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edit approved Mar 4, 2018 at 18:45
Avyansh Katiyar
edit approved Mar 4, 2018 at 18:45
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This is a question that appeared in Atkin's Physical Chemistry Book in the Chapter 3 exercise questions. It is question 3.3(a):

Question

Calculate $\Delta S$ when the state of $\mathrm{3~mol}$ of perfect gas atoms, for which $C_{p,m}$ = $\frac{5}{2}R$, is changed from $\mathrm{298~K}$ and $\mathrm{1~atm}$ to $\mathrm{398~K}$ and $\mathrm{5~atm}$. How do you rationalize the sign of $\Delta S$?

Atkin's Physical Chemistry Book; Chapter 3; question 3.3(a):
Question
Calculate $\Delta S$ when the state of $\mathrm{3~mol}$ of perfect gas atoms, for which $C_{p,m}$ = $\frac{5}{2}R$, is changed from $\mathrm{298~K}$ and $\mathrm{1~atm}$ to $\mathrm{398~K}$ and $\mathrm{5~atm}$. How do you rationalize the sign of $\Delta S$?

Attempt

Since this is at a constant pressure, $dq = dH = C_pdT$. Then I used the equation $$dS = \frac{C_pdT}{T}$$ I integrated both sides to get the following equation: $$\Delta S = C_p \ln\frac{398}{298}$$

Subbing in the values I got that $\Delta S = \mathrm{+18.3~J}$. However, the answer in the book is $\mathrm{-21~J}$. Shouldn't that be definitely wrong because the temperature of the system increased and hence the total entropy should have increased?

This is a question that appeared in Atkin's Physical Chemistry Book in the Chapter 3 exercise questions. It is question 3.3(a):

Question

Calculate $\Delta S$ when the state of $\mathrm{3~mol}$ of perfect gas atoms, for which $C_{p,m}$ = $\frac{5}{2}R$, is changed from $\mathrm{298~K}$ and $\mathrm{1~atm}$ to $\mathrm{398~K}$ and $\mathrm{5~atm}$. How do you rationalize the sign of $\Delta S$?

Attempt

Since this is at a constant pressure, $dq = dH = C_pdT$. Then I used the equation $$dS = \frac{C_pdT}{T}$$ I integrated both sides to get the following equation: $$\Delta S = C_p \ln\frac{398}{298}$$

Subbing in the values I got that $\Delta S = \mathrm{+18.3~J}$. However, the answer in the book is $\mathrm{-21~J}$. Shouldn't that be definitely wrong because the temperature of the system increased and hence the total entropy should have increased?

Atkin's Physical Chemistry Book; Chapter 3; question 3.3(a):
Question
Calculate $\Delta S$ when the state of $\mathrm{3~mol}$ of perfect gas atoms, for which $C_{p,m}$ = $\frac{5}{2}R$, is changed from $\mathrm{298~K}$ and $\mathrm{1~atm}$ to $\mathrm{398~K}$ and $\mathrm{5~atm}$. How do you rationalize the sign of $\Delta S$?

Attempt

Since this is at a constant pressure, $dq = dH = C_pdT$. Then I used the equation $$dS = \frac{C_pdT}{T}$$ I integrated both sides to get the following equation: $$\Delta S = C_p \ln\frac{398}{298}$$

Subbing in the values I got that $\Delta S = \mathrm{+18.3~J}$. However, the answer in the book is $\mathrm{-21~J}$. Shouldn't that be definitely wrong because the temperature of the system increased and hence the total entropy should have increased?

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This is a question that appeared in Atkin's Physical Chemistry Book in the Chapter 3 exercise questions. It is question 3.3(a):

Question

Calculate $\Delta S$ when the state of 3mol$\mathrm{3~mol}$ of perfect gas atoms, for which $C_{p,m}$ = $\frac{5}{2}R$, is changed from 298K$\mathrm{298~K}$ and 1atm$\mathrm{1~atm}$ to 398K$\mathrm{398~K}$ and 5atm$\mathrm{5~atm}$. How do you rationalize the sign of $\Delta S$?

Attempt

Since this is at a constant pressure, $dq = dH = C_pdT$. Then I used the equation $$dS = \frac{C_pdT}{T}$$ I integrated both sides to get the following equation: $$\Delta S = Cp ln\frac{398}{298}$$$$\Delta S = C_p \ln\frac{398}{298}$$

Subbing in the values I got that $\Delta S$ is +18.3J$\Delta S = \mathrm{+18.3~J}$. However, the answer in the book is -21J$\mathrm{-21~J}$. Shouldn't that be definitely wrong because the temperature of the system increased and hence the total entropy should have increased?

This is a question that appeared in Atkin's Physical Chemistry Book in the Chapter 3 exercise questions. It is question 3.3(a):

Question

Calculate $\Delta S$ when the state of 3mol of perfect gas atoms, for which $C_{p,m}$ = $\frac{5}{2}R$, is changed from 298K and 1atm to 398K and 5atm. How do you rationalize the sign of $\Delta S$?

Attempt

Since this is at a constant pressure, $dq = dH = C_pdT$. Then I used the equation $$dS = \frac{C_pdT}{T}$$ I integrated both sides to get the following equation: $$\Delta S = Cp ln\frac{398}{298}$$

Subbing in the values I got that $\Delta S$ is +18.3J. However the answer in the book is -21J. Shouldn't that be definitely wrong because the temperature of the system increased and hence the total entropy should have increased?

This is a question that appeared in Atkin's Physical Chemistry Book in the Chapter 3 exercise questions. It is question 3.3(a):

Question

Calculate $\Delta S$ when the state of $\mathrm{3~mol}$ of perfect gas atoms, for which $C_{p,m}$ = $\frac{5}{2}R$, is changed from $\mathrm{298~K}$ and $\mathrm{1~atm}$ to $\mathrm{398~K}$ and $\mathrm{5~atm}$. How do you rationalize the sign of $\Delta S$?

Attempt

Since this is at a constant pressure, $dq = dH = C_pdT$. Then I used the equation $$dS = \frac{C_pdT}{T}$$ I integrated both sides to get the following equation: $$\Delta S = C_p \ln\frac{398}{298}$$

Subbing in the values I got that $\Delta S = \mathrm{+18.3~J}$. However, the answer in the book is $\mathrm{-21~J}$. Shouldn't that be definitely wrong because the temperature of the system increased and hence the total entropy should have increased?

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Calculating Entropy

This is a question that appeared in Atkin's Physical Chemistry Book in the Chapter 3 exercise questions. It is question 3.3(a):

Question

Calculate $\Delta S$ when the state of 3mol of perfect gas atoms, for which $C_{p,m}$ = $\frac{5}{2}R$, is changed from 298K and 1atm to 398K and 5atm. How do you rationalize the sign of $\Delta S$?

Attempt

Since this is at a constant pressure, $dq = dH = C_pdT$. Then I used the equation $$dS = \frac{C_pdT}{T}$$ I integrated both sides to get the following equation: $$\Delta S = Cp ln\frac{398}{298}$$

Subbing in the values I got that $\Delta S$ is +18.3J. However the answer in the book is -21J. Shouldn't that be definitely wrong because the temperature of the system increased and hence the total entropy should have increased?