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.

  • 1
    $\begingroup$ Thank you very very much....and +1 and...green check mark...:-) $\endgroup$ – Sebastiano Feb 12 at 21:09
  • 1
    $\begingroup$ Thank you! Grazie mille! :-) $\endgroup$ – annAB Feb 12 at 21:10

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