Question

Consider the reversible isothermal expansion of an ideal gas in a closed system at two different temperatures $T_1$ and $T_2 (T_1 < T_2)$. The correct graphical depiction of the dependence of work done $(w)$ on the final volume $(V)$ is:

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The Correct Option is B

To solve the problem of graphically depicting the dependence of work done \((w)\) on the final volume \((V)\) for the reversible isothermal expansion of an ideal gas, we need to use the concepts of thermodynamics, specifically those concerning ideal gases and isothermal processes.

For a reversible isothermal expansion of an ideal gas, the work done \((w)\) can be calculated using the formula:

\(w = nRT \ln \left( \frac{V_f}{V_i} \right)\)

Where:

• \(n\) is the number of moles of gas.
• \(R\) is the universal gas constant.
• \(T\) is the temperature in Kelvin (constant during the process).
• \(V_f\) is the final volume.
• \(V_i\) is the initial volume.

Since we are comparing two temperatures, \(T_1\) and \(T_2\) where \(T_1 < T_2\), we note:

• The work done is directly proportional to the temperature \((T)\). This can be inferred from the formula since \(T\) is a multiplicative factor.
• As the final volume \((V_f)\) increases, the term \(\ln \left( \frac{V_f}{V_i} \right)\) becomes larger, thus increasing the work done.
• Hence, for a given final volume, the work done at higher temperature \((T_2)\) will be more than at the lower temperature \((T_1)\).

This results in two curves on the graph, both increasing but with the curve for \(T_2\) above the one for \(T_1\) at any point, as more work is done at the higher temperature for the same change in volume.

The correct graph will show two such curves, one for each temperature, illustrating the relationship where the curve for \(T_2\) lies above the curve for \(T_1\). The correct graphical depiction from the given options is:

This graph accurately depicts how work done \((w)\) increases with the final volume \((V)\) for both temperatures, and confirms that with \(T_2 > T_1\), the work done is always higher at \(T_2\).