Chapter 17, Problem 17.50E

Interpretation Introduction

(a)

Interpretation:

The value of $S$ for $\text{C}$ atoms at $\text{1000}\text{K}$ is to be calculated.

Concept introduction:

The measure of the total amount of randomness or disorder of molecules of the system is known as entropy. It is denoted by $S$. It is an extensive property of the thermodynamic system.

Answer to Problem 17.50E

The value of $S$ for $\text{C}$ atoms at $\text{1000}\text{K}$ is $15.766\text{J}/\left(\text{mol}\cdot \text{K}\right)$.

Explanation of Solution

The given value of temperature is $\text{1000}\text{K}$.

The standard temperature is $298\text{K}$.

The given pressure is $\text{1}\text{atm}$.

The value of universal gas constant is $8.3145\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The standard entropy of carbon atom is $5.7\text{J}/\left(\text{mol}\cdot \text{K}\right)$.

The expression to calculate the value of entropy is given below.

$S=S°+R\ln \frac{T_2}{T_1}$

Where,

• $S°$ is the standard temperature.

• $R$ is the universal gas constant.

• $T_1$ is the initial temperature equals to $298\text{K}$.

• $T_2$ is the final temperature.

Substitute the values of $S°$, $R$, $T_1$ and $T_2$ for carbons atoms in the above expression.

$\begin{array}{rll}S& =& 5.7\text{J}/\left(\text{mol}\cdot \text{K}\right)+8.3145\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times \ln \frac{\text{1000}\text{K}}{298\text{K}}\\ & =& 5.7\text{J}/\left(\text{mol}\cdot \text{K}\right)+8.3145\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 1.21066\\ & =& 15.766\text{J}/\left(\text{mol}\cdot \text{K}\right)\end{array}$

The given value $S$ for $\text{C}$ atoms at $\text{1000}\text{K}$ is $183.2\text{J}/\left(\text{mol}\cdot \text{K}\right)$ but the calculated value is totally deviated from the given value that is $15.766\text{J}/\left(\text{mol}\cdot \text{K}\right)$.

Conclusion

The value of $S$ for $\text{C}$ atoms at $\text{1000}\text{K}$ is $15.766\text{J}/\left(\text{mol}\cdot \text{K}\right)$.

Interpretation Introduction

(b)

Interpretation:

The value of $S$ for $\text{Fe}$ atoms at $\text{3500}\text{K}$ is to be calculated.

Concept introduction:

The measure of the total amount of randomness or disorder of molecules of the system is known as entropy. It is denoted by $S$. It is an extensive property of the thermodynamic system.

Answer to Problem 17.50E

The value of $S$ for $\text{Fe}$ atoms at $\text{3500}\text{K}$ is $22.782\text{J}/\left(\text{mol}\cdot \text{K}\right)$.

Explanation of Solution

The given value of temperature is $\text{3500}\text{K}$.

The standard temperature is $298\text{K}$.

The given pressure is $\text{1}\text{atm}$.

The value of universal gas constant is $8.3145\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The standard entropy of iron atom is $2.3\text{J}/\left(\text{mol}\cdot \text{K}\right)$.

The expression to calculate the value of entropy is given below.

$S=S°+R\ln \frac{T_2}{T_1}$

Where,

• $S°$ is the standard temperature.

• $R$ is the universal gas constant.

• $T_1$ is the initial temperature equals to $298\text{K}$.

• $T_2$ is the final temperature.

Substitute the values of $S°$, $R$, $T_1$ and $T_2$ for carbons atoms in the above expression.

$\begin{array}{rll}S& =& 2.3\text{J}/\left(\text{mol}\cdot \text{K}\right)+8.3145\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times \ln \frac{\text{3500}\text{K}}{298\text{K}}\\ & =& 2.3\text{J}/\left(\text{mol}\cdot \text{K}\right)+8.3145\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 2.4634\\ & =& 22.782\text{J}/\left(\text{mol}\cdot \text{K}\right)\end{array}$

The given value $S$ for $\text{Fe}$ atoms at $\text{3500}\text{K}$ is $239.6\text{J}/\left(\text{mol}\cdot \text{K}\right)$ but the calculated value is totally deviated from the given value that is $22.782\text{J}/\left(\text{mol}\cdot \text{K}\right)$.

Conclusion

The value of $S$ for $\text{Fe}$ atoms at $\text{3500}\text{K}$ is $239.6\text{J}/\left(\text{mol}\cdot \text{K}\right)$.

Interpretation Introduction

(c)

Interpretation:

The value of $S$ for $\text{Hg}$ atoms at $\text{298}\text{K}$ is to be calculated. The trend in agreement between calculation and experiment is to be stated.

Concept introduction:

The measure of the total amount of randomness or disorder of molecules of the system is known as entropy. It is denoted by $S$. It is an extensive property of the thermodynamic system.

Answer to Problem 17.50E

The value of $S$ for $\text{Hg}$ atoms at $\text{298}\text{K}$ is $174.97\text{J}/\left(\text{mol}\cdot \text{K}\right)$.

Explanation of Solution

The given value of temperature is $298\text{K}$.

The standard temperature is $298\text{K}$.

The given pressure is $\text{1}\text{atm}$.

The value of universal gas constant is $8.3145\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The standard entropy of mercury atom is $174.97\text{J}/\left(\text{mol}\cdot \text{K}\right)$.

The expression to calculate the value of entropy is given below.

$S=S°+R\ln \frac{T_2}{T_1}$

Where,

• $S°$ is the standard temperature.

• $R$ is the universal gas constant.

• $T_1$ is the initial temperature equals to $298\text{K}$.

• $T_2$ is the final temperature.

Substitute the values of $S°$, $R$, $T_1$ and $T_2$ for carbons atoms in the above expression.

$\begin{array}{rll}S& =& 174.97\text{J}/\left(\text{mol}\cdot \text{K}\right)+8.3145\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times \ln \frac{298\text{K}}{298\text{K}}\\ & =& 174.97\text{J}/\left(\text{mol}\cdot \text{K}\right)+8.3145\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 0\\ & =& 174.97\text{J}/\left(\text{mol}\cdot \text{K}\right)\end{array}$

The given value $S$ for $\text{Hg}$ atoms at $\text{298}\text{K}$ is $174.97\text{J}/\left(\text{mol}\cdot \text{K}\right)$ which is equal to the given value that is $174.97\text{J}/\left(\text{mol}\cdot \text{K}\right)$.

According to the calculations, there is some major difference in the carbon and iron atoms entropy values whereas the entropy of mercury is same as given.

Conclusion

The value of $S$ for $\text{Hg}$ atoms at $\text{298}\text{K}$ is $174.97\text{J}/\left(\text{mol}\cdot \text{K}\right)$.