An ideal monatomic gas at initial temperature${\mathit{T}}_{\mathbf{0}}$(in Kelvins) expands from initial volume${\mathit{V}}_{\mathbf{0}}$ to volume$\mathbf{2}{\mathit{V}}_{\mathbf{0}}$by each of the five processes indicated in the T-Vdiagram of Fig. 20-20. In which process is the expansion (a) isothermal, (b) isobaric (constant pressure), and (c) adiabatic? Explain your answers. (d) In which processes does the entropy of the gas decrease?

An ideal monatomic gas expands from initial volume $V_0$to$2V_0$

From the condition, for an isothermal expansion, we can predict which path shows isothermal expansion. We can use the ideal gas law to determine which graph is showing isobaric expansion. Using the first law of thermodynamics we can predict which path shows adiabatic expansion. Then using the formula for entropy change we can predict the path which shows a decrease in entropy change.

Formulae:

The change in temperature during an isothermal expansion, $∆T=\mathrm{constant}$ …(i)

The change in entropy of a gas, $∆S=\frac{Q}{T}$ …(ii)

The change in internal energy due to first law of thermodynamics, $∆E_{int}=Q-W$ …(iii)

The ideal gas equation, $PV=nRT$ …(iv)

The change in work done at constant pressure, $\int PdV$ …(v)

The change in internal energy at constant process, $∆E_{int}=nC∆T$ …(vi)

During an isothermal expansion, the temperature of the system remains constant considering equation (i). In given $T-V$plot, the temperature remains constant along path AE.

The pressure of the system remains constant in an isobaric expansion. During this process, the temperature of the system increases.

Now from equation (iv), we get the following condition as follows:

$$\begin{gathered} \frac{T}{V}=\frac{P}{nR} \\ =\mathrm{Constant} \end{gathered}$$

Thus, the ratio relation of temperatures and volumes can be given using the above value as follows:

$$\begin{gathered} \frac{T_1}{V_1}=\frac{T_2}{V_2} \\ \frac{T_1}{T_2}=\frac{V_1}{V_2} \end{gathered}$$

From the graph, we can use$V_1=V_0$and$V_2=2V_0$

Thus, the above equation gives

$\frac{T_1}{T_2}=\frac{1}{2}$

Here,$\frac{1}{2}$ represents the slope of the graph.

From graph, we can see that only AC has slope $\frac{1}{2}$.

Therefore, AC corresponds to an isobaric expansion.

The system does not add or leave heat in an adiabatic expansion.

According to the first law of thermodynamics, if$Q=0$

Thus, the change in internal energy can be given using equation (iii) as follows:

$$\begin{gathered} ∆E_{int}=-W \\ =-\int PdV\left(\mathrm{from}\mathrm{equation}\left(\mathrm{v}\right)\right) \end{gathered}$$

But, from equation (vi), we can say that$∆E_{int}\alpha ∆T$

It indicates that the temperature of the gas decreases with increase in the volume of the gas.

Therefore, the adiabatic expansion takes place along path AF.

The change of entropy is inversely proportional to the temperature of the ideal gas from equation (ii), with decrease in temperature, randomness(entropy) decreases because the motion of the particles decreases and their velocity decreases so they have less entropy at a lower temperature thus along the path AF, entropy of the gas decreases.

Hence along path AF, the entropy of the gas decreases.