Volume-temperature graph at atmospheric pressure for a monoatomic gas ($V \mathrm{i}\mathrm{n}\mathrm{ }{\mathrm{m}}^3, T\mathrm{ in }\mathrm{}\mathrm{celcius}$) is

$\begin{array}{l}PV=nRt\\ V=\frac{nR}{P}(T_c+273.15)\\ V=\left(\frac{nR}{P}\right)T_c+\frac{273.15nR}{P}\end{array}$
Positive slope positive intercept

The volume of gas is zero at absolute zero temperature ($0\text{K}$ )
Thus, the volume has a positive value at $0°\text{C}$ since the volume is directly proportonal to the absolute temperature and $0\text{K}=-273.15°\text{C}$