1. Freezing point of the solution

For a non-electrolyte, $\Delta T_f = K_f m$ Given:

• $K_f = 2.0\ \text{K kg mol}^{-1}$
• $m = 0.5\ \text{mol kg}^{-1}$

So, $\Delta T_f = 2.0 \times 0.5 = 1.0\ \text{K}$

Hence the freezing point of the aqueous ethylene glycol solution is $T = 273 - 1 = 272\ \text{K}$

So initially the gas is in thermal equilibrium at $T_1 = 272\ \text{K}$

2. Initial pressure of the gas

Given:

• $n = 0.1\ \text{mol}$
• $V_1 = 1.0\ \text{dm}^3$
• $R = 0.08\ \text{dm}^3\text{ atm K}^{-1}\text{ mol}^{-1}$
• $T_1 = 272\ \text{K}$

Using ideal gas equation, $P_1V_1 = nRT_1$ $P_1 = \frac{nRT_1}{V_1} = \frac{0.1\times 0.08\times 272}{1.0}$ $P_1 = 2.176\ \text{atm}$

3. What happens when stoppers are withdrawn?

The solution is at its freezing point and is in contact with a large thermal reservoir. When the stoppers are withdrawn, the gas expands against atmospheric pressure. Since the system remains in contact with a large bath at constant temperature, the final equilibrium temperature of the gas remains $T_2 = 272\ \text{K}$

At mechanical equilibrium, the final pressure of the gas becomes atmospheric pressure: $P_2 = 1\ \text{atm}$

4. Final volume of the gas

Use ideal gas law at final state: $P_2V_2 = nRT_2$ $V_2 = \frac{nRT_2}{P_2} = \frac{0.1\times 0.08\times 272}{1}$ $V_2 = 2.176\ \text{dm}^3$

Since $1\ \text{dm}^3 = 1\ \text{L}$, $V_2 = 2.176\ \text{L}$

5. Final integer-style/numerical result

Thus the volume after equilibrium is achieved is approximately $\boxed{2.18\ \text{L}}$

6. Comparison with stored correct answer

Stored correct answer: $2.17$ to $2.23$

Our derived value is $2.176\ \text{L}$ which lies within the given range. So the answer agrees.