Kvant Physics Problem 85
An ideal gas in equilibrium is described by states $(p,V)$ with $p>0$, $V>0$ and equation of state $pV=\nu RT$.
Verified: no
Verdicts: FAIL + FAIL
Solve time: 9m00s
Source on kvant.digital
Problem
F85. Is it possible to carry out a closed process (cycle) with an ideal gas such that points A and B (Fig. 1) lie on the same isotherm?
The temperatures $T_1$ and $T_2$ corresponding to the isotherms drawn in the figure, and the volume $V_1$ (or $V_3$), are given.
Fig. 1: A closed curve passing through points A and B is shown on a $p$–$V$ diagram. Two isotherms of an ideal gas with temperatures $T_1$ and $T_2$ are also shown, and the values $V_1$ and $V_3$ are marked on the volume axis.
Setup and Assumptions
An ideal gas in equilibrium is described by states $(p,V)$ with $p>0$, $V>0$ and equation of state $pV=\nu RT$. The temperature is uniquely determined by $T(p,V)=\frac{pV}{\nu R}$.
The diagram represents not the full $(p,V)$-plane but a restricted physically relevant domain bounded by two isotherms $T_1$ and $T_2$ and by the vertical lines $V=V_1$ and $V=V_3$. This defines an admissible region
$$\mathcal{D}={(p,V)\mid V_1\le V\le V_3,; \nu RT_1 \le pV \le \nu RT_2}.$$
Points $A(p_A,V_A)$ and $B(p_B,V_B)$ lie on the same isotherm, hence
$p_A V_A = p_B V_B = \nu R T_*.$
Since both points are shown inside the figure, they belong to $\mathcal{D}$, which implies
$$T_1 \le T_* \le T_2, \quad V_1 \le V_A,V_B \le V_3.$$
A quasistatic process must therefore be constructed entirely inside $\mathcal{D}$, not merely in the abstract $(p,V)$ state space.
The admissible elementary processes remain isobaric and isochoric transformations, provided that the entire trajectory stays within $\mathcal{D}$.
Physical Principles
A quasistatic process for an ideal gas is a continuous curve in the equilibrium manifold where at each point the state satisfies $pV=\nu RT$. Realizability within the diagram requires an additional constraint: the trajectory must remain inside the domain $\mathcal{D}$ determined by the isotherms $T_1,T_2$ and the volume bounds $V_1,V_3$.
An isobaric segment $p=p_0$ is physically admissible inside $\mathcal{D}$ only as long as the corresponding temperature satisfies
$$T_1 \le \frac{p_0 V}{\nu R} \le T_2,$$
which is equivalent to
$$\frac{\nu R T_1}{p_0} \le V \le \frac{\nu R T_2}{p_0}.$$
An isochoric segment $V=V_0$ is admissible only when
$$T_1 \le \frac{pV_0}{\nu R} \le T_2,$$
which gives
$$\frac{\nu R T_1}{V_0} \le p \le \frac{\nu R T_2}{V_0}.$$
These inequalities ensure that intermediate states remain inside the region represented in the figure.
Derivation
From $A(p_A,V_A)$, consider an isobaric transformation at $p=p_A$ toward volume $V_B$. Along this path,
$$T(V)=\frac{p_A V}{\nu R}.$$
At the endpoints,
$$T(V_A)=T_, \quad T(V_B)=T_,$$
so both endpoints lie in the admissible interval $[T_1,T_2]$. Since $T(V)$ is an affine function of $V$, it attains all intermediate values between $T(V_A)$ and $T(V_B)$, hence for all $V$ between $V_A$ and $V_B$,
$$T_1 \le T(V) \le T_2.$$
The volume also satisfies $V_1 \le V \le V_3$ by construction of $A$ and $B$ inside the diagram. This ensures the entire isobaric segment $A \to C$ remains in $\mathcal{D}$, where $C=(p_A,V_B)$.
From $C$ to $B(p_B,V_B)$, the process is isochoric with $V=V_B$. Along this segment,
$$T(p)=\frac{pV_B}{\nu R}.$$
At the endpoints,
$$T(p_A)=T_, \quad T(p_B)=T_,$$
and since $T(p)$ depends linearly on $p$, all intermediate pressures satisfy
$$T_1 \le T(p) \le T_2.$$
Because $p$ varies between $p_A$ and $p_B$ while $V_B$ is fixed with $V_1 \le V_B \le V_3$, the entire isochoric segment remains inside $\mathcal{D}$. This completes a physically admissible path from $A$ to $B$.
The same verification applies to the reverse construction. From $B$ to $(p_B,V_A)$ the isobaric segment stays in $\mathcal{D}$ because the temperature varies linearly between the common endpoint value $T_*$ and remains within $[T_1,T_2]$, and the volume remains inside $[V_1,V_3]$. The subsequent isochoric step from $(p_B,V_A)$ back to $A$ also remains inside $\mathcal{D}$ for the same linearity reason.
The concatenation of these two admissible paths produces a closed loop entirely contained in $\mathcal{D}$.
Result
A closed quasistatic cycle passing through $A$ and $B$ exists because both states lie inside the bounded domain defined by $T_1,T_2,V_1,V_3$, and because isobaric and isochoric segments connecting them preserve the inequalities defining this domain through linear variation of $T$ along each segment.
Therefore,
$$\boxed{\text{Yes, such a closed cycle is possible.}}$$
Sanity Checks
Each segment satisfies $pV=\nu RT$ at every intermediate point, ensuring equilibrium. The admissibility conditions derived from the diagram, expressed as inequalities in $T$ and $V$, remain satisfied because temperature varies linearly with the single controlled variable in both isobaric and isochoric processes. This guarantees that no part of the constructed cycle leaves the region bounded by the isotherms $T_1$ and $T_2$ and the volume limits $V_1$ and $V_3$.
The construction depends only on the placement of $A$ and $B$ inside the admissible domain and not on additional geometric assumptions about the shape of the boundary beyond those encoded by the isotherms and volume constraints.