Chapter 4, Problem 4.60E

Interpretation Introduction

(a)

Interpretation:

The value of change in pressure required to get a value of $\Delta G$ equal to $+1.000\text{kJ}$ is to be calculated.

Concept introduction:

The Gibbs free energy of the system represents the maximum amount of non-expansion work achieved by a thermodynamic system at isothermal and isobaric conditions. The change in Gibbs free energy is used to predict the spontaneity of the process. The change in Gibbs energy of the condensed system is represented as shown below.

$\Delta G=V\Delta p$

Answer to Problem 4.60E

The value of change in pressure required to get a value of $\Delta G$ equal to $+1.000\text{kJ}$ is $547.7092\text{atm}$.

Explanation of Solution

The number of moles of water is $1.00\text{mol}$.

The molar volume is $18.02{\text{cm}}^3$ which means that the volume of one mole of water is $18.02{\text{cm}}^3$.

The required value $\Delta G$ is $+1.000\text{kJ}$.

The change in Gibbs energy of the condensed system is represented as shown below.

$\Delta G=V\Delta p$…(1)

Where,

• $V$ represents the volume of the system.

• $\Delta p$ represents the change in pressure of the system.

Rearrange the equation (1) for the value of $\Delta p$.

$\Delta p=\frac{\Delta G}{V}$…(2)

Substitute the value of $V$ and $\Delta G$ in the equation (2).

$\begin{array}{rll}\Delta p& =& \left(\frac{+1.000\text{kJ}}{18.02{\text{cm}}^3}\right)\left(\frac{1000{\text{cm}}^3}{1\text{L}}\right)\left(\frac{1000\text{J}}{1\text{kJ}}\right)\left(\frac{1\text{L}\cdot \text{atm}}{101.32\text{J}}\right)\\ & =& 547.7092\text{atm}\end{array}$

Therefore, the value of change in pressure required to get a value of $\Delta G$ equal to $+1.000\text{kJ}$ is $547.7092\text{atm}$.

Conclusion

The value of change in pressure required to get a value $\Delta G$ equal to $+1.000\text{kJ}$ is $547.7092\text{atm}$.

Interpretation Introduction

(b)

Interpretation:

The value of change in pressure for $1\text{mol}$ of ideal gas at $25°\text{C}$ required to get a value $\Delta G$ equal to $+1.000\text{kJ}$ is to be calculated.

Concept introduction:

The Gibbs free energy of the system represents the maximum amount of non-expansion work achieved by a thermodynamic system at isothermal and isobaric conditions. The change in Gibbs free energy is used to predict the spontaneity of the process. The change in Gibbs energy of the gaseous system isrepresented as shown below.

$\Delta G=nRT\ln \frac{p_{\text{f}}}{p_{\text{i}}}$

Answer to Problem 4.60E

The value of change in pressure for $1\text{mol}$ of ideal gas at $25°\text{C}$ required to get a value $\Delta G$ equal to $+1.000\text{kJ}$ is $0.4972\text{atm}$.

Explanation of Solution

The temperature of the ideal gas is $25°\text{C}$.

The temperature of the ideal gas in Kelvin is calculated as shown below.

$\begin{array}{rll}T& =& \left(273+25.0°\text{C}\right)\text{K}\\ & =& \text{298}\text{K}\end{array}$

The required value of $\Delta G$ is $+1.000\text{kJ}$.

The number of moles of the ideal gas is $1.00\text{mol}$.

The change in Gibbs energy of the gaseous system is represented as shown below.

$\Delta G=nRT\ln \frac{p_{\text{f}}}{p_{\text{i}}}$…(3)

Where,

• $p_{\text{f}}$ represents the final pressure of the ideal gas.

• $p_{\text{i}}$ represents the initial pressure of the ideal gas.

• $n$ represents the number of moles of the ideal gas.

• $T$ represents the temperature of the ideal gas.

• $R$ represents the ideal gas constant with value $8.314\text{J}/\text{K}\cdot \text{mol}$.

Rearrange the equation (3) for the value of $\ln \frac{p_{\text{f}}}{p_{\text{i}}}$.

$\ln \frac{p_{\text{f}}}{p_{\text{i}}}=\frac{\Delta G}{nRT}$

Substitute the value of $\Delta G$, $n$, $T$, and $R$ in the above equation.

$\begin{array}{rll}\ln \frac{p_{\text{f}}}{p_{\text{i}}}& =& \frac{\left(+1.000\text{kJ}\right)\left(\frac{1000\text{J}}{1\text{kJ}}\right)}{\left(1.00\text{mol}\right)\left(\text{298}\text{K}\right)\left(8.314\text{J}/\text{K}\cdot \text{mol}\right)}\\ \ln \frac{p_{\text{f}}}{p_{\text{i}}}& =& 0.4036\end{array}$

The above expression is further solved as shown below.

$\begin{array}{rll}\exp \left(\ln \frac{p_{\text{f}}}{p_{\text{i}}}\right)& =& \exp \left(0.4036\right)\\ \frac{p_{\text{f}}}{p_{\text{i}}}& =& 1.4972\end{array}$

One is subtracted from both sides of the above expression.

$\begin{array}{rll}\frac{p_{\text{f}}}{p_{\text{i}}}-1& =& 1.4972-1\\ \frac{p_{\text{f}}-p_{\text{i}}}{p_{\text{i}}}& =& 0.4972\\ \frac{\Delta p}{p_{\text{i}}}& =& 0.4972\end{array}$

When the initial pressure of an ideal gas is assumed to be $1\text{atm}$, the above expression becomes as follows.

$\begin{array}{rll}\frac{\Delta p}{1\text{atm}}& =& 0.4972\\ \Delta p& =& 0.4972\text{atm}\end{array}$

Therefore, the value of change in pressure for $1\text{mol}$ of ideal gas at $25°\text{C}$ required to get a value of $\Delta G$ equal to $+1.000\text{kJ}$ is $0.4972\text{atm}$.

Conclusion

The value of change in pressure for $1\text{mol}$ of ideal gas at $25°\text{C}$ required to get a value $\Delta G$ equal to $+1.000\text{kJ}$ is $0.4972\text{atm}$.

Interpretation Introduction

(c)

Interpretation:

The difference between the two corresponding values of change in pressure is to be explained.

Concept introduction:

The Gibbs free energy of the system represents the maximum amount of non-expansion work achieved by a thermodynamic system at isothermal and isobaric conditions. The change in Gibbs free energy is used to predict the spontaneity of the process. The change in Gibbs energy of the gaseous system isrepresented as shown below.

$\Delta G=nRT\ln \frac{p_{\text{f}}}{p_{\text{i}}}$

Answer to Problem 4.60E

The value of change in pressure for $1\text{mol}$ of ideal gas at $25°\text{C}$ required to obtain a value of $\Delta G$ equal to $+1.000\text{kJ}$ is higher in case of water.

Explanation of Solution

The value of change in pressure for $1\text{mol}$ of ideal gas at $25°\text{C}$ required to get a value of $\Delta G$ equal to $+1.000\text{kJ}$ is $0.4972\text{atm}$.

The value of change in pressure for $1\text{mol}$ of water required to get a value $\Delta G$ equal to $+1.000\text{kJ}$ is $547.7092\text{atm}$.

The value of change in pressure for $1\text{mol}$ of ideal gas at $25°\text{C}$ required to get a value $\Delta G$ equal to $+1.000\text{kJ}$ is higher in case of water.

The gases are easily compressible than a condensed phase system. A small increase in pressure is required for compression of gas and a large increase in pressure is required to compress water. Therefore, the compression of gas requires less pressure than the compression of water.

Conclusion

The value of change in pressure for $1\text{mol}$ of ideal gas at $25°\text{C}$ required to get a value $\Delta G$ equal to $+1.000\text{kJ}$ higher in case of water. The reason for this difference is that compression of gas requires less pressure than compression of water.