Q.vii. Calculate the maximum work when 24 g of O$_2$ are expanded isothermally and reversibly from the pressure of 1.6 bar to 1 bar at 298 K. Ans. : (-873.4 J)

$w_{max}=-nRT\ln(V_2/V_1)$, 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 $dw=-P_{ext}\,dV$:

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

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

…

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

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

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 $V_f$:

…