# JEE Main & Advanced Chemistry States of Matter Gay-Lussacs Law (Amonton's law)

Gay-Lussacs Law (Amonton's law)

Category : JEE Main & Advanced

(1) In 1802, French chemist Joseph Gay-Lussac studied the variation of pressure with temperature and extende the Charle’s law so, this law is also called Charle’s-Gay Lussac’s law.

(2) It states that, “The pressure of a given mass of a gas is directly proportional to the absolute temperature $(={{\,}^{o}}C+273)$ at constant volume.”

Thus, $P\propto T$ at constant volume and mass

or $P=KT=K(t{{(}^{o}}C)+273.15)$          (where K is constant)

$K=\frac{P}{T}$ or $\frac{{{P}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}}{{{T}_{2}}}=K$ (For two or more gases)

(3) If $t={{0}^{o}}C$, then $P={{P}_{0}}$

Hence, ${{P}_{0}}=K\times 273.15$

$\therefore$   $K=\frac{{{P}_{0}}}{273.15}$              $P=\frac{{{P}_{0}}}{273.15}[t+273.15]={{P}_{0}}\left[ 1+\frac{t}{273.15} \right]={{P}_{0}}[1+\alpha t]$

where ${{\alpha }_{P}}$ is the pressure coefficient,

${{\alpha }_{P}}=\frac{P-{{P}_{0}}}{t{{P}_{0}}}=\frac{1}{273.15}=3.661\times {{10}^{-3}}{{\,}^{o}}{{C}^{-1}}$

Thus, for every ${{1}^{o}}$ change in temperature, the pressure of a gas changes by $\frac{1}{273.15}\left( \approx \frac{1}{273} \right)$ of the pressure at ${{0}^{o}}C$.

(4) This law fails at low temperatures, because the volume of the gas molecules be come significant.

(5) Graphical representation of Gay-Lussac's law : A graph between P and T at constant V is called isochore. You need to login to perform this action.
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