One mole of an ideal gas is contained in a vertical cylinder under a massless piston moving without friction. The piston is slowly raised so that the gas expands isothermally at temperature `T_0` = 300 K. Find the amount of work done increasing the volume to `eta`=2 times. The outside pressure is atmospheric.

Answer

To solve the problem of finding the work done during the isothermal expansion of one mole of an ideal gas in a vertical cylinder with a massless piston, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**:
- We have one mole of an ideal gas in a cylinder with a massless piston.
- The gas expands isothermally at a constant temperature \( T_0 = 300 \, \text{K} \).
- The volume of the gas increases to \( \eta = 2 \) times its initial volume \( V_i \).

2. **Identify Initial and Final Volumes**:
- Let the initial volume be \( V_i \).
- The final volume after expansion is \( V_f = 2V_i \).

3. **Work Done in Isothermal Expansion**:
- The work done \( W \) during an isothermal expansion of an ideal gas can be expressed as:
\[ W = \int_{V_i}^{V_f} P \, dV \]
- For an ideal gas, the pressure \( P \) can be expressed using the ideal gas law:
\[ P = \frac{nRT}{V} \]
where \( n \) is the number of moles (1 mole), \( R \) is the universal gas constant, and \( T \) is the temperature.

4. **Substituting Pressure in the Work Formula**:
- Substitute \( P \) in the work formula:
\[ W = \int_{V_i}^{2V_i} \frac{nRT}{V} \, dV \]
- Since \( n = 1 \) and \( T = 300 \, \text{K} \), we have:
\[ W = \int_{V_i}^{2V_i} \frac{RT}{V} \, dV \]

5. **Evaluate the Integral**:
- The integral of \( \frac{1}{V} \) is \( \ln V \):
\[ W = RT \left[ \ln V \right]_{V_i}^{2V_i} \]
- This becomes:
\[ W = RT \left( \ln(2V_i) - \ln(V_i) \right) \]
- Simplifying this gives:
\[ W = RT \ln(2) \]

6. **Calculate the Work Done**:
- Substitute \( R = 8.314 \, \text{J/(mol K)} \) and \( T = 300 \, \text{K} \):
\[ W = 8.314 \times 300 \times \ln(2) \]
- Calculate \( \ln(2) \approx 0.693 \):
\[ W \approx 8.314 \times 300 \times 0.693 \approx 1730.6 \, \text{J} \]

7. **Final Result**:
- The work done during the isothermal expansion of the gas is approximately:
\[ W \approx 1730.6 \, \text{J} \]