Question

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

Answer 1

`T_2=697K`

Explanation:

1. 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?`

2. 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`

3. 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`