91Ó°ÊÓ

The number of moles,$\mathrm{n}=4.00\mathrm{mol}$.

The working substance is an ideal gas.

The process 1 is adiabatic and reversible.

An adiabatic expansion gives,$\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}=\frac{1}{2}$

The process 2 is aconstant volume process.

The process 3 is an isothermal compression back to the initial state of the gas.

If the process is reversible, then we can find the net change in entropy. By using Equation 20-1 for the change in the entropy, we can find the entropy change for processes 1, 2, and 3.

Formulae:

The change in the entropy (Equation 20-1) of a cycle,$∆\mathrm{S}={∫}_{\mathrm{i}}^{\mathrm{f}}\frac{\mathrm{dQ}}{\mathrm{T}}$ (1)

For a reversible set of processes, the net change in entropy,

$∆{\mathrm{S}}_{\mathrm{net}}=0$ (2)

It is a reversible set of processes returning the system to its initial state. So, its net change in entropy using equation (2) is$∆{\mathrm{S}}_{\mathrm{net}}=0$

The process is adiabatic and reversible, that is, $\mathrm{dQ}=0$. So, from equation (1),the change in the entropy is given by:

$$\begin{gathered} \left(\mathrm{If}\mathrm{dQ}=\mathrm{pdV}\right) \\ ∆{\mathrm{S}}_1=0 \end{gathered}$$

Hence, the change in entropy for process 1 is zero.

Since the working substance is an ideal gas, an isothermal process implies

$$\begin{gathered} ∫\frac{\mathrm{dQ}}{\mathrm{T}}=∫\frac{\mathrm{pdV}}{\mathrm{T}} \\ =∫\frac{\mathrm{pdV}}{\left({\displaystyle \frac{\mathrm{pV}}{\mathrm{nR}}}\right)} \\ =\mathrm{nR}∫\frac{\mathrm{dV}}{\mathrm{V}} \end{gathered}$$

Thus, the change in entropy change for process 3 is given as:

$$\begin{gathered} ∆{\mathrm{S}}_3=\mathrm{nRIn}\left(\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\right) \\ =\mathrm{nRIn}\left(\frac{1}{2}\right) \\ =4.00\mathrm{mol}×8.31\mathrm{J}/\mathrm{mol}·\mathrm{K}×\mathrm{In}\left(\frac{1}{2}\right) \\ =-23.0\mathrm{J}/\mathrm{K} \end{gathered}$$.

Hence, the value of the change in internal energy is$-23.0\mathrm{J}/\mathrm{K}$

From the solution of (a), we can get that

$$\begin{gathered} ∆{\mathrm{S}}_1+∆{\mathrm{S}}_2+∆{\mathrm{S}}_3=0 \\ ∆{\mathrm{S}}_2=-\left(∆{\mathrm{S}}_1+∆{\mathrm{S}}_3\right) \\ ∆{\mathrm{S}}_2=-\left(-23.0\mathrm{J}/\mathrm{K}\right) \\ ∆{\mathrm{S}}_2=23.0\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the change in entropy is$23.0\mathrm{J}/\mathrm{K}$