to convert ppm gas (N2O) concentrations in microgram nitrogen per gram dry soil basis?
I assume the ppm value you are reporting is ppm by volume (or equivalently for ideal gases, ppm by mole).  If that is true, then:


The total moles of all gas in the container can be calculated from the ideal gas law:


$$ N_{tot} = \frac{PV}{RT}=\frac{(101325 \rm{~Pa})(\frac{80}{1000^2} \rm{~m^3})}{{(8.314 \rm{~\frac{J}{mol\cdot K}})}{(298 \rm{~K}})}$$

Here I have assumed that the temperature was 298 K, or about 25 &deg;C.  Note that the units work out because pascals ($\rm{Pa}$) are equivalent to joules per cubic meter, $\rm{\frac{J}{m^3}}$.  Also, I assume your experiment was done at approximately atmospheric pressure, which is 101325 Pa.


The moles of $\ce{NO2}$ in this 80 cc of gas can be calculated from the ppm measurement:


$$N_{\ce{NO2}}=\frac{\rm{ppm~value}}{1,000,000}\times N_{tot}$$


The mass of the nitrogen, in grams, in the $\ce{NO2}$ gas can be calculated from the formula weight of a nitrogen atom, $\rm{FW}_{\ce{N}}$:


$$m_{\ce{NO2}}={\rm{FW}_{\ce{N}}}\times N_{\ce{NO2}}=14\frac{\rm{g}}{\rm mol} \times N_{\ce{NO2}}$$


The mass of soil used was 5 grams, so the mass ratio of $\ce{NO2}$ to soil, $x_{\ce{NO2}}$  can be calculated by simply dividing the masses:


$$x_{\ce{NO2}} = \frac{m_{\ce{NO2}}}{5 \rm{~grams}}=\frac{m_{\ce{NO2}}\times \frac{1,000,000 \rm{~\mu g}}{\rm{g}}}{5 \rm{~grams}}$$

The last part of the equation above means that if you want the mass ratio in units of micrograms (of $\ce{NO2}$) per gram of soil, you have to multiply by 1,000,000.

Also, the question says you want a "per gram dry soil basis", but you didn't mention if the 5 g of soil you used is "dry".  If it isn't, then you will first have to figure out how many grams of dry soil you added to your reaction, if it wasn't 5.