At $27°C$, one mole of an ideal gas compressed isothermally and reversibly from a pressure of 2 $atm$ to 1.0 $atm$. Choose the correct option from the following. 

In an isothermal reversible process, the work done is

$w=-2.303nRT\log \left(\frac{P_1}{P_2}\right)$

Here, $n=1$ and $R=2\mathrm{cal}/\mathrm{Kmol}$

And $P_1=2\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}$ and $P_2=1.0\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}$

Temperature, $T=27+273=300K$

Now, $w=-2.303\times 1\times 2\times 300\log \left(\frac{2}{1}\right)$

$$\Rightarrow w=-415.92$$

For an isothermal process,

$$\begin{gathered} \Delta U=0 \\ \text{ So, }q=\Delta U-w \\ q=0-(-415.92)=+415.92\mathrm{cal} \end{gathered}$$

The heat is positive and change in internal energy is zero in an isothermal expansion and the work done is - $415.92\mathrm{cal}$.

Therefore, option (D) is correct.