Question
Answer the following question.Derive the expression for the maximum work.
✓
Answer
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 $V_1$ to final volume $V_2 $ 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 $V_1$ to $V_2$ 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,
$\int_{\text {initial }}^{\text {final }} dW =-\int_{ V _1}^{ V _2} PdV$
Hence,
$W _{\max }=- \int_{ V _1}^{ V _2} \operatorname{PdV} \cdots(4)$
5. Using the ideal gas law, $P V=n R T$,
$W _{\max }=- \int_{ V _1}^{ V _2} nRT \frac{ dV }{ V }$
$= -nRT \int_{ V _1}^{ V _2} \frac{ dV }{ V }$
$\ldots(\because T$ is constant.)
$ =-n R T \ln (V)_{V_1}^{V_2}$
$=-n R T\left(\ln V_2-\ln V_1\right)$
$=-n R t \ln \frac{V_2}{V_1}$
$=-2.303 \ n R T \ \log _{10} \frac{V_2}{V_1} \cdots(5) $
6. At constant temperature, $P _1 V _1= P _2 V _2$ 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,
$W_{\max }=-2.303 n R T \log \frac{P_1}{P_2} \cdots(6)$
Equations (5) and (6) are expressions for work done in reversible isothermal process.
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