Sodium nitrate decomposes above-800°C to give 

Sodium nitrate on decomposition upto 500°C to give 

$NaN{O}_2$ and given

$2{\mathrm{NaNO}}_3\stackrel{3}{⟶}2{\mathrm{NaNO}}_2+{\mathrm{O}}_2\uparrow$

while at higher temperature (i.e .. above to 800°C) this 

$NaN{O}_3$ is decomposed into $N{a}_{2}O.N_2$ and $O_2.$

$2{\mathrm{NaNO}}_3\stackrel{{800}^{\circ }\mathrm{C}}{\to }{\mathrm{Na}}_2\mathrm{O}+3/2{\mathrm{O}}_2\uparrow +{\mathrm{N}}_2\uparrow$