Respuesta :
Answer: 1.439 × [tex]10^{-23}[/tex] J/K
Explanation:
initial rotational kinetic energy Er1 = 22ε
final rotational kinetic energy Er2 = 62ε
rotational entropy for nonlinear polyatomic gas molecule = Sr
Sr = K1n ( ) + 3/2 K ………eq1
Er = the rotational kinetic energy is:
E_r = ( [tex]\frac{8π^2 (8π^3 )^(1/2) (KT)^(3/2}{h^3 }[/tex] )
inserting this equation into eq1
= K1n (/σr) +3/2 K
∆S_r= [tex]S_{r2}[/tex] - [tex]S_{r1}[/tex]
=[K1n * (/σr) +3/2 K] - [K1n * (/σr) +3/2 K]
=K1n ([tex]E_{r2}[/tex]/[tex]E_{r1}[/tex])
Now calculate the change in entropy
∆S_r= K1n (/)
Where K is Boltzmann constant
K = 1.389 × 10^(-23) J/K
= 1.389 × 10^(-23) J/K × 1n (62/22)
= 1.439 × 10^(-23) J/K
Answer:
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[tex]_f[/tex] - K×㏑W[tex]_i[/tex]
Where
W[tex]_i[/tex] = Total initial micro-states = 22
W[tex]_f[/tex] = Final total micro-states = 62
K = Boltzmann constant = 1.38064852 × 10⁻²³ m²·kg·s⁻²·K⁻¹
we have ΔS = K×(㏑W[tex]_f[/tex] - ㏑W[tex]_i[/tex])
= 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.
