One mole of a gas is put under a weightless piston of a vertical cylinder at temperature T. The space over the piston opens into atmosphere. Initially, piston was in equilibrium. How much work should be performed by some external force to increase isothermally the volume under the piston to twice the volume? (Neglect friction of piston).

Explanation

To solve the problem of how much work should be performed by some external force to increase isothermally the volume under the piston to twice the volume, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- We have 1 mole of an ideal gas at temperature \( T \).
- The initial volume of the gas is \( V_0 \).
- The final volume after expansion is \( V_f = 2V_0 \).
- The pressure inside the cylinder initially equals the atmospheric pressure \( P_0 \).

2. **Identify the Work Done**:
- The work done \( W \) during an isothermal expansion can be calculated using the formula:
\[ W = \int_{V_0}^{V_f} P \, dV \]
- Since the process is isothermal, we can express the pressure \( P \) using the ideal gas law:
\[ P = \frac{nRT}{V} \]
- For 1 mole of gas (\( n = 1 \)), this simplifies to:
\[ P = \frac{RT}{V} \]

3. **Set Up the Integral**:
- Substitute \( P \) into the work integral:
\[ W = \int_{V_0}^{2V_0} P_0 \, dV - \int_{V_0}^{2V_0} \frac{RT}{V} \, dV \]
- The first term represents the work done against the atmospheric pressure, and the second term represents the work done by the gas.

4. **Evaluate the Work Done Against Atmospheric Pressure**:
- The work done against the atmospheric pressure is:
\[ W_{atm} = P_0 (V_f - V_0) = P_0 (2V_0 - V_0) = P_0 V_0 \]

5. **Evaluate the Work Done by the Gas**:
- The integral \( \int \frac{RT}{V} \, dV \) is:
\[ W_{gas} = RT \ln \left( \frac{V_f}{V_0} \right) = RT \ln \left( \frac{2V_0}{V_0} \right) = RT \ln(2) \]

6. **Combine the Results**:
- The total work done \( W \) is:
\[ W = P_0 V_0 - RT \ln(2) \]

7. **Final Expression**:
- Thus, the work performed by the external force to increase the volume under the piston to twice the volume is:
\[ W = P_0 V_0 - RT \ln(2) \]