Two nitro [tex]$\left( NO _2\right)$[/tex] 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 [tex]$NO _2$[/tex] group can have is required to be a multiple of [tex]$\varepsilon$[/tex], where [tex]$\varepsilon=1.0 \times 10^{-24} J$[/tex]. In other words, each [tex]$NO _2$[/tex] group could have [tex]$\varepsilon$[/tex] of rotational kinetic energy, or [tex]$2 \varepsilon$[/tex], or [tex]$3 \varepsilon$[/tex], 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 [tex]$12 \varepsilon$[/tex]. Then, some heat is added to the system, and the total rotational kinetic energy rises to [tex]$87 \varepsilon$[/tex]. Calculate the change in entropy.
Two rotating [tex]$NO _2$[/tex] bonded to a surfac
Round your answer to 3 significant digits, and be sure it has the correct unit symbol.
Answer (2)
Calculate the initial number of microstates: W 1 = 13 .
Calculate the final number of microstates: W 2 = 88 .
Calculate the change in entropy: Δ S = k ln ( W 2 / W 1 ) .
The change in entropy is: 2.64 × 1 0 − 23 J / K .
Explanation
Problem Analysis and Entropy Formula We are given a system of two N O 2 groups that can rotate. The rotational kinetic energy of each group is quantized in multiples of ε , where ε = 1.0 × 1 0 − 24 J . The total initial rotational kinetic energy is 12 ε , and after adding heat, it becomes 87 ε . We need to calculate the change in entropy.\The entropy S is related to the number of microstates W by the Boltzmann formula: S = k ln W where k is the Boltzmann constant, k = 1.38 × 1 0 − 23 J / K .
Calculate Initial Microstates First, we need to find the number of microstates W 1 corresponding to the initial energy 12 ε . This is equivalent to finding the number of ways to distribute 12 ε between the two N O 2 groups. Let x 1 and x 2 be the energy levels of the two groups in units of ε . Then we have x 1 + x 2 = 12 where x 1 and x 2 are non-negative integers. The number of solutions is given by W 1 = 12 + 1 = 13 .
Calculate Final Microstates Next, we need to find the number of microstates W 2 corresponding to the final energy 87 ε . Similarly, we have x 1 + x 2 = 87 where x 1 and x 2 are non-negative integers. The number of solutions is given by W 2 = 87 + 1 = 88 .
Calculate Change in Entropy Now we can calculate the initial and final entropies: S 1 = k ln W 1 = k ln 13 S 2 = k ln W 2 = k ln 88 The change in entropy is given by Δ S = S 2 − S 1 = k ln W 2 − k ln W 1 = k ln W 1 W 2 = k ln 13 88
Final Calculation and Rounding Plugging in the value of k , we have Δ S = ( 1.38 × 1 0 − 23 J / K ) ln 13 88 Using a calculator, we find ln 13 88 ≈ 1.9095 Therefore, Δ S ≈ ( 1.38 × 1 0 − 23 J / K ) × 1.9095 ≈ 2.635 × 1 0 − 23 J / K Rounding to 3 significant digits, we get Δ S ≈ 2.64 × 1 0 − 23 J / K
Final Answer The change in entropy is approximately 2.64 × 1 0 − 23 J / K .
Examples
This type of problem is relevant in statistical mechanics, where we analyze systems with many particles and quantized energy levels. For example, consider a collection of molecules adsorbed on a surface, each with quantized rotational energy levels. When heat is added to the system, the molecules can transition to higher energy levels, increasing the number of accessible microstates and thus the entropy of the system. Understanding how entropy changes with energy is crucial for predicting the behavior of such systems, such as their thermal stability and reactivity.
The change in entropy for the system of two N O 2 groups as the rotational kinetic energy increases from 12 ε to 87 ε is approximately 2.64 × 1 0 − 23 J / K . This is calculated by determining the number of accessible microstates for both energy levels and applying the Boltzmann formula. The method highlights how energy quantization impacts entropy in a rotational system.
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