This problem involves a mixture of gases in a container with a movable piston under constant external pressure. We are given 0.1 mol of He (an ideal gas) and 1.0 mol of an unknown compound. The key information is that this unknown compound has a vapour pressure of 0.68 atm at 0°C. This vapour pressure is a crucial clue. For a pure substance, the vapour pressure is the pressure exerted by its vapour when it is in equilibrium with its liquid (or solid) phase at a given temperature. Here, at 0°C, the vapour pressure is 0.68 atm. This means if this compound were alone in a closed container at 0°C, the pressure above its liquid would be 0.68 atm.

Now, the vessel is evacuated and has a movable piston under a fixed external pressure of 1 atm. When we introduce the gases, the piston will move until the pressure inside the vessel equals the external pressure applied on the piston. Therefore, the total pressure of the gas mixture inside the vessel at equilibrium will be $P_{\text{total}}=1\text{atm}$.

We have two components:
1. Helium (He): 0.1 mol. This is an ideal gas and will contribute to the total pressure.
2. Unknown compound: 1.0 mol. However, we cannot assume all of it is in the vapour phase. The vapour pressure tells us the maximum partial pressure this compound can have at this temperature before it starts condensing.

Since the total pressure we need to achieve (1 atm) is greater than the compound's vapour pressure (0.68 atm), the system will adjust. The partial pressure of the unknown compound in the vapour phase cannot exceed its vapour pressure. If it tried to, the excess would condense into its liquid phase.

Therefore, at equilibrium, the partial pressure of the unknown compound will be exactly equal to its vapour pressure:
$P_{\text{unknown}}=0.68\text{atm}$

According to Dalton's Law of partial pressures, the total pressure is the sum of the partial pressures of all gases:
$P_{\text{total}}=P_{\text{He}}+P_{\text{unknown}}$
We know $P_{\text{total}}=1\text{atm}$ and $P_{\text{unknown}}=0.68\text{atm}$.
So, the partial pressure of Helium must be:
$P_{\text{He}}=P_{\text{total}}-P_{\text{unknown}}=1\text{atm}-0.68\text{atm}=0.32\text{atm}$

Now, we treat the helium gas as ideal. The ideal gas law is $PV=nRT$. We will use this to find the volume of the container. The volume is the same for all gases in the mixture. We know the pressure, moles, and temperature for helium.

Given:
For Helium: $P_{\text{He}}=0.32\text{atm}$, $n_{\text{He}}=0.1\text{mol}$
Temperature, $T=0°C=273K$
Gas constant, $R=0.0821L·\mathrm{atm}·{\mathrm{mol}}^{-1}·K^{-1}$

Rearrange the ideal gas law to solve for volume $V$:
$V=\frac{n_{\text{He}}RT}{P_{\text{He}}}$

Substitute the values:
$V=\frac{\left(0.1\text{mol}\right)\times (0.0821L·\mathrm{atm}·{\mathrm{mol}}^{-1}·K^{-1})\times \left(273K\right)}{0.32\text{atm}}$

Calculate step-by-step:
First, calculate $nRT=0.1\times 0.0821\times 273$
$0.1\times 0.0821=0.00821$
$0.00821\times 273\approx 2.24133$
Now, divide by $P_{\text{He}}$:
$V=\frac{2.24133}{0.32}\approx 7.004L$

So, the total volume of the gases (which is the volume of the container) is approximately 7 litres. This volume is occupied by the vapour of the unknown compound and the helium gas. The rest of the unknown compound (the amount beyond what is needed to create a partial pressure of 0.68 atm) exists as a liquid in the container, but the question asks for the volume of the gases, which is this calculated volume.

Related Concepts and Formulae

Dalton's Law of Partial Pressures: In a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.
$P_{\text{total}}=P_1+P_2+P_3+\dots +P_n$

Ideal Gas Law: The relationship between pressure ($P$), volume ($V$), number of moles ($n$), and temperature ($T$) for an ideal gas.
$PV=nRT$

Vapour Pressure: The pressure exerted by a vapour in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. It is a property of the substance and depends only on temperature.