0
$\begingroup$

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 $\Delta t=\Delta T$?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.