The graphical representation of the pressure law is always a straight line passing through the origin, only if the temperature scale is

The pressure law (Gay-Lussac's law) states that for a fixed mass of gas at constant volume the pressure is directly proportional to the absolute temperature: \[P \propto T \quad\Rightarrow\quad \frac{P}{T} = \text{constant}.\] A graph of \(P\) against \(T\) can only be a straight line through the origin if the temperature axis is zeroed at the point where the pressure itself would be zero, that is at absolute zero. The scale defined that way, independent of any particular substance, is the thermodynamic (absolute, kelvin) scale, so that is the scale the question is after.

Plotting the same experimental data on the Celsius scale gives a straight line of the same gradient, but its zero of temperature is displaced: the line cuts the temperature axis at \(-273\,^\circ\text{C}\) and cuts the pressure axis at a positive intercept, so it does not pass through the origin.

Note that Fahrenheit shares the Celsius problem in a worse form, since its zero lies at about \(-459\,^\circ\text{F}\) below the pressure-zero point. Rankine is genuinely an absolute scale as well (\(0\,^\circ\text{R}\) is absolute zero, with degrees the size of Fahrenheit degrees), so a Rankine plot would also pass through the origin; it is not, however, the scale physics defines the gas laws on, and "thermodynamic scale" is the standard name for the absolute scale used in \(P \propto T\). The examination point to carry away is that every gas-law calculation and graph requires temperature in kelvin: convert with \(T/\text{K} = \theta/^\circ\text{C} + 273\) before substituting.