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Using the statistical definition of entropy, what is the entropy of a system where [tex]W=4[/tex]?

A. [tex]1.87 \times 10^{-23}[/tex] joules/kelvin
B. [tex]1.56 \times 10^{-23}[/tex] joules/kelvin
C. [tex]1.91 \times 10^{-23}[/tex] joules/kelvin
D. [tex]2.07 \times 10^{-23}[/tex] joules/kelvin


Sagot :

To determine the entropy [tex]\( S \)[/tex] of a system where [tex]\( W = 4 \)[/tex] using the statistical definition of entropy, we use the formula:

[tex]\[ S = k_B \ln(W) \][/tex]

where:
- [tex]\( S \)[/tex] is the entropy,
- [tex]\( k_B \)[/tex] is the Boltzmann constant (approximately [tex]\( 1.38 \times 10^{-23} \)[/tex] joules/kelvin),
- [tex]\( W \)[/tex] is the number of possible microstates.

Let's proceed with the calculation:

1. Identify the given values:
- [tex]\( W = 4 \)[/tex]
- [tex]\( k_B = 1.38 \times 10^{-23} \)[/tex] joules/kelvin

2. Determine the natural logarithm of [tex]\( W \)[/tex]:
[tex]\[ \ln(4) \][/tex]

3. Multiply [tex]\( k_B \)[/tex] by [tex]\( \ln(4) \)[/tex]:
[tex]\[ S = 1.38 \times 10^{-23} \times \ln(4) \][/tex]

Performing this calculation, we find the entropy [tex]\( S \)[/tex]. The result of this calculation is approximately:

[tex]\[ S \approx 1.91 \times 10^{-23} \text{ joules/kelvin} \][/tex]

Therefore, the correct answer is:

C. [tex]\( 1.91 \times 10^{-23} \)[/tex] joules/kelvin