Question

Nitrogen gas is filled in an insulated container. If $\alpha$ fraction of moles dissociates without the exchange of any energy, then the fractional change in its temperature is:

Answer 1

Hint: Use the formula for internal energy. Find the internal energy of nitrogen before dissociation and after dissociation. Using the formula for total internal energy before dissociation finds the temperature before dissociation and using the formula for total internal energy after dissociation to find temperature after dissociation. Subtract both the temperatures and divide them by initial temperature. This will give a fractional change in temperature.

Formula used:
$E=n{ C }_{ V }T$

Complete step-by-step solution:
Let the initial temperature of Nitrogen gas be ${ T }_{ 1 }$
The number of moles of ${ N }_{ 2 }$ is n
Internal energy is given by,
$E=n{ C }_{ V }T$ …(1)
Nitrogen ${ N }_{ 2 }$ is a diatomic molecule. Hence, its ${ C }_{ V }$ is $\dfrac { 5R }{ 2 }$
Substituting values in the equation. (1) we get,
$E={n}_{1}\dfrac { 5R }{ 2 } { T }_{ 1 }$
Then, Nitrogen ${ N }_{ 2 }$ dissociates to 2N
Therefore, number of moles of N = 0
After dissociation by $\alpha$ fraction,
Number of moles of ${ N }_{ 2 }$= n(1-$\alpha$)
Number of moles of N=2n $\alpha$
Let the temperature after dissociation be ${ T }_{ 2 }$
N is a monatomic molecule. Hence, its ${ C }_{ V }$ is $\dfrac { 3R }{ 2 }$
Substituting values in equation.(1) we get,
$E={n}_{2}\dfrac { 3R }{ 2 } { T }_{ 2 }$
Therefore, total internal energy (E)= Internal energy of ${ N }_{ 2 }$+ Internal energy of N
Substituting the values we get,
$E=n\left( 1-\alpha \right) \times \dfrac { 5R }{ 2 } \times { T }_{ 2 }+2n\alpha \times \dfrac { 3R }{ 2 } \times { T }_{ 2 }$
After simplifying the above equation we get,
${ T }_{ 2 }=\dfrac { 2E }{ nR\left[ 5+\alpha \right] }$
Similarly, ${ T }_{ 1 }$ can be calculated by substituting the value of n before dissociation,
$\therefore { T }_{ 1 }=\dfrac { 2E }{ 5nR }$
Fractional change in temperature is given by,
$\dfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 } }$
$\therefore$ Fractional change= $\dfrac { \dfrac { 2E }{ Rn\left( 5+\alpha \right) } -\dfrac { 2E }{ 5nR } }{ \dfrac { 2E }{ 5nR } }$
$\therefore$ Fractional change=$\dfrac { \dfrac { 1 }{ \left( 5+\alpha \right) } -\dfrac { 1 }{ 5 } }{ \dfrac { 1 }{ 5 } }$
$\therefore$ Fractional change=$\dfrac { -\alpha }{ 5+\alpha }$
Therefore, the fractional change in temperature is $\dfrac { -\alpha }{ 5+\alpha }$.

Note: Be careful while writing the specific heat of molecules. Monoatomic moles have specific heat as $\dfrac { 3R }{ 2 }$, diatomic molecules have $\dfrac { 5R }{ 2 }$, it is $\dfrac { 6R }{ 2 }$ for angular triatomic molecules and $\dfrac { 7R }{ 2 }$ for linear triatomic molecules. The numerator value of these specific heats implies the degree of freedom. Such as a monatomic molecule has 3 degrees of freedom, a diatomic molecule has 5 and, etc.