The value on a cyclinder containing initially `1` litres of an ideal gas at `7 atm` and `25^(@)C` is opened to the atmosphere, Whose the pressure is `760` torr and the temperature is `25^(@)C`. Assuming that the process is isothermal, how much work ( in `L atm`) is done on hte atmosphere by the action of expansion ?

Explanation

To solve the problem step by step, we will follow the principles of thermodynamics related to isothermal processes and the work done during the expansion of an ideal gas.

### Step 1: Identify Initial and Final Conditions
- **Initial Conditions**:
- Initial volume (V1) = 1 liter
- Initial pressure (P1) = 7 atm
- Temperature (T) = 25°C (constant)

- **Final Conditions**:
- Final pressure (P2) = 1 atm (atmospheric pressure)

### Step 2: Use the Ideal Gas Law
Since the process is isothermal, we can use the relationship \( P_1 V_1 = P_2 V_2 \) to find the final volume (V2).

### Step 3: Calculate Final Volume (V2)
Using the equation:
\[ P_1 V_1 = P_2 V_2 \]
Substituting the known values:
\[ 7 \, \text{atm} \times 1 \, \text{L} = 1 \, \text{atm} \times V_2 \]
Solving for \( V_2 \):
\[ V_2 = \frac{7 \, \text{atm} \times 1 \, \text{L}}{1 \, \text{atm}} = 7 \, \text{L} \]

### Step 4: Calculate Work Done (W)
The work done by the gas during expansion against the atmospheric pressure is given by:
\[ W = -P_{atm} \Delta V \]
Where:
- \( \Delta V = V_2 - V_1 \)
- \( P_{atm} = 1 \, \text{atm} \)

Calculating \( \Delta V \):
\[ \Delta V = V_2 - V_1 = 7 \, \text{L} - 1 \, \text{L} = 6 \, \text{L} \]

Now substituting into the work equation:
\[ W = -1 \, \text{atm} \times 6 \, \text{L} = -6 \, \text{L atm} \]

### Final Answer
The work done on the atmosphere by the action of expansion is:
\[ W = -6 \, \text{L atm} \]