When the volume of a gas is changed from 250 mL to 425 mL, the temperature will change from 137° C. What is temperature 2? Using Charles law

Question

4918 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-13T22:43:34

$T_2=697K$

Set up the formula and identify data as provided in the problem.
https://en.wikipedia.org/wiki/Charles%27s_law

$V_1T_2=V_2T_1$
$where$ :

$V_1=250ml$
$V_2=425ml$
$T_1=137^oC$
$T_2=ul?$
Make sure to convert $137^oC-> K$ . This is the most important point to remember because the volume of a fixed mass of a given gas is directly proportional to the temperature expressed in $K$ ; not $C$ ; i.e.,
$137^oC->ul?K$
$K=*^oC+273$
$K=137+273=410$

Note: If ever you forgot the conversion factor; just set up the following to derive the formula: ( The values reflected here are that of water. )*
$(C-"freezing pt.")/("boiling pt.-freezing pt.")=(K-"freezing pt")/("boiling pt-freezing pt")$
$(C-0)/(100-0)=(K-273)/(373-273)$ ; simplify
$C/100=(K-273)/100$ ; the scales have same size unit so cancel out 100
$C/cancel(100)=(K-273)/cancel(100)$ ; therefore
$C=K-273$ ; isolate K by adding 273 both sides of the equation
$C+273=Kcancel(-273+273)$
$K=C+273$
Now plug in data to the given formula.
$V_1T_2=V_2T_1$
$(250ml)(T_2)=(425ml)(410K)$ ; divide both sides by 250 to isolate $T_2$
$(cancel((250ml))(T_2))/cancel(250ml)=((425cancel(ml))(410K))/(250cancel(ml))$
$T_2=697K$