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 of calculating the work done on the atmosphere by the expansion of an ideal gas in a cylinder, we can follow these steps:

### Step 1: Understand the Initial Conditions
We have an ideal gas in a cylinder with the following initial conditions:
- Initial volume (V1) = 1 liter
- Initial pressure (P1) = 7 atm
- Temperature = 25°C (which is constant during the isothermal process)

### Step 2: Convert Atmospheric Pressure to the Same Units
The atmospheric pressure (P2) is given as 760 torr. We need to convert this pressure to atm for consistency:
- 1 atm = 760 torr
- Therefore, P2 = 760 torr = 1 atm

### Step 3: Use the Ideal Gas Law
Since the process is isothermal, we can use the relationship between the initial and final states of the gas:
\[ P_1 V_1 = P_2 V_2 \]
Where:
- P1 = 7 atm
- V1 = 1 liter
- P2 = 1 atm (atmospheric pressure)
- V2 = final volume (unknown)

### Step 4: Solve for Final Volume (V2)
Using the equation from Step 3:
\[ 7 \, \text{atm} \times 1 \, \text{L} = 1 \, \text{atm} \times V_2 \]
\[ V_2 = \frac{7 \, \text{atm} \times 1 \, \text{L}}{1 \, \text{atm}} = 7 \, \text{L} \]

### Step 5: Calculate the Change in Volume (ΔV)
Now, we can find the change in volume (ΔV):
\[ \Delta V = V_2 - V_1 = 7 \, \text{L} - 1 \, \text{L} = 6 \, \text{L} \]

### Step 6: Calculate the Work Done (W)
The work done by the gas during expansion against the atmospheric pressure is given by:
\[ W = -P_{\text{external}} \Delta V \]
Where:
- \( P_{\text{external}} = P_2 = 1 \, \text{atm} \)
- \( \Delta V = 6 \, \text{L} \)

Substituting the values:
\[ W = -1 \, \text{atm} \times 6 \, \text{L} = -6 \, \text{L atm} \]

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

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