Question

Work done in reversible isothermal process by an ideal gas is given by:
(A) $2.303\text{ nRT log}\dfrac{{{V}_{2}}}{{{V}_{1}}}$
(B) $\dfrac{nR}{(\gamma -1)}({{T}_{2}}-{{T}_{1}})$
(C) $2.303\text{ nRT log}\dfrac{{{V}_{1}}}{{{V}_{2}}}$
(D) none

Answer 1

Hint: Understand the conditions prevalent in a reversible reaction. Isothermal indicates no variation in temperature. So, apply Charles law to understand the variation of volume and pressure and the relation between the two. With this you can devise a method to find the work done applying the laws of thermodynamics.

Complete step by step answer:
Consider 'n' moles of an ideal gas enclosed in a cylinder fitted with a weightless, frictionless, airtight movable position. Let the pressure of the gas be P which is equal to external atmospheric pressure P. Let the external pressure be reduced by an infinitely small amount $dP$ and the corresponding small increase in volume be $dV$.
Therefore, the small work done in the expansion process will be-
\[dW=-{{P}_{ext}}.\text{ }dV\]
\[\Rightarrow \text{ }dW=-\left( P-dP \right)dV\]
\[\Rightarrow \text{ }dW=-P.dV + dP.dV\]

Since both $dP$ and $dV$ are very small, the product \[dP.dV\] will be very small in comparison with \[P.dV\] and thus can be neglected.
\[dW=-P.dV\]
When the expansion of the gas is carried out reversibly there shall exist a series of such \[P.dV\] terms. Thus, the total maximum work \[{{W}_{\max }}\] can be obtained by integrating this equation between the limits ${{V}_{1}}$ to ${{V}_{2}}$.
$W\ =\ \int\limits_{\ {{V}_{1}}}^{\ {{V}_{2}}}{dW}$
Where,
${{V}_{1}}$ is initial volume,
${{V}_{2}}$ is final volume

Substituting the values, we get
\[W\ =\ \int\limits_{\ {{V}_{1}}}^{\ {{V}_{2}}}{(-P.dV)}\]
As we know that, from ideal gas equation,
\[PV=nRT\]
$\Rightarrow \ P=\dfrac{nRT}{V}$
\[W\ =\ \int\limits_{\ {{V}_{1}}}^{\ {{V}_{2}}}{(-\dfrac{nRT}{V}.dV)}\]
$\Rightarrow W\text{ = }-nRT\ \int\limits_{\ {{V}_{1}}}^{\ {{V}_{2}}}{(\dfrac{dV}{V})}$​
$\Rightarrow W\text{ = }-nRT\ [\ln V]_{{{V}_{1}}}^{{{V}_{2}}}$
$\Rightarrow W\text{ = }-nRT\ (\ln {{V}_{2}}-\ln {{V}_{1}})$
$\Rightarrow W\text{ = }nRT\text{ ln}\left( \dfrac{{{V}_{1}}}{{{V}_{2}}} \right)$
$\Rightarrow W\text{ = }nRT\text{ }\left( \text{2}\text{.303 x log}\left( \dfrac{{{V}_{1}}}{{{V}_{2}}} \right) \right)$
$\Rightarrow W\text{ = 2}\text{.303}nRT\text{ }\left( \text{log}\left( \dfrac{{{V}_{1}}}{{{V}_{2}}} \right) \right)$

Hence work done in reversible isothermal process by an ideal gas is given by-
$W\text{ = 2}\text{.303}nRT\text{ }\left( \text{log}\left( \dfrac{{{V}_{1}}}{{{V}_{2}}} \right) \right)$
So, the correct answer is “Option C”.

Note: It is important to know that although we calculate the work done in a reversible reaction accurately, it fails to tell us the time taken for the reaction to complete. By experimental calculation, it is found that reversible reactions usually complete at infinite amounts of time. This is because the reaction is always at equilibrium and thus takes place very slow.