# A temperature variation in one of the two Gay-Lussac formulas

The first law of Gay-Lussac, also called first law - Gay-Lussac, known abroad as Charles law or as Charles - Volta - Gay-Lussac law, at the constant pressure, the volume of an ideal gas is directly proportional to its absolute temperature.

Indicating with $$V_{0}$$ the volume of a fixed quantity of gas at the temperature of $$0\,°C$$ and with $$V$$ its volume at the temperature $$t$$ (measured in degrees Celsius), the law is expressed mathematically by the relation:

$$\bbox[5px,border:2px solid #09C5A3]{V=V_{0}(1+\alpha t)} \iff \bbox[5px,border:2px solid #C509A6 ]{V=V_0\alpha \Biggl(\frac{1}{\alpha} +t\Biggr)} \tag 1$$

If I use

$$T=t+\frac{1}{\alpha}$$ where $$T$$ it is the Kelvin temperature and $$t$$ in Celsius degree, I have

$$V=V_0\alpha T$$

The law says that an ideal gas has zero volume at temperature $$T=0\, \mathrm{K}$$ hence:

$$t=-\frac{1}{\alpha}=-273.15\, °C \iff \alpha=\frac1{273.15}\;^{\circ}C^{-1}$$

My question:

If in the $$(1)$$, $$\bbox[5px,border:2px solid #09C5A3]{V=V_{0}(1+\alpha t)}$$ I wrote,

$$\bbox[5px,border:2px solid #56D24C]{V=V_{0}(1+\alpha \color{red}{\Delta t})}$$ in Celsius degree the variation of temperature $$\Delta t$$ or equivalently

$$\bbox[5px,border:2px solid #C70039]{V=V_{0}(1+\alpha \color{blue}{\Delta T})}$$ in Kelvin degree $$\Delta T$$,

would it change anything, where $$t$$ ($$T$$) is the final temperature and $$t_0$$ ($$T_0$$) is the initial temperature remembering that $$\Delta t=\Delta T$$?

Nothing would change, provided that $$V$$ has to be the volume at $$t(T)$$ and as $$V_0$$ is fixed as the volume at $$0$$°C, $$t_0(T_0)$$ has to be $$0$$ in Celsius and $$273.15$$ in Kelvin, to use your notation.
In particular if you bring as $$\Delta t$$ a random one, if you fix $$V_0$$ as before the law will only give the volume $$V$$ at $$\Delta t$$ °C, or equivalently the $$V$$ at $$(273.15+\Delta T)$$ K. If you pass completely to the differential law, you have $$\Delta V=\alpha \Delta T$$ and in this case you can do all you want provided that $$\Delta T$$ is the variation of temperature between the extremes of $$\Delta V$$.
In short if you meant if you can differentiate only $$t$$ the answer is that it cannot be do, but if you meant if you can see a variable $$t$$ as $$\Delta t := t-0$$, then yes, you can.