Question

A three-step cycle is undergone reversibly by 4.00 mol. of an ideal gas: (1) an adiabatic expansion that gives the gas 20times its initial volume, (2) a constant-volume process, (3) an isothermal compression back to the initial state of the gas. We do not know whether the gas is monatomic or diatomic; if it is diatomic, we do not know whether the molecules are rotating or oscillating. What are the entropy changes for (a) the cycle, (b) process 1, (c) process 3, and (d) process 2?

Short answer

The entropy change for

1. The cycle is 0.
2. The process 1 is 0 .
3. The process 3 is$-23.0\mathrm{J}/\mathrm{K}$ .
4. The process 2 is $23.0\mathrm{J}/\mathrm{K}$.

Step-by-step solution

The given data

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.

Understanding the concept of thermodynamics relations

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,$∆\mathrm{S}={\int }_{\mathrm{i}}^{\mathrm{f}}\frac{\mathrm{dQ}}{\mathrm{T}}$ (1)

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

$∆{\mathrm{S}}_{\mathrm{net}}=0$ (2)

a) Calculation of entropy change for the full cycle

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

b) Calculation of entropy change for process 1

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.

c) Calculation of entropy change for process 3

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

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

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

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

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

d) Calculation of entropy change for process 2

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$23.0\mathrm{J}/\mathrm{K}$