Question

To what temperature must a balloon, initially at 25°c and 2.00 l, be heated in order to have a volume of 6.00 l?

Answer 1

V1 = 2.00 L
T1 = 25 + 273 = 298 K
V2 = 6.00 L
T2 = ?
Assuming the pressure is to remain constant, then
V1/T1 = V2/T2
T2 = T1V2/V1 = (298)(6)/(2) = 894 deg K

Answer 2

Answer: 894 K

Solution :

Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

`V\propto T` (At constant pressure and number of moles)

The combined gas equation is,

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

where,

`V_1` = initial volume of gas = 2.00 L

`V_2` = final volume of gas = 6.00 L

`T_1` = initial temperature of gas =
`25^oC=273+25=298K`

`T_2` = final temperature of gas = ?

Now put all the given values in the above equation, we get the final temperature of the gas

`(2.00L)/(298K)=(6.00L)/(T_2K)`

`T_2=894K`

Therefore, the temperature must be 894 K in order for balloon to have a volume of 6.00 L.