In an isothermal reversible expansion, if the volume of $96\;gm$ of oxygen at ${27^ \circ }C$ is increased from $70\,\text{litres}$ to $140\,\text{litres}$ , then the work done by the gas will be
A. $300\,R{\log _{10}}2$
B. $81\,R{\log _{10}}2$
C. $900\,R{\log _{10}}2$
D. $2.3 \times 900{\log _{10}}2$

Hint: In an Isothermal process in a thermodynamic system, temperature is constant and at constant temperature, the formula for work done is defined as $W = 2.306nRT{\log _{10}}\left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)\,$. Hence by substituting the required values in expression and simplifying it, we will get the required answer.

Formula used:
Work done in isothermal process,
$\text{Workdone} = 2.306nRT{\log _{10}}\left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)\,$
where n = Number of moles
R = Universal Gas Constant
T = Constant Absolute Temperature of the gas
${V_2}$ = Final Volume of the gas
${V_1}$ = Initial Volume of the gas
$\text{Number of Moles (n)} = \dfrac{\text{Mass of substance(m)}}{\text{Mass of one mole(M)}}$

Complete step by step solution:
It is given that the volume of oxygen of mass $m = 96\;gm$ at temperature $T = {27^ \circ }C$ is increased from $70{\text{ }}litres({V_1})$ to $140{\text{ }}litres({V_2})$.
Temperature of oxygen,
$T = {27^ \circ }C = 300K$................$\left( {^ \circ C + 273 = K} \right)$

In the Isothermal process, the formula for work done can be stated as: -
At constant pressure,
$\text{Workdone} = W = 2.306nRT{\log _{10}}\left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1) \\ $
Also, we know that
$\text{Number of Moles (n)} = \dfrac{\text{Mass of substance(m)}}{\text{Mass of one mole(M)}}\,\,\,\,\,\,\,\,\,\,\,\,...\,(2)$
Mass of one mole of oxygen ${O_2}$ is $M = 32\,gm$.

Now, from equations $(1)$ and $(2)$, we get
$ \Rightarrow W = 2.306\left( {\dfrac{m}{M}} \right)RT{\log _{10}}\left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)\,\,\,\,\,\,\,\,...\,(3) \\ $
Substitute all the required values from the question in equation $(3)$, and we get
$ \Rightarrow W = 2.306\left( {\dfrac{{96}}{{32}}} \right)R(300){\log _{10}}\left( {\dfrac{{140}}{{70}}} \right) \\ $
$ \therefore W = 2.306\left( 2 \right)R(300){\log _{10}}\left( 2 \right) = 2.306 \times 900R{\log _{10}}2 \\ $
Thus, the work done by the gas in this isothermal process will be $2.3 \times 900R{\log _{10}}2$.

Hence, the correct option is D.

Note: In isothermal contraction, the gas does the work whereas in isothermal expansion, the gas does the work. This suggests that W>0, or work that is positive, is at work when heat is absorbed by the gas.