Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The heat capacity of a solid material is measured at different temperatures, and the data is as follows: <img alt="" src="https://cdn.entrance360.com/media/uploads/2023/08/17/thermodyna-q

The heat capacity of a solid material is measured at different temperatures, and the data is as follows:

Use this data to estimate the entropy change (S) for the solid material as it is cooled from 400 K to
100 K. Assume that the heat capacity is constant over this temperature range.

Option: 1

34.09 J/mol-K


Option: 2

41.6 J/mol-K


Option: 3

34.5 J/mol-K


Option: 4

39 J/mol-K


Answers (1)

best_answer

We can estimate the change in entropy (?S) for this solid material as it is cooled from 400K to 100 K by calculating the integral of the heat capacity over the temperature range.
Step 1: Identify the given data: - Heat capacity\mathrm{ (Cp)} at different temperatures (K):

\mathrm{\begin{array}{ll} T_1=100 \mathrm{~K}, & C_{p 1}=30 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\ T_2=200 \mathrm{~K}, & C_{p 2}=40 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\ T_3=300 \mathrm{~K}, & C_{p 3}=50 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\ T_4=400 \mathrm{~K}, & C_{p 4}=60 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \end{array}}

Step 2: Apply the formula to estimate ?S: The change in entropy (?S) can be estimated as the integral of the heat capacity over the temperature range (from 400 K to 100 K):

\mathrm{\Delta S=\int_{T_1}^{T_4} \frac{C_p}{T} d T}

Step 3: Perform the calculation:

\mathrm{\Delta S=\int_{100 \mathrm{~K}}^{400 \mathrm{~K}} \frac{C_p}{T} d T}

We need to integrate \mathrm{C_p} over the temperature range from 100 K to 400 K, but we need to know how \mathrm{C_p} changes with temperature. If the heat capacity is constant (independent of temperature) over this range, the integral simplifies:

\mathrm{\Delta S=C_p \cdot \ln \left(\frac{T_4}{T_1}\right)}

Substitute the values:

\mathrm{\Delta S=30 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \cdot \ln \left(\frac{400 \mathrm{~K}}{100 \mathrm{~K}}\right)}

Step 4: Perform the calculation:

?S ≈ 41.6 J/mol·K

So, the estimated entropy change (?S) for the solid material as it is cooled from 400 K to 100 K is approximately 41.6 J/mol·K.
So, correct option is B

Posted by

SANGALDEEP SINGH

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 session 2 application correction deadline today; edit fields at jeemain.nta.nic.in
anam-khan

JEE Mains 2025 application correction link out at jeemain.nta.nic.in; how to make changes?
anam-khan

JEE Mains 2025 session 2 application form correction window opens tomorrow; edit details by February 28

Latest Question