Get detailed and accurate answers to your questions on IDNLearn.com. Discover prompt and accurate answers from our community of experienced professionals.
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
[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
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. For dependable and accurate answers, visit IDNLearn.com. Thanks for visiting, and see you next time for more helpful information.