Q.i. Derive the expression for the maximum work.

$w_{max}=-nRT\ln(V_2/V_1)$, derived by integrating $dw=-\frac{nRT}{V}dV$ over the reversible isothermal path.

For $n$ moles of an ideal gas undergoing reversible, isothermal expansion at temperature $T$ from $V_1$ to $V_2$, the external pressure at every instant equals the gas pressure $P = \dfrac{nRT}{V}$ (this is what makes it reversible and gives the maximum possible work).

Work done, using $dw=-P_{ext}\,dV$:

$w = -\displaystyle\int_{V_1}^{V_2} P\,dV = -\int_{V_1}^{V_2}\dfrac{nRT}{V}\,dV = -nRT\Big[\ln V\Big]_{V_1}^{V_2}$

$w = -nRT\ln\dfrac{V_2}{V_1} = -2.303\,nRT\log\dfrac{V_2}{V_1}$

Since $V_2>V_1$ for expansion, $w$ comes out negative — work is done by the gas on the surroundings, and its magnitude is the maximum obtainable work for that expansion (reversible work is the theoretical maximum).

Reversible isothermal work for an expanding ideal gas is $-nRT\ln(V_2/V_1)$, giving about −1.01 kJ here.

For a reversible isothermal expansion of an ideal gas, the work done (IUPAC convention, work done on the system) is:

$w = -nRT\ln\left(\frac{V_2}{V_1}\right)$

Substituting $n=1\ mol$, $R=8.314\ J\,K^{-1}mol^{-1}$, $T=300\ K$, $V_1=10\ L$, $V_2=15\ L$:

$w = -(1)(8.314)(300)\ln\left(\frac{15}{10}\right) = -2494.2 \times \ln(1.5) = -2494.2 \times 0.4055$

$w \approx -1011.5\ J \approx -1.01\ kJ$

The negative sign indicates that as the gas expands, it does work on the surroundings (energy leaves the system as work).

Work done by a system is maximised in a reversible process, where the system stays essentially in equilibrium with its surroundings throughout.

For an expansion/compression, work done depends on the external pressure applied: $w = -\int P_{ext}\,dV$. If the process is carried out irreversibly (a sudden, large pressure difference), the work exchanged is less than the theoretical maximum. Work is maximised only when the process is reversible, i.e. carried out in an infinite series of infinitesimal steps such that $P_{ext} = P_{int} \mp dP$ at every instant, keeping the system arbitrarily close to equilibrium throughout. For a reversible isothermal expansion of an ideal gas from $V_i$ to $V_f$:

$w_{max} = -nRT\ln\dfrac{V_f}{V_i} = -2.303\,nRT\log_{10}\dfrac{V_f}{V_i}$

This is the largest magnitude of work obtainable between the same initial and final states — any real (irreversible, finite-rate) path between the same two states yields less work.