`10` litre of a monoatomic ideal gas at `0^(@)C` and `10 atm` pressure is suddenly released to `1 atm` pressure and the gas expands adiabatically against this constant pressure. The final volume `(L)` of the gas.

Explanation

To solve the problem step by step, we will follow the principles of thermodynamics, particularly focusing on the adiabatic expansion of an ideal gas.

### Step 1: Identify the Given Data
- Initial volume, \( V_1 = 10 \, \text{liters} \)
- Initial temperature, \( T_1 = 0^\circ C = 273 \, \text{K} \)
- Initial pressure, \( P_1 = 10 \, \text{atm} \)
- Final pressure, \( P_2 = 1 \, \text{atm} \)
- The gas is monoatomic, so \( \gamma = \frac{5}{3} \)

### Step 2: Use the Adiabatic Condition
For an adiabatic process, the work done can be expressed in two ways:
1. Work done by the gas against external pressure:
\[ W = -P_{\text{external}} \Delta V = -P_{\text{external}} (V_2 - V_1) \]
Here, \( P_{\text{external}} = P_2 = 1 \, \text{atm} \).

2. Work done in terms of initial and final states:
\[ W = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1} \]

### Step 3: Set the Two Expressions for Work Equal
Since both expressions represent the work done during the adiabatic expansion, we can set them equal to each other:
\[ -P_{\text{external}} (V_2 - V_1) = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1} \]
Substituting the known values:
\[ -1(V_2 - 10) = \frac{1 \cdot V_2 - 10 \cdot 10}{\frac{5}{3} - 1} \]

### Step 4: Simplify the Equation
First, calculate \( \gamma - 1 \):
\[ \gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3} \]
Now substituting this back into the equation:
\[ -(V_2 - 10) = \frac{V_2 - 100}{\frac{2}{3}} \]
Multiplying both sides by \( \frac{2}{3} \):
\[ -\frac{2}{3}(V_2 - 10) = V_2 - 100 \]
Distributing on the left side:
\[ -\frac{2}{3}V_2 + \frac{20}{3} = V_2 - 100 \]

### Step 5: Rearranging the Equation
Bringing all terms involving \( V_2 \) to one side:
\[ -\frac{2}{3}V_2 - V_2 = -100 - \frac{20}{3} \]
Converting \( V_2 \) to a common denominator:
\[ -\frac{2}{3}V_2 - \frac{3}{3}V_2 = -100 - \frac{20}{3} \]
This simplifies to:
\[ -\frac{5}{3}V_2 = -\frac{300}{3} - \frac{20}{3} = -\frac{320}{3} \]

### Step 6: Solve for \( V_2 \)
Multiplying both sides by -1:
\[ \frac{5}{3}V_2 = \frac{320}{3} \]
Now, multiplying both sides by \( \frac{3}{5} \):
\[ V_2 = \frac{320}{5} = 64 \, \text{liters} \]

### Final Answer
The final volume \( V_2 \) of the gas after adiabatic expansion is:
\[ \boxed{64 \, \text{liters}} \]