If 1.5 L of an ideal gas at a pressure of 20 atm expands isothermally and reversibly to a final volume of 15 L. The work done by the gas in L atm is 

Thermodynamic terms

In an isothermal reversible process, the work done is

$w=-2.303nRT\log \left(\frac{V_x}{V_1}\right)$

And we know, $PV=nRT$

Here it is given that pressure, $P=20\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}$

And volume, $V$ or $V_1=1.5L,V_2=15$.

$$\begin{gathered} w=-2.303PV\log \left(\frac{V_2}{V_1}\right) \\ w=-2.303\times 20\times 1.5\log \left(\frac{1.5}{1.5}\right)=69.09\mathrm{Latm} \end{gathered}$$

Hence, $w=-69.09\mathrm{ }\mathrm{L}$ atm

Therefore, option (D) is correct.