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,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

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

$\frac{2.00L}{298K}=\frac{6.00L}{T_2K}$

$T_2=894K$

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