20 mL of liquid pyridine ( C_5H_5N , specific gravity 0.79 ) is...

Mass of pyridine ($C_5H_5N$) $= \text{Volume} \times \text{Density} = 20 \times 0.79 = 15.8 \text{ g}$

Molar mass of $C_5H_5N = (5 \times 12) + (5 \times 1) + 14 = 79 \text{ g/mol}$

Moles of pyridine $= \dfrac{15.8}{79} = 0.2 \text{ mol}$

The balanced chemical equation for the combustion of pyridine is:
$C_5H_5N(l) + \dfrac{25}{4} O_2(g) \rightarrow 5 CO_2(g) + \dfrac{5}{2} H_2O(l) + \dfrac{1}{2} N_2(g)$

Since water is in the liquid state, the gaseous products are $CO_2$ and $N_2$.

Moles of gaseous products per mole of pyridine $= 5 + \dfrac{1}{2} = 5.5 \text{ mol}$

Total moles of gaseous products formed $= 0.2 \times 5.5 = 1.1 \text{ mol}$

Volume of gaseous products at STP $= 1.1 \times 22.4 \text{ L} = 24.64 \text{ L}$

Answer: $24.64 \text{ L}$