Question

Derive the expression for maximum work.

Accepted Answer

Consider $n$ moles of an ideal gas enclosed in a cylinder fitted with frictionless movable rigid piston. It is expanded isothermally and reversibly from initial volume $V_{1}$ to final volume $V_{2}$ at temperature $T$ . The expansion takes place in a number of infinitesimally small steps, few of which are shown in figure.

Volume of the gas will be increased by an infinitesimal quantity $dV$ in a single step. The process is repeated in such a way that every time $p_{ex}$ is lowered infinitesimally and the gas undergoes a series of infinitesimal increments in volume until the final volume $V_{2}$ is reached.

When the volume of the gas increases by an infinitesimal amount $dV$ in a single step, the small quantity of work done is given by.

$dW = - p_{ex} dV . . . (1)$

Because the expansion is reversible, the pressure of the gas is greater by a very small amount $dP$ than $p_{ex}$ .

Hence, $P - p_{ex} = dP$ or $P_{ex} = P - dP . . . . (2)$

By combination of equations $(1)$ and $(2)$ , we write

$dW = - (P - dP) dV = - PdV + dP . dV$

Nelglecting the product $dP . dV$ which is very small, we get

$dW = - PdV . . . (3)$

The total amount of work done during the entire expansion from volume $V_{1}$ to volume $V_{2}$ will be the sum of all the infinitesimal contributions of all the series of steps. Mathematically, the total amount of work is obtained by integrating the equation $(3)$ between the limits of initial and final states. This work is the maximum work because the expansion is reversible. Thus,

$\int_{1}^{2} dW = - \int_{V_{1}}^{V_{2}} PdV$

Hence, $W_{\max} = - \int_{V_{1}}^{V_{2}} PdV . . . (4)$

The ideal gas equation for $n$ moles is $PV = nRT$ or $P = \frac{nRT}{V} . . . (5)$

Substitution of equation ( $5$ ) into equation ( $4$ ) gives

$W_{\max} = - \int_{V_{1}}^{V_{2}} nRT \frac{dV}{V}$

Because expansion is isothermal, $T$ is constant. It follows, then, that

$W_{\max} = - nRT \int_{V_{1}}^{V_{2}} \frac{dV}{V}$

$= - nRTIn {(V)}_{V_{1}}^{V_{2}}$

$= - nRT ({InV}_{2} - {InV}_{1})$

$= - nRTIn \frac{V_{2}}{V_{1}}$

$= - 2 .303 {nRTlog}_{10} \frac{V_{2}}{V_{1}} . . . (6)$

At constant temperature, according to Boyle's law,

$P_{1} V_{1} = P_{2} V_{2}$ or $\frac{V_{2}}{V_{1}} = \frac{P_{1}}{P_{2}}$ and

Replacing $\frac{V_{2}}{V_{1}}$ in equation $6$ by $\frac{P_{1}}{P_{2}}$ , we have

$W_{\max} = - 2 .303 {nRTlog}_{10} \frac{P_{1}}{P_{2}}$

(A)	Isothermal process	(i)	$q = ∆ U$
(B)	Adiabatic process	(ii)	$W = - P \times ΔV$
(C)	Isobaric process	(iii)	$W = ∆ U$
(D)	Isochoric process	(iv)	$W = - nRTln (\frac{V_{f}}{V_{i}})$

Derive the expression for maximum work.

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