Question

At what temperature will a balloon have a volume of 6.08 L if the temperature is 41.0 ℃ when its volume is 4.87 L? Answer in Kelvin.

P.S please show the equation steps.

Answer 1

392.20K

Step-by-step explanation:

-We apply Charles' Law which states that for an ideal gas at constant pressure, its volume is directly proportional to it's temperature:

`(V_1)/(T_1)=(V_2)/(T_2)`

-Given that:

`V_1=4.87L, V_2=6.08\\\\T_1=41.0\textdegree C=314.15\ K\\T_2`

#We substitute in the ratio formula above and calculate for
`T_2`:

`(V_1)/(T_1)=(V_2)/(T_2)\\\\T_2=(V_2T_1)/(V_1)\\\\=(6.08\ l* 314.15\ K)/(4.87\ L)\\\\\\=392.20\ K`

Hence, the temperature of the balloon at a volume of 6.08L is 392.20K