Consider the reversible isothermal expansion of an ideal gas in a closed system at two different temperatures `T_(1)` and `T_(2)(T_(1)ltT_(2))`. The correct graphical depiction of the dependence of work done (w) on the final volume (V) is:

Options

A.
B.
C.
D.

Explanation

To solve the problem of the reversible isothermal expansion of an ideal gas at two different temperatures \( T_1 \) and \( T_2 \) (where \( T_1 < T_2 \)), we need to analyze the relationship between the work done (w) and the final volume (V) of the gas.

### Step-by-Step Solution:

1. **Understand the Work Done in Isothermal Expansion**:
The work done \( W \) during the isothermal expansion of an ideal gas can be expressed using the formula:
\[ W = -nRT \ln \left(\frac{V_f}{V_i}\right) \]
where:
- \( n \) = number of moles of gas
- \( R \) = universal gas constant
- \( T \) = absolute temperature (in Kelvin)
- \( V_f \) = final volume
- \( V_i \) = initial volume

2. **Magnitude of Work Done**:
Since we are interested in the magnitude of work done, we can ignore the negative sign:
\[ |W| = nRT \ln \left(\frac{V_f}{V_i}\right) \]

3. **Graphical Representation**:
- The work done \( |W| \) is dependent on the natural logarithm of the ratio of final volume to initial volume.
- As \( V_f \) increases, \( |W| \) also increases, indicating a positive correlation between work done and final volume.

4. **Comparing Two Temperatures**:
- For two different temperatures \( T_1 \) and \( T_2 \):
- At a higher temperature \( T_2 \), the work done will be greater for the same change in volume compared to \( T_1 \) because \( |W| \) is directly proportional to \( T \).
- Therefore, the graph of work done versus final volume will show that at \( T_2 \), the curve will be above that of \( T_1 \).

5. **Identifying the Correct Graph**:
- The graph will show two curves, one for \( T_1 \) and one for \( T_2 \). Both curves will be increasing, with the curve for \( T_2 \) lying above the curve for \( T_1 \).
- The intercepts on the work axis will be negative because as \( V_f \) approaches \( V_i \), the work done approaches zero, but remains negative due to the nature of expansion.

### Conclusion:
The correct graphical depiction of the dependence of work done (w) on the final volume (V) will show two curves: one for \( T_1 \) and one for \( T_2 \), with the curve for \( T_2 \) lying above that of \( T_1 \) and both having negative intercepts.