Question

Exactly 1 mol ice is heated reversibly at atmospheric pressurefrom $\mathbf{-}\mathbf{20}\mathbf{°}\mathbf{C}$to $\mathbf{0}\mathbf{°}\mathbf{C}$, melted reversibly at $\mathbf{0}\mathbf{°}\mathbf{C}$, and then heated reversibly at atmospheric pressure to 20°C. ${\mathbf{\Delta H}}_{\mathbf{fus}}\mathbf{=}\mathbf{6007}{\mathbf{Jmol}}^{\mathbf{-}\mathbf{1}}$: ${\mathbf{C}}_{\mathbf{p}}\left(\mathbf{ice}\right)\mathbf{=}\mathbf{38}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}$; and ${\mathbf{C}}_{\mathbf{p}}\left(\mathbf{water}\right)\mathbf{=}\mathbf{75}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}$. Calculate$\mathbf{∆}\mathbf{S}$for the system, the surroundings, and the thermodynamic universe for this process.

Short answer

The entropy of the system, surrounding, and the universe is , $30.2J{K}^{-1}$,$-30.2J{K}^{-1}$ and 0.

Step-by-step solution

Gibbs Free Energy.

Gibbs free energy is known as the maximum reversible work that is done on a thermodynamic system at a constant temperature and pressure. It is symbolized as G.

The concept.

For the reversible isobaric process, the expression for entropy change is:

$\Delta S={\int }_{T_1}^{T_2}\frac{n{c}_p}{T}dT$

*$∆\mathrm{S}$ is the entropy change.

*n is the number of moles, i.e., 1 mol.

*${\mathrm{c}}_{\mathrm{p}}$is the heat capacity at constant pressure.

*T is the temperature.

The entropy changes in the reversible isobaric process.

Theentropy change of the ice in the reversible isobaric process is calculated as:
$\Delta S=n{c}_p\ln \frac{T_2}{T_1}$
n is 1, $c_p\left(ice\right)=38J{K}^{-1}mo{l}^{-1}$,${\mathrm{T}}_2=0°\mathrm{C}=273\hspace{0.17em}\mathrm{K}$and,${\mathrm{T}}_1=-20°\mathrm{C}=253\mathrm{K}$

On substituting in the above equation, we get
$$\begin{gathered} \Delta S=n{c}_p\ln \frac{T_2}{T_1} \\ =1mol\times 38J{K}^{-1}mo{l}^{-1}\times \ln \frac{273 K}{253K} \\ =2.89 J{K}^{-1} \end{gathered}$$
Thus, the entropy change of the ice in the reversible isobaric process is .

Theentropy change of the water in the reversible isobaric process is calculated as:
$\Delta S=n{c}_p\ln \frac{T_2}{T_1}$
Here, n is 1, $c_p\left(water\right)=75J{K}^{-1}mo{l}^{-1}$, ${\mathrm{T}}_1=-0°\mathrm{C}=273\mathrm{K}$and${\mathrm{T}}_2=20°\mathrm{C}=293\mathrm{K}$

On substituting all the known values, we get
$$\begin{gathered} \Delta S=n{c}_p\ln \frac{T_2}{T_1} \\ =1mol\times 75J{K}^{-1}mo{l}^{-1}\times \ln \frac{293 K}{273K} \\ =5.30 J{K}^{-1} \end{gathered}$$
Thus, the entropy change of the water in the reversible isobaric process is $5.30 J{K}^{-1}$.

The entropy of system, surrounding and universe.

The$∆\mathrm{S}$is calculated by using the formula:
$\Delta S_{fus}=\frac{\Delta H_{fus}}{T_f}$
The$\Delta H_{fus}$ is 6007 J and the final temperature in K is 273 K. On substituting the given values, we get
$$\begin{gathered} \Delta S_{fus}=\frac{\Delta H_{fus}}{T_f} \\ =\frac{6007J}{273K} \\ =22.00J{K}^{-1} \end{gathered}$$
The$\Delta S_{sys}$ is calculated by using the formula as:
$\Delta S_{sys}=\Delta S_{ice}+\Delta S_{water}+\Delta S_{fus}$
Here, $\Delta S_{ice}$is $2.89J{K}^{-1}$, $\Delta S_{water}$is $5.30J{K}^{-1}$and $\Delta S_{fusion}$is$22.00J{K}^{-1}$

Substituting the given values, we get

$$\begin{gathered} \Delta S_{sys}=\Delta S_{ice}+\Delta S_{water}+\Delta S_{fus} \\ =2.89 J{K}^{-1}+5.30J{K}^{-1}+22.00J{K}^{-1} \\ =30.2J{K}^{-1} \end{gathered}$$

Thus, the entropy of the system is $30.2J{K}^{-1}$
In a reversible process total entropy of the system plus its surrounding is zero. The entropy of the surrounding is calculated as:
$$\begin{gathered} \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\Delta S_{sys}+\Delta S_{surr}=0 \\ 30.2J{K}^{-1}+\Delta S_{surr}=0 \\ \Delta S_{surr}=-30.2J{K}^{-1} \end{gathered}$$
Thus, theentropy of the surroundingis . The entropy of the universe is calculated as:
$$\begin{gathered} \Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr} \\ =30.2J{K}^{-1}+\left(-30.2J{K}^{-1}\right) \\ =0 \end{gathered}$$
Thus, theentropy of the universe is 0.