A cylinder containing an ideal gas (0.1 mol of 1.0 dm^3) is in thermal equilibrium with a large volu

Q: Why is the final pressure of the gas 1 atm?

The liquid is in an open container, meaning it is exposed to the atmospheric pressure of 1 atm. When the stoppers are removed, the frictionless piston moves until the forces on both sides balance out. Thus, the gas pressure must equal the external atmospheric pressure.

Q: Why do we use the freezing point of the solution as the temperature of the gas?

The problem states that the gas is in "thermal equilibrium" with the solution at its freezing point. Thermal equilibrium means both systems have reached the exact same temperature.

Q: Does the volume of the liquid change when the piston moves?

Liquids are generally incompressible. While the piston moves, the liquid is simply displaced upwards in its open container, but its volume remains essentially constant. The change in the gas volume is accommodated by the movement of the piston.

Q: Why did we use $R = 0.08$ instead of $0.0821$?

Always use the values provided in the question! The problem explicitly gives $R = 0.08 \text{ dm}^3 \text{ atm K}^{-1} \text{mol}^{-1}$ to simplify calculations. Using $0.0821$ would lead to a slightly different answer and might cause you to lose marks.

Q: What is the significance of $1 \text{ dm}^3$?

$1 \text{ dm}^3$ is exactly equal to $1 \text{ Liter}$. The gas constant is given in $\text{dm}^3$, so the volume calculated from the ideal gas equation will be in $\text{dm}^3$, which directly translates to Liters.

Question

A cylinder containing an ideal gas ($0.1 	ext{ mol}$ of $1.0 	ext{ dm}^3$) is in thermal equilibrium with a large volume of $0.5 	ext{ molal}$ aqueous solution of ethylene glycol at its freezing point. If the stoppers $S_1$ and $S_2$ (as shown in the figure) are suddenly withdrawn, the volume of the gas in litres after equilibrium is achieved will be ………
(Given, $K_f 	ext{ (water)} = 2.0 	ext{ K kg mol}^{-1}$, $R = 0.08 	ext{ dm}^3 	ext{ atm K}^{-1} 	ext{mol}^{-1}$)

Sigma Percentile · Accepted Answer

The problem presents a fascinating interplay between two seemingly distinct worlds: the thermodynamics of an ideal gas and the colligative properties of a solution. Imagine a cylinder divided by a frictionless piston. On the left, an ideal gas is trapped, eager to expand. On the right, a large volume of an aqueous ethylene glycol solution sits quietly, open to the atmosphere. The piston is locked in place by two stoppers, $S_1$ and $S_2$.

The key to unlocking this problem lies in the phrase **"thermal equilibrium."** This means the gas and the liquid share the exact same temperature. But what is that temperature?

### Phase 1: The Chilling Effect of Ethylene Glycol

The problem states that the aqueous solution is at its freezing point. Pure water freezes at $273 	ext{ K}$, but adding a solute like ethylene glycol lowers this freezing point—a phenomenon known as **freezing point depression**.

We can calculate this depression using the formula:
$$ \Delta T_f = K_f 	imes m $$

We are given the cryoscopic constant for water, $K_f = 2.0 	ext{ K kg mol}^{-1}$, and the molality of the solution, $m = 0.5 	ext{ molal}$. Substituting these values:
$$ \Delta T_f = 2.0 	imes 0.5 = 1.0 	ext{ K} $$

This means the freezing point is depressed by exactly $1 	ext{ K}$. Therefore, the freezing point of our solution is:
$$ T_f = 273 	ext{ K} - 1.0 	ext{ K} = 272 	ext{ K} $$

Because the gas is in thermal equilibrium with the solution, the temperature of the ideal gas is also **$272 	ext{ K}$**.

### Phase 2: Unleashing the Piston

Now, imagine we suddenly pull out the stoppers $S_1$ and $S_2$. The frictionless piston is no longer restrained. It will slide left or right until the forces on both sides perfectly balance out. This state is called **mechanical equilibrium**.

On the right side, the liquid is open to the atmosphere. This means the atmosphere is constantly pushing down on the liquid with a pressure of $1 	ext{ atm}$. According to Pascal's principle, this pressure is transmitted through the liquid to the piston. For the piston to stop moving, the gas on the left must push back with the exact same pressure.

Thus, at equilibrium, the final pressure of the gas is:
$$ P = 1 	ext{ atm} $$

### Phase 3: The Final Expansion

We now know everything about the gas in its final state except its volume. We have the number of moles ($n = 0.1 	ext{ mol}$), the temperature ($T = 272 	ext{ K}$), and the pressure ($P = 1 	ext{ atm}$). We also have the gas constant ($R = 0.08 	ext{ dm}^3 	ext{ atm K}^{-1} 	ext{mol}^{-1}$).

It's time to bring in the trusty ideal gas equation:
$$ PV = nRT $$

Rearranging for volume, we get:
$$ V = \frac{nRT}{P} $$

Let's plug in our numbers:
$$ V = \frac{0.1 	imes 0.08 	imes 272}{1} $$
$$ V = 0.008 	imes 272 $$
$$ V = 2.176 	ext{ dm}^3 $$

Since $1 	ext{ dm}^3$ is exactly equal to $1 	ext{ L}$, our final volume is $2.176 	ext{ L}$. Rounding to two decimal places, we arrive at our final answer: **$2.18 	ext{ L}$**.

A beautiful problem that elegantly bridges the gap between colligative properties and gas laws!

Enter Numerical Value:

Visualized Solution

\text{Analyzing the System}

\text{Temperature of the System}

\Delta T_f = K_f \times m

T_f = 273 \text{ K} - \Delta T_f

\text{Mechanical Equilibrium}

V = \frac{nRT}{P}

V = 2.176 \text{ L}

\text{Final Answer}

The Sigma Insight: Colligative Properties

Phase 1

Phase 2

Phase 3

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