The Reasons Why Ideal Gas Always Use Kelvin Temperature

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when we are studying about the ideal gas, the unit of temperature should use a unit kelvin whereas the unit of temperature can be in celcius or in farenheit. in this article we will discuss the reasons why the temperature of the unit gas should be in kelvin.

Determining the units physics in the kinetic theory of gas, temperature should be stated in kelvin. Based on Gay-Lussac experiments that the coefficient expansion of any gas is always 1/273 ⁰C^-1. Here are the mathematical derivations and mathematical descriptions of the ideal gas equations in the fixed pressure. Derivations starts from Volume gas expansion equation.

$V={{V}_{i}}(1+\gamma \Delta t)$

Vi = Initial volume

Δt = Temperature difference

$\gamma =\frac{1}{273}{}^{0}{{C}^{-1}}$ is coefficient expansion of any gas

$V={{V}_{i}}(1+\frac{1}{273}\Delta t)$

we change the equation to be

$V=\frac{{{V}_{i}}}{273}\left( 273+\left( t-{{t}_{i}} \right) \right)$

The initial temperature t_i is zero 0⁰C the equation becomes

$V=\frac{{{V}_{i}}}{273}\left( 273+t \right)$

we can change (373 + t) = T this is why the ideal gas temperature should be in kelvin

$V=\frac{{{V}_{i}}}{273}T$

$\frac{V}{T}=\frac{{{V}_{i}}}{273}$

we find the Gay-Lussac equation for constant temperature

$\frac{V}{T}=C$

we can change (373 + t) = T this is why the ideal gas temperature should be in kelvin

$V=\frac{{{V}_{i}}}{273}T$

$\frac{V}{T}=\frac{{{V}_{i}}}{273}$

we find the Gay-Lussac equation for constant temperature

$\frac{V}{T}=C$

$\frac{{{V}_{1}}}{{{T}_{1}}}=\frac{{{V}_{2}}}{{{T}_{2}}}$

The Reasons Why Ideal Gas Always Use Kelvin Temperature

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