. The internal energy for a particular type of gas depends upon how many degrees of freedom the gas has. Monoatomic gas has three degrees of freedom and diatomic gas has five degrees of freedom.
FORMULA USED: $Q=\Delta U={{U}_{f}}-{{U}_{i}}$,
$U=\dfrac{5}{2}NKT$ for diatomic gas
$U=\dfrac{3}{2}NKT$ for monoatomic gas
Where ${{U}_{f}}$ → Final Internal Energy
$N$ → No. of Moles
${{U}_{i}}$ → Initial Internal Energy
$K$ → Boltzmann Constant
$T$ → Temperature

Complete Step By Step solution
Given$Q=\left( {}^{3}/{}_{2} \right)RT$, Moles of diatomic gas = 4
Let the moles dissociated be ‘$x$’
1 mole of diatomic gas becomes 2 moles of monatomic gas when the gas is dissociated into atoms.
So, if $x$moles of diatomic gas are dissociated then they will form $2x$moles of monatomic gas.
So, ${{U}_{f}}$ (final internal energy) $=\left( 2x\times \dfrac{3}{2}RT \right)+\left( \left( 4-x \right)\times \dfrac{5}{2}RT \right)$ $[R=NK]$
$R$→ ideal gas constant for monoatomic gas for diatomic gas
And ${{U}_{i}}$ (initial internal energy) $=4\times \dfrac{5}{2}RT$ (as all moles are of diatomic gas before dissociation)
So, $Q={{U}_{f}}-{{U}_{i}}$
$\Rightarrow$ $\dfrac{3}{2}RT=\left( 2x\times \frac{3}{2}RT+\left( 4-x \right)\times \dfrac{5}{2}RT \right)-4\times \dfrac{5}{2}RT$
$\Rightarrow$ $\dfrac{3}{2}RT=3xRT+10RT-\frac{5x}{2}RT-10RT$
$\Rightarrow$ $\dfrac{3}{2}RT=3xRT-\dfrac{5x}{2}RT$
Dividing both sides by$RT$we get, $\dfrac{3}{2}=3x-\dfrac{5x}{2}$
$\Rightarrow$ $\dfrac{3}{2}=\dfrac{6x-5x}{2}$
$\Rightarrow$ $\dfrac{3}{2}=\dfrac{x}{2}$
So,
$\Rightarrow$ $x=3$
So, 3 moles of the gas are dissociated into atoms if temperature is kept constant.

So, option (i) is correct.

Additional Information Kinetic Energy is the only one, contributing to the internal energy. Each degree of freedom contributes, ${}^{1}/{}_{2}KT$ per atom to the internal energy. For monatomic ideal gases with $N$atoms, its total internal energy $U$ is given as, $U=\frac{3}{2}NKT$ and for diatomic gases, $U=\dfrac{5}{2}NKT$
The degrees of freedom ($F$) for monatomic gas is 3 and for diatomic gas is 5.
Degree of Freedom: It is the no. of values in the final calculation of a statistic that are free to vary or we can say that it is the no. of independent ways by which a dynamic system can move, without violating any constraint imposed on it.
Monoatomic gas has three translational degrees of freedom. These are due to the motion of the centre of mass of that molecule along$x-axis$, $y-axis$and $z-axis$
Diatomic molecule can rotate about any axis at right angles to its own axis. Hence it has two degrees of freedom of rotational motion in addition to three degrees of freedom of translational motion.

Note Concepts of thermodynamics and its laws should be studied well. Degrees of Freedom of various types of molecules should be known along with their internal energies. The role of temperature can be studied and its effect on the variation of internal energy.