Two nitro NO2 groups are chemically bonded to a patch of surface. They can't move to another location on the surface, but they can rotate (see sketch at right). It turns out that the amount of rotational kinetic energy each NO2 group can have is required to be a multiple of ε, where =ε×1.010−24 J. In other words, each NO2 group could have ε of rotational kinetic energy, or 2ε, or 3ε, and so forth — but it cannot have just any old amount of rotational kinetic energy. Suppose the total rotational kinetic energy in this system is initially known to be 22ε. Then, some heat is added to the system, and the total rotational kinetic energy rises to 62ε. Calculate the change in entropy. Round your answer to 3 significant digits, and be sure it has the correct unit symbol.

The answer to the question is;

The change in entropy ΔS of the system  ≈ 1.43 × 10⁻²³ J·K⁻¹.

Explanation:

The Boltzmann Entropy Equation is

S = K×㏑W

Where W = number of different micro-states

Since the total initial Kinetic energy of the system is 22ε where the number of possible states are ε, 2ε, 3ε,...., 22ε we have 22 micro-states

Similarly since the final rotational kinetic energy rises is 62ε we have 62 states

The Change in entropy then is

ΔS = Final entropy - Initial Entropy

ΔS = K×㏑W -  K×㏑W

Where

W = Total initial micro-states = 22

W  = Final total micro-states = 62

K = Boltzmann constant = 1.38064852 × 10⁻²³ m²·kg·s⁻²·K⁻¹

we have ΔS = K×(㏑W  -  ㏑W)

= 1.38064852 × 10⁻²³ m²·kg·s⁻²·K⁻¹×(㏑(62) - ㏑(22))

= 1.38064852 × 10⁻²³ m²·kg·s⁻²·K⁻¹×(4.13 - 3.091)

ΔS =  1.430480636 × 10⁻²³ m²·kg·s⁻²·K⁻¹.

Where 1 m²·kg·s⁻² = 1 J

ΔS ≈  1.43 × 10⁻²³ m²·kg·s⁻²·K⁻¹ = 1.43 × 10⁻²³ J·K⁻¹.

3 hours because it is a decent distance.

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