Question

A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the final number of microstates is 0.833 times that of the initial number of microstates? Express your answer numerically in joules per kelvin per particle.

Answer 1

Explanation:

Entropy means the amount of randomness present within the molecules of the body of a substance.

Relation between entropy and microstate is as follows.

           S = $K_{b} \times ln \Omega$

where,      S = entropy

             $K_{b}$ = Boltzmann constant

             $\Omega$ = number of microstates

This equation only holds good when the system is neither losing or gaining energy. And, in the given situation we assume that the system is neither gaining or losing energy.

Also, let us assume that $\Omega$ = 1, and $\Omega'$ = 0.833

Therefore, change in entropy will be calculated as follows.

     $\Delta S = K_{b} \times ln \Omega' - K_{b} \times ln \Omega$

                 = $1.38 \times 10^{-23} \times ln(0.833) - 1.38 \times 10^{-23} \times \times ln(1)$

                 = $1.38 \times 10^{-23} \times (-0.182)$

                 = $-0.251 \times 10^{-23}$

or,             = $-2.51 \times 10^{-24}$

Thus, we can conclude that the entropy change for a particle in the given system is $-2.51 \times 10^{-24}$ J/K particle.

Answer 2

The entropy change for a particle in this system should be $-2.51 \times 10^-24$ J/K particle.

Calculation of change in entropy:

Entropy represents the amount of randomness i.e. available within the molecules of the body with respect to the substance.

Here the Relation between entropy and microstate is

$S = k_b \times ln\Omega$

Let us presume here $\Omega = 1, and\ \Omega' = 0.833$

Now

$= 1.38 \times 10^{-23}\times ln(0.833) - 1.38 \times 10^{-23}\times ln(1)\\\\= 1.38 \times 10^{-23} \times (-0.182)\\\\= -0.251 \times 10^{-23}\\\\= -2.51\times 10^{-24}$

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