(A) We know that the change in entropy is given by $\Delta S = \frac{q_{rev}}{T}$.Given that the quantity of heat $q_{rev}$ is the same in both cases,

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(A) We know that the change in entropy is given by $\Delta S = \frac{q_{rev}}{T}$.Given that the quantity of heat $q_{rev}$ is the same in both cases,we have $\Delta S_{1} = \frac{q_{rev}}{T_{1}}$ and $\Delta S_{2} = \frac{q_{rev}}{T_{2}}$.Since $T_{1} > T_{2}$,the denominator in the expression for $\Delta S_{1}$ is larger than the denominator in the expression for $\Delta S_{2}$.Therefore,$\Delta S_{1} < \Delta S_{2}$.

Class 11 Chemistry Thermodynamics 2 nd Law of thermodynamics and Entropy

Medium

When the same quantity of heat is absorbed by a system at two different temperatures $T_{1}$ and $T_{2}$,such that $T_{1} > T_{2}$,the changes in entropy are $\Delta S_{1}$ and $\Delta S_{2}$ respectively. Then:

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A
$\Delta S_{1} < \Delta S_{2}$
B
$\Delta S_{1} = \Delta S_{2}$
C
$\Delta S_{1} > \Delta S_{2}$
D
$\Delta S_{2} < \Delta S_{1}$

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2 nd Law of thermodynamics and Entropy

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When the same quantity of heat is absorbed by a system at two different temperatures $T_{1}$ and $T_{2}$,such that $T_{1} > T_{2}$,the changes in entropy are $\Delta S_{1}$ and $\Delta S_{2}$ respectively. Then:

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Assertion : Entropy of ice is less than water.Reason : Ice has a cage-like structure.

Calculate the increase in entropy (approximate) of $3 \ mol$ of hydrogen gas as it changes from $300 \ K$ at $0.1 \ atm$ to $1000 \ K$ at $1 \ atm$ $(C_p = 7 \ cal \ K^{-1} \ mol^{-1})$. $[\log \ 3.33 \approx 0.52, \ \ln \ x = 2.3 \ \log \ x]$ .......$cal \ K^{-1}$

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Identify the reaction/process in which the entropy increases.

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Assertion : Entropy of ice is less than water.
Reason : Ice has a cage-like structure.