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प्रश्न
Answer the following question.
Derive the expression for the maximum work.
उत्तर
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 = -Pext dV ....(1)
3. As the expansion is reversible, P is greater by a very small quantity dp than Pext.
Thus P - Pext = dP or Pext = 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,
Wmax = - \[\int\limits_{\text{V}_1}^{\text{V}_2} \text{PdV}\] .....(4)
5. Using the ideal gas law, PV = nRT,
Wmax = - \[\int\limits_{\text{V}_1}^{\text{V}_2} \text{nRT} \frac{\text{dV}}{\text{V}}\]
= -\[\text{nRT} \int\limits_{\text{V}_1}^{\text{V}_2} \frac{\text{dV}}{\text{V}}\] ...(∵ 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,
Wmax = - 2.303 nRT log `"P"_1/"P"_2` ....(6)
Equations (5) and (6) are expressions for work done in reversible isothermal process.
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