Kvant Physics Problem 47

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Problem

If an ideal gas undergoes the process $ABC$ (Fig. 1), it is supplied with a quantity of heat $Q$. What quantity of heat is supplied to the gas in the process $ADC$?

Fig. 1

Moscow City Physics Olympiad (1970)

Setup and Assumptions

An ideal gas is taken through two different quasi-static processes between the same initial state $A$ and final state $C$. The initial state is $A(V_1,p_1)$ and the final state is $C(V_2,p_2)$. The two paths are $ABC$ and $ADC$ on a $pV$ diagram, forming a rectangle.

From the figure, the states are

$A(V_1,p_1), \quad B(V_2,p_1), \quad D(V_1,p_2), \quad C(V_2,p_2),$

with $V_2 = 2V_1$ and $p_2 = 2p_1$.

The gas is ideal with constant heat capacities. The internal energy depends only on temperature. Heat capacities are not required explicitly because the difference of heats will be eliminated through state functions.

The quantity of heat supplied along $ABC$ is given as $Q$. The task is to determine the heat supplied along $ADC$.

Physical Principles

The first law of thermodynamics is written as

$Q = \Delta U + W,$

where $Q$ is the heat supplied to the gas, $\Delta U$ is the change in internal energy, and $W$ is the work done by the gas.

For an ideal gas, the internal energy depends only on temperature, hence $\Delta U$ depends only on the initial and final states and is independent of the path.

The work along a quasi-static process on a $pV$ diagram is

$W = \int p , dV.$

Along an isochoric process, $dV = 0$, so $W = 0$. Along an isobaric process, $W = p \Delta V$.

For a monatomic ideal gas,

$U = \frac{3}{2}nRT = \frac{3}{2}pV,$

so the change in internal energy between states $A$ and $C$ is

$\Delta U = \frac{3}{2}(p_C V_C - p_A V_A).$

Derivation

Along the path $ABC$, the segment $AB$ is isobaric at pressure $p_1$ and the volume changes from $V_1$ to $V_2$. The work is

$W_{AB} = p_1(V_2 - V_1).$

The segment $BC$ is isochoric, hence

$W_{BC} = 0.$

The total work along $ABC$ is

$W_{ABC} = p_1(V_2 - V_1).$

The heat supplied along $ABC$ is given as

$Q = \Delta U + p_1(V_2 - V_1).$

Along the path $ADC$, the segment $AD$ is isochoric, so

$W_{AD} = 0.$

The segment $DC$ is isobaric at pressure $p_2$, so

$W_{DC} = p_2(V_2 - V_1).$

The total work along $ADC$ is

$W_{ADC} = p_2(V_2 - V_1).$

The heat supplied along $ADC$ is

$Q_{ADC} = \Delta U + p_2(V_2 - V_1).$

Subtracting the expressions for $Q$ and $Q_{ADC}$ eliminates $\Delta U$:

$Q_{ADC} - Q = (p_2 - p_1)(V_2 - V_1).$

Substituting $V_2 = 2V_1$ and $p_2 = 2p_1$ gives

$V_2 - V_1 = V_1,$

$p_2 - p_1 = p_1.$

Hence,

$Q_{ADC} - Q = p_1 V_1,$

$Q_{ADC} = Q + p_1 V_1.$

To express $p_1 V_1$ in terms of $Q$, we use the expression for $Q$ in terms of state variables. First compute $\Delta U$:

$\Delta U = \frac{3}{2}(p_2V_2 - p_1V_1).$

With $p_2 = 2p_1$ and $V_2 = 2V_1$,

$p_2V_2 = 4p_1V_1,$

$\Delta U = \frac{3}{2}(4p_1V_1 - p_1V_1) = \frac{3}{2}\cdot 3p_1V_1 = \frac{9}{2}p_1V_1.$

Then

$Q = \frac{9}{2}p_1V_1 + p_1V_1 = \frac{11}{2}p_1V_1.$

Solving for $p_1V_1$ gives

$p_1V_1 = \frac{2}{11}Q.$

Substituting into $Q_{ADC}$ yields

$Q_{ADC} = Q + \frac{2}{11}Q = \frac{13}{11}Q.$

Result

$Q_{ADC} = \frac{13}{11}Q.$

Substituting the given value $Q$ (in joules),

$Q_{ADC} = \frac{13}{11}Q ;\text{J}.$

Sanity Checks

The expression depends only on the ratio of pressures and volumes, consistent with the fact that both processes connect identical endpoints so $\Delta U$ cancels in the difference. The result is dimensionally consistent because it is a dimensionless factor multiplying $Q$.

If $p_2 \to p_1$, the rectangle collapses vertically and both paths coincide in work, giving $Q_{ADC} \to Q$, consistent with the formula since $\frac{13}{11} \to 1$ is not relevant in this limit because the assumed geometry $p_2=2p_1$ is essential to the derivation.

A sign error would most easily occur in the work terms, since reversing $V_2 - V_1$ or swapping the high-pressure and low-pressure branches changes the correction term $p_1V_1$ that determines the final ratio.