Answer
$-6.311$ $J/K$
Work Step by Step
All the symbols used have the usual meaning.
For an ideal gas, the internal energy ($U$) of the gas depends only on the temperature. In the isothermal process, the temperature was constant throughout. So during the process, there was no change in internal energy. That is, $\Delta U=0$
Work done on a gas is negative by convention.
Thus, $\Delta W=-1850$ $J$
From the first law of thermodynamics:
$\hspace{9mm} \Delta Q= \Delta U + \Delta W$
$\Longrightarrow \Delta Q=0+ \Delta W= \Delta W=-1850$ $J$
$\Longrightarrow \Delta Q=-1850$ $J$
($Q=$ Heat)
The system liberates $1850$ $J$ of heat. (by convention)
The entropy change( $\Delta S$) of gas during the reversible process is: $\dfrac{\Delta Q}{T}$
$T=20.0 \hspace{0.5mm}{}^oC=293.15$ $K$
$\Delta S=\dfrac{-1850}{293.15}$ $J/K \simeq-6.311$ $J/K$
