# About Charles's Law I have a confusion?

Aug 13, 2017

The equation for Charles's law is

ulbar(|stackrel(" ")(" "(T_1)/(V_1) = (T_2)/(V_2)" ")|)" " (constant pressure and quantity of gas)

The problem says that the volume of the gas increases by $\frac{1}{273}$ times its original volume at $0$ $\text{^"o""C}$ for every temperature degree it rises.

If we make the final temperature ${T}_{2}$ be equal to $1 \pm$ the original temperature, then we have

${T}_{2} = {T}_{1} \pm 1$

Or

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

Solving for the final volume, ${V}_{2}$, we get

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

The temperature $0$ $\text{^"o""C} = 273$ $\text{K}$, so if we make that be the original temperature:

color(red)(ulbar(|stackrel(" ")(" "V_2 = V_1 +-(1/273)V_1" ")|)

(the units canceled out when we solved for ${V}_{2}$)

Or, in words, the final volume is equal to the original volume plus or minus $\frac{1}{273}$ times the original volume.