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,

\(\int\limits_{\text{initial}}^{\text{final}} dW = - \int\limits_{V_1}^{V_2} PdV\)

Hence,

W_max = \(- \int\limits_{V_1}^{V_2} PdV\)   .....(4)

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

W_max = \(- \int\limits_{V_1}^{V_2} nRT\frac{dV}V\)

\(=-nRT \int\limits_{V_1}^{V_2} \frac{dV}V\)   ...(∵ T is constant.)

= - nRT ln \((V)_{V_1}^{V_2}\)

= - nRT (ln V₂ - ln V₁)

= - nRt ln \(\frac {V_2}{V_1}\)

= - 2.303 nRT log₁₀ \(\frac {V_2}{V_1}\)     ....(5)

6. At constant temperature, P₁V₁ = P₂V₂  or \(\frac {V_2}{V_1} = \frac{P_1}{P_2}\)

Replacing \(\frac {V_2}{V_1}\) in in equation (5) by \( \frac{P_1}{P_2}\), we get,

W_max = - 2.303 nRT log \( \frac{P_1}{P_2}\)   .....(6)

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

Answer 2

Consider n moles of an ideal gas enclosed in an ideal cylinder fitted with a massless and frictionless movable rigid piston. Let V be the volume of the gas at a pressure P and a temperature T.

If in an infinitesimal change pressure changes from P to P – dP and volume increases from V to V + dV. Then the work obtained is, dW = - (P-dP) dV

= -PdV + dPdV

Since dP.dV is negligibly small relative to PdV 

dW = -PdV

Let the state of the system change from A(P₁, L₁) to B (P₂, V₂) isothermally and reversibly, at temperature T involving number of infinitesimal steps.

Then the total work or maximum work in the process is obtained by integrating above equation.

At constant temperature,

where n, P, V and T represent number of moles, pressure, volume and temperature respectively. For the process,

ΔU = 0, ΔH = 0.

The heat absorbed in reversible manner Q_rev, is completely converted into work.

Q_rev = -W_max.

Hence work obtained is maximum.