Question

Answer the following question.

Derive the expression for the maximum work.

Answer 1

1. Consider n moles of an ideal gas enclosed in a cylinder fitted with a frictionless movable rigid piston. It expands isothermally and reversibly from the initial volume V₁ to final volume V₂ at temperature T. The expansion takes place in a number of steps as shown in the figure.

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

dW = -P_ext dV        ....(1)

3. As the expansion is reversible, P is greater by a very small quantity dp than P_ext.

Thus P - P_ext = dP or P_ext = P - dP     ....(2)

Combining equations (1) and (2),

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

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

dW = - PdV         .....(3)

4. The total amount of work done during the entire expansion from volume V₁ to V₂ would be the sum of the infinitesimal contributions of all the steps. The total work is obtained by integration of Equation (3) between the limits of initial and final states. This is the maximum work, the expansion being reversible.

Thus,

\[\int\limits_{\text{initial}}^{\text{final}}\text{dW} = - \int\limits_{\text{V}_1}^{\text{V}_2} \text{PdV}\]

Hence,

W_max = - \[\int\limits_{\text{V}_1}^{\text{V}_2} \text{PdV}\]      .....(4)

5. Using the ideal gas law, PV = nRT,

W_max = - \[\int\limits_{\text{V}_1}^{\text{V}_2} \text{nRT} \frac{\text{dV}}{\text{V}}\]

= -\[\text{nRT} \int\limits_{\text{V}_1}^{\text{V}_2} \frac{\text{dV}}{\text{V}}\]     ...(∵ T is constant.)

= - nRT ln `("V")_("V"_1)^("V"_2)`

= - nRT (ln V₂ - ln V₁)

= - nRt ln `"V"_2/"V"_1`

= - 2.303 nRT log₁₀ `"V"_2/"V"_1`     ....(5)

6. At constant temperature, P₁V₁ = P₂V₂  or `"V"_2/"V"_1 = "P"_1/"P"_2`

Replacing `"V"_2/"V"_1` in equation (5) by `"P"_1/"P"_2`, we get,

W_max = - 2.303 nRT log `"P"_1/"P"_2`     ....(6)

Equations (5) and (6) are expressions for work done in reversible isothermal process.