Please tell the answer for this

Solution
Pressure due to dry air at 14^∘C or 287 K = Pressure of gas at 14^∘C − Vapour pressure of water at 14^∘C
Pressure due to dry air at 14^∘C or 287 K = 750 − 12 = 738 mm Hg
Volume of dry air at 14^∘C or 287 K = 28 cm³
STP conditions imply Standard Temperature and Pressure, i.e., P is 760 mm Hg and T is 273 K.
According to gas equation,
​​​​​​​$$\begin{gathered} \frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2} \\ \text{Putting all the given values, we get} \\ \frac{738\times 28}{287}=\frac{760\times V_2}{273} \\ {V}_2=25.9 {\mathrm{cm}}^3 \end{gathered}$$

Hence, the volume of dry air at STP is 25.9 cm³​​​​​​​.