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To calculate the amount of work done in joules during reversible isothermal expansion of an ideal gas, the volume must be expressed in:
A. ${{m}^{3}}$only
B. $d{{m}^{3}}$only
C. $c{{m}^{{{3}^{{}}}}}$only
D. Any one of them

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Last updated date: 24th Jul 2024
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Answer
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Hint: As a very first step, one could read the question well and hence understand the important points. Then one could recall the concept of reversible isothermal expansion and hence the expression for work done in that process. Thereby, you will be able to answer the question.

Complete step by step answer:
In the question we are discussing a case where we calculate the amount of work done in joules during a reversible isothermal expansion of certain ideal gas. We are supposed to find the unit in which the volume of the ideal gas should be expressed while doing the calculation.
In order to answer this question, let us recall the concept of reversible isothermal expansion and also the expression for work done in that particular process.
This process could be defined as the infinitely slow increase in volume at a constant temperature. Here we have the temperature constant. Work done could be given by,
$dW=-PdV$
From ideal gas equation, $PV=nRT$
$\Rightarrow dW=-\left( \dfrac{nRT}{V} \right)dV$
Integrating on both sides,
$\int{dW=-\int{\left( \dfrac{nRT}{V} \right)}}dV$
$\Rightarrow W=-nRT\ln \left( \dfrac{{{V}_{2}}}{{{V}_{1}}} \right)$
So, in the expression for work done in this process, we have the ratio of the final and initial volume of the ideal gas and hence, the unit in which volume is expressed doesn’t affect the unit of work done.

So, the correct answer is “Option D”.

Note: The definition of reversible isothermal expansion lies within the name of the process. Any process that is done infinitely slowly such that the final state regenerates the initial state is reversible. Isothermal process is that process where the temperature remains constant and expansion simply refers to the increase in volume.