Statement

Revisit the two adiabatic-compression experiments from lesson 5, recalling the useful results here. A thermally insulated vertical cylinder of cross-sectional area $\mathcal S$ contains $n$ moles of ideal gas. Its piston is also insulated, moves without friction and has negligible mass. The molar heat capacities $c_V$ and $c_P$ are constant, with

\gamma=\frac{c_P}{c_V}>1,\qquad c_V=\frac{R}{\gamma-1},\qquad c_P=c_V+R.

Initially, only atmospheric pressure $P_0$ acts on the piston: the gas is in state $A=(P_0,V_0,T_0)$ , with $P_0V_0=nRT_0$ . An added load of mass $M$ raises the external pressure to

P_1=P_0+\frac{Mg}{\mathcal S}=xP_0,\qquad x>1.

Both experiments start from $A$ . In the first, the load is added and then removed one grain of sand at a time: the gas follows the reversible adiabat $A\to B\to A$ . In the second, the load is applied all at once, equilibrium $B'$ is reached, then the load is removed all at once and equilibrium $B"$ is reached. External pressure is constant during each sudden step: $P_1$ during loading and $P_0$ during unloading. Assume that oscillations are damped by dissipation in the gas; the piston is at rest in the initial and final states and stores no energy.

Both compressions start from A (centre). On the left, the load is gradually added until B; gradual removal returns the gas to A. On the right, the load is applied suddenly until B'; sudden removal leads to another state B”. The diagram shows the compressions; the four states are specified below. — Figure 1. Both compressions start from $A$ (centre). On the left, the load is gradually added until $B$ ; gradual removal returns the gas to $A$ . On the right, the load is applied suddenly until $B'$ ; sudden removal leads to another state $B"$ . The diagram shows the compressions; the four states are specified below.

The adiabatic relations along the reversible path and the first law along the sudden paths gave the following results. Define

\begin{aligned} \theta=\frac{1+(\gamma-1)x}{\gamma},\qquad \phi=\frac{1+(\gamma-1)/x}{\gamma}. \begin{array}{c|ccc} \text{State} & P/P_0 & V/V_0 & T/T_0\\ \hline A & 1 & 1 & 1\\[3pt] B & x & x^{-1/\gamma} & x^{(\gamma-1)/\gamma}\\[3pt] B' & x & \theta/x & \theta\\[3pt] B" & 1 & \theta\phi & \theta\phi \end{array} \end{aligned}

In particular,

\theta>x^{(\gamma-1)/\gamma},\qquad \phi>x^{-(\gamma-1)/\gamma},\qquad \theta\phi-1=\frac{\gamma-1}{\gamma^2}\frac{(x-1)^2}{x}>0.

All steps are adiabatic: $Q=0$ and $W=\Delta U=nc_V(T_f-T_i)$ , with energy received by the gas counted as positive. Thus,

\begin{aligned} W_{AB}&=nc_VT_0\left[x^{(\gamma-1)/\gamma}-1\right], & W_{BA}&=-W_{AB},\\ W_{AB'}&=nc_VT_0(\theta-1), & W_{B'B"}&=nc_VT_0\theta(\phi-1). \end{aligned}

The reversible round trip has zero net work, whereas the sudden path receives $W_{AB'B"}=nc_VT_0(\theta\phi-1)>0$ .

Calculate the entropy changes along $A\to B$ and $B\to A$ , then establish the entropy balance for the reversible round trip.
Calculate $\Delta S_{AB'}$ and show that it is strictly positive. How much of this change is exchanged and how much is produced?
Do the same for sudden unloading $B'\to B"$ .
Calculate $\Delta S_{AB"}$ directly and verify that it equals the sum of the two preceding entropy productions. Does the path $A\to B'\to B"$ constitute a cycle for the gas?
Compare the energy and entropy balances of both round trips. Explain why the absence of heat exchange does not imply entropy conservation.

Hint

Between two equilibrium states of an ideal gas,

\Delta S=nc_V\ln\frac{T_f}{T_i}+nR\ln\frac{V_f}{V_i}.

For $B'\to B"$ , use $T_{B"}/T_{B'}=\phi$ and $V_{B"}/V_{B'}=x\phi$ . The inequalities recalled in the problem determine the signs. At each step, $S_{\mathrm{e}}=0$ .

Detailed solution

Solution

Question 1. The table gives

T_B/T_0=x^{(\gamma-1)/\gamma}

and

V_B/V_0=x^{-1/\gamma}

. Substituting into the entropy formula gives

\begin{aligned} \Delta S_{AB} &=nc_V\ln\left(x^{(\gamma-1)/\gamma}\right) +nR\ln\left(x^{-1/\gamma}\right)\\ &=nc_V\frac{\gamma-1}{\gamma}\ln x-\frac{nR}{\gamma}\ln x\\ &=\frac n\gamma[c_V(\gamma-1)-R]\ln x=0, \end{aligned}

because $c_V(\gamma-1)=R$ . The return follows the same path in the opposite direction, so $\Delta S_{BA}=0$ . For each step and the full round trip, $S_{\mathrm{e}}=S_{\mathrm{i}}=\Delta S=0$ . The gas returns to its state $A$ .

Question 2. For the sudden compression,

T_{B'}/T_0=\theta

and

V_{B'}/V_0=\theta/x

. Using

R=c_V(\gamma-1)

followed by the logarithm rules gives

\begin{aligned} \Delta S_{AB'} &=nc_V\ln\theta+nR\ln\frac{\theta}{x}\\ &=nc_V[\ln\theta+(\gamma-1)(\ln\theta-\ln x)]\\ &=nc_V[\gamma\ln\theta-(\gamma-1)\ln x]\\ &=nc_V\ln\left(\frac{\theta^\gamma}{x^{\gamma-1}}\right). \end{aligned}

The given inequality $\theta>x^{(\gamma-1)/\gamma}$ implies $\theta^\gamma>x^{\gamma-1}$ : the ratio inside the logarithm exceeds 1, hence $\Delta S_{AB'}>0$ . No entropy is exchanged, so $S_{\mathrm{i},AB'}=\Delta S_{AB'}$ . The gas's entropy increases through production without receiving heat.

Question 3. During sudden unloading, the ratios between this step's final and initial states are

\frac{T_{B"}}{T_{B'}}=\frac{\theta\phi}{\theta}=\phi, \qquad \frac{V_{B"}}{V_{B'}}=\frac{\theta\phi}{\theta/x}=x\phi.

Thus,

\begin{aligned} \Delta S_{B'B"} &=nc_V\ln\phi+nR\ln(x\phi)\\ &=nc_V[\ln\phi+(\gamma-1)(\ln x+\ln\phi)]\\ &=nc_V[\gamma\ln\phi+(\gamma-1)\ln x]\\ &=nc_V\ln\left(\phi^\gamma x^{\gamma-1}\right)>0. \end{aligned}

Indeed, $\phi>x^{-(\gamma-1)/\gamma}$ implies $\phi^\gamma x^{\gamma-1}>1$ . Again $S_{\mathrm{e}}=0$ and $S_{\mathrm{i},B'B"}=\Delta S_{B'B"}$ . The gas cools during this expansion ( $\phi<1$ ), but its entropy increases: the volume-increase term dominates.

Question 4. States

A

and

B"

have the same pressure and

T_{B"}/T_0=V_{B"}/V_0=\theta\phi

, hence

\Delta S_{AB"}=nc_V\ln(\theta\phi)+nR\ln(\theta\phi) =nc_P\ln(\theta\phi)>0.

We used $c_P=c_V+R$ and $\theta\phi>1$ . Now add the two preceding results:

\begin{aligned} \Delta S_{AB'}+\Delta S_{B'B"} &=nc_V[\gamma\ln\theta-(\gamma-1)\ln x\\ &\qquad+\gamma\ln\phi+(\gamma-1)\ln x]\\ &=nc_V\gamma(\ln\theta+\ln\phi)\\ &=nc_P\ln(\theta\phi)=\Delta S_{AB"}. \end{aligned}

The $\ln x$ terms cancel. Since each step is adiabatic, this sum also equals the sum of the two entropy productions. Returning the external pressure to $P_0$ is not enough to complete a gas cycle: $B"\ne A$ , since $T_{B"}>T_0$ and $V_{B"}>V_0$ .

Question 5. For

A\to B\to A

,

Q_{\mathrm{tot}}=W_{\mathrm{tot}}=\Delta U_{\mathrm{tot}}=0

and

S_{\mathrm{i},\mathrm{tot}}=0

. For

A\to B'\to B"

,

Q_{\mathrm{tot}}=0,\qquad W_{\mathrm{tot}}=\Delta U_{\mathrm{tot}}=nc_VT_0(\theta\phi-1)>0, S_{\mathrm{e},\mathrm{tot}}=0,\qquad S_{\mathrm{i},\mathrm{tot}}=\Delta S_{\mathrm{tot}}=nc_P\ln(\theta\phi)>0.

The first law describes the net work retained as internal energy; the second quantifies irreversibility through entropy production. An adiabatic process is isentropic only if it produces no entropy, as in the reversible case here.