To an evacuated vessel with movable piston under external pressure of 1 atm and 0.1mole of He and 1 mole of an unknown compound (vapour pressure 0.68 atm at `0^(@)`C) are introduced. Considering the ideal behaviour, the volume (in litre) of the gases at `0^(@)`C is close to:

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have:
- 0.1 moles of Helium (He)
- 1 mole of an unknown compound with a vapor pressure of 0.68 atm at 0°C
- External pressure is 1 atm

### Step 2: Calculate the partial pressure of Helium
Since the total pressure (P_total) is 1 atm and the vapor pressure of the unknown compound (P_vapor) is 0.68 atm, we can find the partial pressure of Helium (P_He) using the formula:
\[ P_{\text{He}} = P_{\text{total}} - P_{\text{vapor}} \]
\[ P_{\text{He}} = 1 \, \text{atm} - 0.68 \, \text{atm} = 0.32 \, \text{atm} \]

### Step 3: Use the Ideal Gas Law to find the volume of Helium
The Ideal Gas Law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure in atm
- \( V \) = volume in liters
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin (0°C = 273 K)

We need to rearrange the equation to solve for volume \( V \):
\[ V = \frac{nRT}{P} \]

### Step 4: Substitute the values for Helium
For Helium:
- \( n = 0.1 \, \text{moles} \)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 273 \, \text{K} \)
- \( P = 0.32 \, \text{atm} \)

Now substituting these values into the equation:
\[ V = \frac{0.1 \, \text{moles} \times 0.0821 \, \text{L·atm/(K·mol)} \times 273 \, \text{K}}{0.32 \, \text{atm}} \]

### Step 5: Calculate the volume
Calculating the numerator:
\[ 0.1 \times 0.0821 \times 273 = 2.24373 \]

Now, divide by the pressure:
\[ V = \frac{2.24373}{0.32} \approx 7.015 \, \text{L} \]

### Conclusion
The volume of the gases at 0°C is approximately 7 liters.

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