New solution

Valter edited Sep 29, 2026, 6:31 PM

View solution History

revision #22155

en/8.2.32.md +143 −0

		@@ -0,0 +1,143 @@
		+	### Statement
		+
		+	$8.2.32^\*.$ When a capacitor with a charge of $q$ is discharged, a mass $m$ of rattlesnake gas is released through an electrolytic bath with acidified water. The mass of the substance released during electrolysis depends only on the passed charge. So, by discharging the capacitor through $k$ series-connected baths, we get the mass $km$ of rattlesnake gas. By burning this gas, we will get a lot of energy. For a sufficiently large $k$, this energy will exceed the original energy of the charged capacitor! Therefore, in some ways our reasoning is wrong. Find this error.
		+
		+	### Solution
		+
		+	Let the capacitance of the capacitor be $C$, and let its initial charge be $q$. The initial voltage across the capacitor is therefore
		+
		+	$$
		+	U_0=\frac{q}{C}.
		+	$$
		+
		+	According to Faraday's law of electrolysis, the mass of a substance produced at an electrode is proportional to the total charge that passes through the electrolyte:
		+
		+	$$
		+	m=\beta Q,
		+	$$
		+
		+	where $\beta$ is the electrochemical equivalent and $Q$ is the charge that has passed through the electrolytic cell.
		+
		+	At first sight, one might argue that if $k$ identical electrolytic cells are connected in series, the same charge $q$ passes through every cell. If one cell produces a mass $m$ of oxyhydrogen gas, then $k$ cells would apparently produce
		+
		+	$$
		+	m_{\Sigma}=km.
		+	$$
		+
		+	For sufficiently large $k$, the chemical energy obtainable by burning this gas would then seem capable of exceeding the initial energy stored in the capacitor.
		+
		+	The flaw in this argument is the implicit assumption that the capacitor can discharge completely through an arbitrarily large number of electrolytic cells.
		+
		+	During electrolysis, each cell develops a chemical back electromotive force directed against the discharge current. Let the effective back EMF of one cell be denoted by $\mathcal{E}$.
		+
		+	For $k$ identical cells connected in series, these back EMFs add:
		+
		+	$$
		+	\mathcal{E}_{\Sigma}=k\mathcal{E}.
		+	$$
		+
		+	At the same time, the voltage across the capacitor decreases as it discharges.
		+
		+	If a charge $Q$ has already passed through the circuit, the remaining charge on the capacitor is
		+
		+	$$
		+	q_C=q-Q,
		+	$$
		+
		+	and the capacitor voltage is
		+
		+	$$
		+	U_C=\frac{q-Q}{C}.
		+	$$
		+
		+	The discharge and the electrolysis can continue only while the capacitor voltage exceeds the total chemical back EMF:
		+
		+	$$
		+	U_C>k\mathcal{E}.
		+	$$
		+
		+	As the capacitor discharges, $U_C$ decreases. Eventually the limiting condition is reached:
		+
		+	$$
		+	\frac{q-Q}{C}=k\mathcal{E}.
		+	$$
		+
		+	At this point the current vanishes and the electrolysis stops.
		+
		+	Therefore, the total charge that actually passes through the cells is
		+
		+	$$
		+	Q=q-Ck\mathcal{E}.
		+	$$
		+
		+	Thus, the charge passing through each cell is not equal to the initial capacitor charge $q$.
		+
		+	By Faraday's law, the mass of gas produced in one cell is
		+
		+	$$
		+	m_1=\beta Q.
		+	$$
		+
		+	Hence the total mass produced in all $k$ cells is
		+
		+	$$
		+	m_{\Sigma}=k\beta Q.
		+	$$
		+
		+	Substituting the expression for $Q$, we obtain
		+
		+	$$
		+	m_{\Sigma}=k\beta\left(q-Ck\mathcal{E}\right).
		+	$$
		+
		+	Thus,
		+
		+	$$
		+	\boxed{m_{\Sigma}=\beta\left(kq-C\mathcal{E}k^2\right)}.
		+	$$
		+
		+	Therefore, the total mass of gas does not increase linearly with the number of cells $k$. The reason is that increasing $k$ increases not only the number of electrolytic cells, but also the total chemical back EMF. As a result, the charge $Q$ that can pass through the circuit before the discharge stops decreases as $k$ increases:
		+
		+	$$
		+	Q=q-Ck\mathcal{E}.
		+	$$
		+
		+	In other words, increasing the number of cells simultaneously increases the number of cells producing gas and decreases the charge passing through each of them.
		+
		+	Moreover, if the number of cells is so large that
		+
		+	$$
		+	k\mathcal{E}\geq U_0,
		+	$$
		+
		+	then the initial capacitor voltage is not sufficient to drive electrolysis at all.
		+
		+	Since
		+
		+	$$
		+	U_0=\frac{q}{C},
		+	$$
		+
		+	this condition may also be written as
		+
		+	$$
		+	k\mathcal{E}\geq\frac{q}{C}.
		+	$$
		+
		+	In this case,
		+
		+	$$
		+	Q=0,
		+	$$
		+
		+	and no gas is produced.
		+
		+	Therefore, there is no contradiction with energy conservation. Faraday's law itself remains valid: the amount of substance produced is indeed determined by the charge that actually passes through the electrolyte. The error lies in assuming that the entire initial charge $q$ of the capacitor passes through every one of the $k$ cells.
		+
		+	As the number of cells increases, their total chemical back EMF increases, and the capacitor ceases to discharge before its charge reaches zero.
		+
		+	#### Answer
		+
		+	The capacitor does not discharge completely because electrolysis produces a chemical back EMF. For $k$ electrolytic cells connected in series, the total back EMF increases approximately in proportion to $k$. The discharge stops when the capacitor voltage falls to the value of this total back EMF.
		+
		+	Therefore, the charge passing through the cells is smaller than the initial charge $q$, and the conclusion that an arbitrarily large mass $km$ of gas can be produced is incorrect.

	@@ -0,0 +1,143 @@
			### Statement

			$8.2.32^\*.$ When a capacitor with a charge of $q$ is discharged, a mass $m$ of rattlesnake gas is released through an electrolytic bath with acidified water. The mass of the substance released during electrolysis depends only on the passed charge. So, by discharging the capacitor through $k$ series-connected baths, we get the mass $km$ of rattlesnake gas. By burning this gas, we will get a lot of energy. For a sufficiently large $k$, this energy will exceed the original energy of the charged capacitor! Therefore, in some ways our reasoning is wrong. Find this error.

			### Solution

			Let the capacitance of the capacitor be $C$, and let its initial charge be $q$. The initial voltage across the capacitor is therefore

			$$
			U_0=\frac{q}{C}.
			$$

			According to Faraday's law of electrolysis, the mass of a substance produced at an electrode is proportional to the total charge that passes through the electrolyte:

			$$
			m=\beta Q,
			$$

			where $\beta$ is the electrochemical equivalent and $Q$ is the charge that has passed through the electrolytic cell.

			At first sight, one might argue that if $k$ identical electrolytic cells are connected in series, the same charge $q$ passes through every cell. If one cell produces a mass $m$ of oxyhydrogen gas, then $k$ cells would apparently produce

			$$
			m_{\Sigma}=km.
			$$

			For sufficiently large $k$, the chemical energy obtainable by burning this gas would then seem capable of exceeding the initial energy stored in the capacitor.

			The flaw in this argument is the implicit assumption that the capacitor can discharge completely through an arbitrarily large number of electrolytic cells.

			During electrolysis, each cell develops a chemical back electromotive force directed against the discharge current. Let the effective back EMF of one cell be denoted by $\mathcal{E}$.

			For $k$ identical cells connected in series, these back EMFs add:

			$$
			\mathcal{E}_{\Sigma}=k\mathcal{E}.
			$$

			At the same time, the voltage across the capacitor decreases as it discharges.

			If a charge $Q$ has already passed through the circuit, the remaining charge on the capacitor is

			$$
			q_C=q-Q,
			$$

			and the capacitor voltage is

			$$
			U_C=\frac{q-Q}{C}.
			$$

			The discharge and the electrolysis can continue only while the capacitor voltage exceeds the total chemical back EMF:

			$$
			U_C>k\mathcal{E}.
			$$

			As the capacitor discharges, $U_C$ decreases. Eventually the limiting condition is reached:

			$$
			\frac{q-Q}{C}=k\mathcal{E}.
			$$

			At this point the current vanishes and the electrolysis stops.

			Therefore, the total charge that actually passes through the cells is

			$$
			Q=q-Ck\mathcal{E}.
			$$

			Thus, the charge passing through each cell is not equal to the initial capacitor charge $q$.

			By Faraday's law, the mass of gas produced in one cell is

			$$
			m_1=\beta Q.
			$$

			Hence the total mass produced in all $k$ cells is

			$$
			m_{\Sigma}=k\beta Q.
			$$

			Substituting the expression for $Q$, we obtain

			$$
			m_{\Sigma}=k\beta\left(q-Ck\mathcal{E}\right).
			$$

			Thus,

			$$
			\boxed{m_{\Sigma}=\beta\left(kq-C\mathcal{E}k^2\right)}.
			$$

			Therefore, the total mass of gas does not increase linearly with the number of cells $k$. The reason is that increasing $k$ increases not only the number of electrolytic cells, but also the total chemical back EMF. As a result, the charge $Q$ that can pass through the circuit before the discharge stops decreases as $k$ increases:

			$$
			Q=q-Ck\mathcal{E}.
			$$

			In other words, increasing the number of cells simultaneously increases the number of cells producing gas and decreases the charge passing through each of them.

			Moreover, if the number of cells is so large that

			$$
			k\mathcal{E}\geq U_0,
			$$

			then the initial capacitor voltage is not sufficient to drive electrolysis at all.

			Since

			$$
			U_0=\frac{q}{C},
			$$

			this condition may also be written as

			$$
			k\mathcal{E}\geq\frac{q}{C}.
			$$

			In this case,

			$$
			Q=0,
			$$

			and no gas is produced.

			Therefore, there is no contradiction with energy conservation. Faraday's law itself remains valid: the amount of substance produced is indeed determined by the charge that actually passes through the electrolyte. The error lies in assuming that the entire initial charge $q$ of the capacitor passes through every one of the $k$ cells.

			As the number of cells increases, their total chemical back EMF increases, and the capacitor ceases to discharge before its charge reaches zero.

			#### Answer

			The capacitor does not discharge completely because electrolysis produces a chemical back EMF. For $k$ electrolytic cells connected in series, the total back EMF increases approximately in proportion to $k$. The discharge stops when the capacitor voltage falls to the value of this total back EMF.

			Therefore, the charge passing through the cells is smaller than the initial charge $q$, and the conclusion that an arbitrarily large mass $km$ of gas can be produced is incorrect.