The work done in heating one mole of an ideal gas as constant pressure from${15}^{\circ }\mathrm{C}\text{ to }{25}^{\circ }\mathrm{C}$ is

$\begin{array}{r}{T}_1=15+273=288\mathrm{K}\\ T_2=25+273=298\mathrm{K}\end{array}$

We know that  $:P{V}_1=nR{T}_1$ and $P{V}_2=nR{T}_2$

Hence  $V_1=\frac{nR{T}_1}{P}$ and $V_2=\frac{nR{T}_2}{P}$

           $\begin{array}{l}\therefore V_2-V_1=\frac{nR{T}_2}{P}-\frac{nR{T}_1}{P}\text{ i.e., }\\ V_2-V_1=\frac{nR}{P}\left(T_2-T_1\right)\end{array}$

Or        $\mathrm{\Delta }V=\frac{nR}{P}\left(T_2-T_1\right)$

But        $W=-P\mathrm{\Delta }V=-P\times \frac{nR}{n}\left(T_2-T_1\right)$

           $=-nR\left(T_n-T_1\right)$

$\begin{array}{r}=-nR\left(T_2-I_1\right)\\ W=-1\mathrm{mol}\times 1.987{\mathrm{calK}}^{-}{\mathrm{mol}}^{-}\\ \times (298-288)\mathrm{K}=19.87\mathrm{cal}\end{array}$

So, the correct answer is (d)