Kvant Physics Problem 47

Verified: yes
Verdicts: PASS + PASS
Solve time: 13m11s
Source on kvant.digital

Problem

If an ideal gas undergoes the process $ABC$ (Fig. 1), it receives an amount of heat $Q$. What amount of heat is transferred to the gas during the process $ADC$?

Fig. 1

Moscow City Physics Olympiad (1970)

Solution to patch

The missing figure prevents numerical evaluation of the heat along either path, but the comparison of the two processes can still be derived exactly from thermodynamics, and this is sufficient to determine how the answer must be constructed once the diagram is specified.

For any quasi-static process of an ideal gas, the first law of thermodynamics gives

$Q = \Delta U + W,$

where $W = \int p,dV$ is the work done by the gas. The internal energy of an ideal gas depends only on temperature, so it depends only on the state points. Both processes $ABC$ and $ADC$ start at $A$ and end at $C$, hence the change in internal energy is identical for both paths,

$\Delta U_{A\to C}^{(ABC)} = \Delta U_{A\to C}^{(ADC)} = U_C - U_A.$

Writing the first law for each path gives

$Q_{ABC} = (U_C - U_A) + W_{ABC},$

$Q_{ADC} = (U_C - U_A) + W_{ADC}.$

Subtracting these expressions eliminates the internal energy term,

$Q_{ABC} - Q_{ADC} = W_{ABC} - W_{ADC}.$

The difference of the works depends only on the geometry of the closed contour formed by the two paths. Combining the forward path $A \to B \to C$ with the reverse of the second path $C \to D \to A$ produces the closed loop $A \to B \to C \to D \to A$, so

$W_{ABC} - W_{ADC} = \oint_{ABCD} p,dV,$

where the sign is fixed by the orientation of the cycle.

Geometrically, $\oint p,dV$ equals the algebraic area enclosed by the curve in the $p$–$V$ plane. Therefore,

$Q_{ADC} = Q_{ABC} - \oint_{ABCD} p,dV.$

At this stage, the problem reduces to evaluating the signed area enclosed by the path $A B C D$ in the provided diagram. Once the coordinates of the vertices or the specific shape of the segments are known, this integral becomes a sum of simple geometric areas, since each segment in a typical Kvant construction is either isobaric or isochoric and contributes either a rectangle or zero work.

Thus the required heat for process $ADC$ is determined completely by subtracting the loop area in the $p$–$V$ diagram from the given value $Q$ for process $ABC$.