Question
Derive an expression for 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
$d W=-P_{\text {ext }} d V....(1)$
$3.$ 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,
$W _{\text {max }}=-\int_{V_1}^{V_2} PdV .....(4)$
$5.$ Using the ideal gas law, $PV = nRT$,
$W_{\max }=-\int_{V_1}^{V_2} n R T \frac{d V}{V}$
$=-n R T \int_{V_1}^{V_2} \frac{d V}{V} \ldots(\because T$ is constant. $)$
$=-n R T \ln (V)_{V_1}^{V_2}$
$= - nRT (ln V_2 - ln V_1)$
$=- nRt \ln \frac{ V _2}{ V _1}$
$=-2.303 nRT \log _{10} \frac{ V _2}{ V _1}....(5)$
$6.$ 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 equation $(5)$ by $\frac{P_1}{P_2}$, we get,
$W_{\max }=-2.303 n R T \log \frac{P_1}{P_2} ....(6)$
Equations $(5)$ and $(6)$ are expressions for work done in reversible isothermal process.
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